Polynomials

Lesson: Anatomy and Classification of Polynomials

Polynomials are one of the most important types of expressions in algebra. To work with them well, you need to be able to recognize their parts, write them neatly, and classify them correctly.

In this lesson, you will learn how to identify the terms, coefficients, variables, and degree of a polynomial. You will also learn how to write a polynomial in standard form and classify it by its degree as linear, quadratic, cubic, or higher-degree.

1. What is a polynomial?

A polynomial is an algebraic expression made of terms added or subtracted together, where the variable has a whole-number exponent.

Examples of polynomials:

• \(3x+5\)
• \(x^2-4x+7\)
• \(2x^3+x-9\)
• \(6\)

Examples that are not polynomials:

• \(\frac{1}{x}+2\) because the variable is in the denominator
• \(x^{-2}+1\) because it has a negative exponent
• \(\sqrt{x}+3\) because \(\sqrt{x}=x^{1/2}\), and the exponent is not a whole number

2. The anatomy of a polynomial

Let us look at the polynomial

$$4x^3-2x^2+7x-5$$

This polynomial has several important parts.

• Terms: The parts separated by addition or subtraction. Here the terms are \(4x^3\), \(-2x^2\), \(7x\), and \(-5\).
• Variable: The letter that represents a number. Here the variable is \(x\).
• Coefficients: The numerical factors of the variable terms. The coefficients are \(4\), \(-2\), and \(7\).
• Constant term: A term with no variable. Here the constant term is \(-5\).
• Exponent: The power on the variable. The exponents here are \(3\), \(2\), and \(1\).

Remember that in \(7x\), the exponent of \(x\) is really \(1\), even though it is not written. In a constant like \(-5\), the variable is not present, so it is just a constant term.

3. Degree of a polynomial

The degree of a polynomial is the greatest exponent of the variable, after the polynomial is simplified.

Examples:

• \(5x+1\) has degree \(1\)
• \(3x^2-7x+4\) has degree \(2\)
• \(8x^3+2x\) has degree \(3\)
• \(9\) is a nonzero constant polynomial, so its degree is \(0\)

If a polynomial has like terms, combine them first before finding the degree.

For example,

$$2x^2+5x-3x^2+1=-x^2+5x+1$$

Now the greatest exponent is \(2\), so the degree is \(2\).

4. Standard form

A polynomial is in standard form when its terms are written from highest degree to lowest degree.

Example:

$$3x-8+2x^3-x^2$$

In standard form, this becomes

$$2x^3-x^2+3x-8$$

Why is standard form useful?

• It makes the degree easy to see.
• It helps you compare polynomials.
• It makes graphing and solving easier later.

5. Classifying polynomials by degree

Polynomials are often named by their degree.

• Degree 0: constant, such as \(7\)
• Degree 1: linear, such as \(2x-3\)
• Degree 2: quadratic, such as \(x^2+4x+1\)
• Degree 3: cubic, such as \(5x^3-2x\)
• Degree 4: quartic, such as \(x^4-1\)
• Degree 5 or more: higher-degree polynomial

So if you are asked to classify \(6x^2-9x+2\), you should say it is a quadratic polynomial because its degree is \(2\).

6. Classifying polynomials by number of terms

Polynomials can also be named by how many terms they have.

• Monomial: 1 term, for example \(4x^2\)
• Binomial: 2 terms, for example \(x^2-9\)
• Trinomial: 3 terms, for example \(x^2+5x+6\)

A polynomial with more than 3 terms is usually just called a polynomial.

You can classify a polynomial using both its degree and its number of terms. For example, \(x^2+5x+6\) is a quadratic trinomial.

7. How to analyze a polynomial step by step

When you are given a polynomial, use this process:

1. Simplify by combining like terms if needed.
2. Write the polynomial in standard form.
3. Identify the terms, coefficients, variable, and constant term.
4. Find the degree by looking at the greatest exponent.
5. Classify the polynomial by degree.
6. If asked, classify it by number of terms too.

Worked Example 1

Analyze the polynomial \(5x^2-3x+7\).

Step 1: Check standard form

The terms are already in descending order of exponents, so it is already in standard form.

Step 2: Identify the terms

• \(5x^2\)
• \(-3x\)
• \(7\)

Step 3: Identify coefficients and constant

• Coefficients: \(5\) and \(-3\)
• Constant term: \(7\)

Step 4: Find the degree

The greatest exponent is \(2\), so the degree is \(2\).

Step 5: Classify

• By degree: quadratic
• By number of terms: trinomial

Answer: \(5x^2-3x+7\) is a quadratic trinomial.

Worked Example 2

Write \(4-2x^3+6x\) in standard form and classify it.

Step 1: Rearrange the terms

Write from highest exponent to lowest:

$$-2x^3+6x+4$$

Step 2: Find the degree

The highest exponent is \(3\), so the degree is \(3\).

Step 3: Classify

• By degree: cubic
• By number of terms: trinomial

Answer: In standard form, the polynomial is \(-2x^3+6x+4\). It is a cubic trinomial.

Worked Example 3

Analyze \(3x^4-5x^2+x-9\).

Step 1: Identify the terms

• \(3x^4\)
• \(-5x^2\)
• \(x\)
• \(-9\)

Step 2: Identify coefficients

• \(3\)
• \(-5\)
• \(1\) for the term \(x\)

Step 3: Find the constant term

The constant term is \(-9\).

Step 4: Find the degree

The greatest exponent is \(4\), so the degree is \(4\).

Step 5: Classify

• By degree: quartic
• By number of terms: polynomial with 4 terms

Answer: \(3x^4-5x^2+x-9\) is a quartic polynomial.

Worked Example 4

Analyze and classify

$$2x-4+x^2-7x+3x^3$$

Step 1: Combine like terms

The like terms are \(2x\) and \(-7x\):

$$2x-7x=-5x$$

So the polynomial becomes

$$x^2-5x-4+3x^3$$

Step 2: Write in standard form

$$3x^3+x^2-5x-4$$

Step 3: Find the degree

The highest exponent is \(3\), so the degree is \(3\).

Step 4: Identify key parts

• Terms: \(3x^3\), \(x^2\), \(-5x\), \(-4\)
• Coefficients: \(3\), \(1\), \(-5\)
• Constant term: \(-4\)

Step 5: Classify

• By degree: cubic
• By number of terms: polynomial with 4 terms

Answer: The simplified polynomial is \(3x^3+x^2-5x-4\), which is a cubic polynomial.

8. Common mistakes to avoid

• Forgetting to simplify first: Always combine like terms before finding the degree.
• Using the number of terms as the degree: The degree depends on the highest exponent, not how many terms there are.
• Ignoring missing exponents: In \(x\), the exponent is \(1\).
• Confusing coefficients and constants: In \(4x^2+3\), the coefficient is \(4\) and the constant is \(3\).
• Not writing in standard form: Standard form makes analysis much easier.

9. Quick check questions

Try these on your own:

• What is the degree of \(7x^5-2x^2+1\)?
• Write \(6-x+2x^2\) in standard form.
• Classify \(9x-4\) by degree and by number of terms.
• What are the coefficients in \(-3x^3+x^2-8\)?

Answers:

• Degree \(5\)
• \(2x^2-x+6\)
• Linear binomial
• \(-3\) and \(1\)

Summary

A polynomial is made of terms with whole-number exponents on the variable. To understand a polynomial, identify its terms, coefficients, variable, and constant term, then write it in standard form.

The degree is the highest exponent, and it tells you how to classify the polynomial: linear for degree \(1\), quadratic for degree \(2\), cubic for degree \(3\), and so on. You can also describe a polynomial by its number of terms, such as monomial, binomial, or trinomial.

Once you can recognize the anatomy and classification of polynomials, you are ready to do more advanced work such as factoring, graphing, and connecting roots to polynomial equations.

Geometric Interpretation of Zeros

When we study a polynomial, we can look at it in two connected ways:

• Algebraically — by solving equations such as \(p(x)=0\)
• Geometrically — by drawing the graph of \(y=p(x)\)

The idea of the geometric interpretation of zeros tells us how these two views are related.

If \(a\) is a zero of the polynomial \(p(x)\), then \(p(a)=0\). On the graph of \(y=p(x)\), this means that when \(x=a\), the value of \(y\) is \(0\). So the point \((a,0)\) lies on the graph.

This means:

• The real zeros of a polynomial are the x-coordinates of the points where the graph meets the x-axis.
• These points are called the x-intercepts of the graph.

So, in simple words:

Zeros of a polynomial = x-intercepts of its graph

Let us understand this carefully.

1. What is a zero of a polynomial?

A polynomial in one variable is an expression like:

\(p(x)=x^2-5x+6\) or \(p(x)=2x^3+x-4\)

A number \(a\) is called a zero of the polynomial if substituting \(x=a\) makes the polynomial equal to zero.

That is, \(a\) is a zero if

$$p(a)=0$$

For example, for \(p(x)=x^2-5x+6\):

$$p(2)=2^2-5(2)+6=4-10+6=0$$

So \(2\) is a zero.

Also,

$$p(3)=3^2-5(3)+6=9-15+6=0$$

So \(3\) is also a zero.

2. What is the graph of a polynomial?

If we write \(y=p(x)\), then every value of \(x\) gives a corresponding value of \(y\). Plotting all such points gives the graph of the polynomial.

For example, if \(y=x^2-5x+6\), then:

• when \(x=0\), \(y=6\)
• when \(x=1\), \(y=2\)
• when \(x=2\), \(y=0\)
• when \(x=3\), \(y=0\)

So the graph passes through \((2,0)\) and \((3,0)\).

Because these points lie on the x-axis, they are x-intercepts.

3. Why do zeros match x-intercepts?

On the x-axis, the value of \(y\) is always \(0\).

So if a graph crosses or touches the x-axis at the point \((a,0)\), then at \(x=a\), the output of the polynomial is \(0\). That means:

$$p(a)=0$$

So \(a\) is a zero.

Similarly, if \(a\) is a zero, then \(p(a)=0\), so the graph contains the point \((a,0)\). Therefore, the graph meets the x-axis there.

This gives the exact relationship:

$$a\text{ is a real zero of }p(x) \iff (a,0)\text{ is an x-intercept of }y=p(x)$$

4. Important note: real zeros

Only real zeros appear as x-intercepts on the usual Cartesian graph.

Some polynomials have no real zeros. In that case, their graph does not meet the x-axis.

For example, consider:

$$y=x^2+1$$

If we try to find zeros, we solve:

$$x^2+1=0$$

$$x^2=-1$$

No real number has square \(-1\), so this polynomial has no real zeros.

Geometrically, its graph never touches or crosses the x-axis.

5. Crossing the x-axis and touching the x-axis

A graph can meet the x-axis in different ways:

• It may cross the x-axis.
• It may just touch the x-axis and turn back.

In both cases, the x-coordinate is still a zero because the y-value there is \(0\).

So whenever the graph has a point on the x-axis, that x-value is a real zero.

6. How to find zeros from an equation and from a graph

You can identify zeros in two main ways:

1. From the polynomial expression: solve \(p(x)=0\).
2. From the graph: look for where the graph meets the x-axis and read the x-coordinates.

These two methods should give the same real zeros.

Worked Example 1: A linear polynomial

Find the zero of \(p(x)=x-4\) and explain its geometric meaning.

Step 1: Set the polynomial equal to zero.

$$x-4=0$$

$$x=4$$

So the zero is \(4\).

Step 2: Interpret on the graph.

The graph is \(y=x-4\), which is a straight line.

At \(x=4\),

$$y=4-4=0$$

So the graph passes through \((4,0)\).

Conclusion: The zero \(4\) corresponds to the x-intercept \((4,0)\).

Worked Example 2: A quadratic with two real zeros

Find the zeros of \(p(x)=x^2-5x+6\) and state the x-intercepts of its graph.

Step 1: Solve \(p(x)=0\).

$$x^2-5x+6=0$$

Factorize:

$$x^2-5x+6=(x-2)(x-3)$$

So:

$$ (x-2)(x-3)=0 $$

Therefore,

$$x=2 \quad \text{or} \quad x=3$$

So the zeros are \(2\) and \(3\).

Step 2: Interpret geometrically.

The graph of \(y=x^2-5x+6\) meets the x-axis where \(y=0\), that is, at:

$$ (2,0) \text{ and } (3,0) $$

Conclusion: The zeros \(2\) and \(3\) are exactly the x-intercepts of the graph.

Worked Example 3: A quadratic with one repeated real zero

Consider \(p(x)=x^2-4x+4\). Find its zero and explain what happens on the graph.

Step 1: Solve \(p(x)=0\).

$$x^2-4x+4=0$$

Factorize:

$$x^2-4x+4=(x-2)^2$$

So:

$$ (x-2)^2=0 $$

$$ x=2 $$

The polynomial has a zero at \(x=2\).

Step 2: Interpret geometrically.

At \(x=2\),

$$p(2)=0$$

So the graph has the x-intercept \((2,0)\).

Here the graph touches the x-axis at \((2,0)\) and turns back, instead of crossing it.

Conclusion: Even if the graph only touches the x-axis, the x-coordinate is still a zero.

Worked Example 4: A polynomial with no real zero

Examine \(p(x)=x^2+4\). Are there any x-intercepts?

Step 1: Solve \(p(x)=0\).

$$x^2+4=0$$

$$x^2=-4$$

No real number satisfies this equation.

So the polynomial has no real zero.

Step 2: Interpret geometrically.

Since there is no real value of \(x\) for which \(y=0\), the graph of \(y=x^2+4\) does not meet the x-axis.

Conclusion: No real zero means no x-intercept.

7. Connecting factor form and x-intercepts

If a polynomial can be written in factor form, the zeros are often easy to see.

For example:

$$p(x)=(x-1)(x+2)$$

To find zeros:

$$ (x-1)(x+2)=0 $$

So:

• \(x-1=0 \Rightarrow x=1\)
• \(x+2=0 \Rightarrow x=-2\)

Therefore the graph of \(y=(x-1)(x+2)\) has x-intercepts at:

$$ (1,0) \text{ and } (-2,0) $$

This is why factorization is very useful when studying graphs of polynomials.

8. Reading a graph carefully

When a graph is given, remember:

• Look only at points where the graph meets the x-axis.
• Read the x-coordinate of each such point.
• Those x-coordinates are the real zeros.

Do not confuse:

• x-intercept — where \(y=0\)
• y-intercept — where \(x=0\)

The y-intercept is not the zero of the polynomial unless it also lies on the x-axis.

9. Key facts to remember

• A zero of \(p(x)\) is a value of \(x\) for which \(p(x)=0\).
• On the graph \(y=p(x)\), this means \(y=0\).
• Points where \(y=0\) lie on the x-axis.
• So real zeros of a polynomial are exactly the x-intercepts of its graph.
• If the graph does not meet the x-axis, the polynomial has no real zero.
• The graph may cross or only touch the x-axis; in both cases, the x-value is a zero.

Quick Check

1. For \(p(x)=x+5\), what is the zero and what is the x-intercept?
2. If the graph of a polynomial meets the x-axis at \((-3,0)\) and \((1,0)\), what are its real zeros?
3. Does \(y=x^2+9\) have any real zeros? Why?

Answers

1. Zero: \(-5\); x-intercept: \((-5,0)\)
2. The real zeros are \(-3\) and \(1\)
3. No. Because \(x^2+9\) cannot be \(0\) for any real value of \(x\)

Summary

The geometric interpretation of zeros connects equations and graphs. If \(a\) is a real zero of a polynomial, then \(p(a)=0\), so the graph of \(y=p(x)\) passes through \((a,0)\). Therefore, the real zeros of a polynomial are exactly the x-intercepts of its graph.

This idea helps you move between algebra and geometry: you can solve \(p(x)=0\) to find x-intercepts, or look at the graph to identify real zeros.

Lesson: The Division Algorithm for Polynomials

When you divide whole numbers, you often get a quotient and sometimes a remainder. For example, dividing 17 by 5 gives:

$$17 = 5(3) + 2$$

Polynomials work in a very similar way. The Division Algorithm for Polynomials tells us that when one polynomial is divided by another, we can write the result in a special form.

If a polynomial \(f(x)\) is divided by a nonzero polynomial \(d(x)\), then:

$$f(x) = d(x)q(x) + r(x)$$

Here:

• \(f(x)\) is the dividend
• \(d(x)\) is the divisor
• \(q(x)\) is the quotient
• \(r(x)\) is the remainder

The most important rule about the remainder is this:

$$\text{degree of } r(x) < \text{degree of } d(x)$$

This means the remainder must have lower degree than the divisor. For example:

• If you divide by a linear polynomial such as \(x-2\), the remainder is a constant.
• If you divide by a quadratic polynomial such as \(x^2+1\), the remainder can be linear or constant.

Why this matters: Polynomial division helps us rewrite polynomials, check factors, and solve many problems involving roots and remainders.

Main Idea of Polynomial Long Division

Polynomial long division is very similar to number long division. The goal is to figure out what must be multiplied by the divisor to match the dividend step by step.

Here is the process:

1. Write the polynomials in descending powers of \(x\).
2. If a term is missing, use a placeholder with coefficient 0.
3. Divide the first term of the dividend by the first term of the divisor.
4. Write that result in the quotient.
5. Multiply the whole divisor by that quotient term.
6. Subtract.
7. Bring down the next term and repeat.
8. Stop when the remainder has degree less than the divisor.

Important note: Always subtract the entire line carefully. Sign mistakes are very common in polynomial division.

Worked Example 1: Dividing by a linear polynomial

Divide \(x^2 + 5x + 6\) by \(x + 2\).

We want to find \(q(x)\) and \(r(x)\) so that:

$$x^2 + 5x + 6 = (x+2)q(x) + r(x)$$

Set up the long division:

First, divide the leading terms:

$$\frac{x^2}{x} = x$$

So the first term of the quotient is \(x\).

Put \(x\) in the quotient, then multiply:

$$x(x+2)=x^2+2x$$

Subtract from the dividend:

$$\left(x^2+5x+6\right)-\left(x^2+2x\right)=3x+6$$

Now divide again:

$$\frac{3x}{x}=3$$

Put \(3\) in the quotient. Multiply:

$$3(x+2)=3x+6$$

Subtract:

$$\left(3x+6\right)-\left(3x+6\right)=0$$

So the quotient is \(x+3\) and the remainder is 0.

Final answer:

$$x^2+5x+6=(x+2)(x+3)$$

This tells us that \(x+2\) divides the polynomial exactly.

Worked Example 2: Division with a nonzero remainder

Divide \(2x^3 + 3x^2 - 5x + 4\) by \(x - 1\).

Start with the leading terms:

$$\frac{2x^3}{x}=2x^2$$

Write \(2x^2\) in the quotient.

Multiply:

$$2x^2(x-1)=2x^3-2x^2$$

Subtract:

$$\left(2x^3+3x^2-5x+4\right)-\left(2x^3-2x^2\right)=5x^2-5x+4$$

Repeat:

$$\frac{5x^2}{x}=5x$$

Put \(5x\) in the quotient.

Multiply:

$$5x(x-1)=5x^2-5x$$

Subtract:

$$\left(5x^2-5x+4\right)-\left(5x^2-5x\right)=4$$

Now the remainder is \(4\), which has degree less than the divisor \(x-1\), so we stop.

Final answer:

$$2x^3+3x^2-5x+4=(x-1)(2x^2+5x)+4$$

So:

$$\text{Quotient}=2x^2+5x, \quad \text{Remainder}=4$$

Worked Example 3: Be careful with missing terms

Divide \(x^3 - 4x + 1\) by \(x - 2\).

The dividend is missing the \(x^2\) term, so write it as:

$$x^3 + 0x^2 - 4x + 1$$

Now divide.

First step:

$$\frac{x^3}{x}=x^2$$

Put \(x^2\) in the quotient.

Multiply:

$$x^2(x-2)=x^3-2x^2$$

Subtract:

$$\left(x^3+0x^2-4x+1\right)-\left(x^3-2x^2\right)=2x^2-4x+1$$

Next step:

$$\frac{2x^2}{x}=2x$$

Put \(2x\) in the quotient.

Multiply:

$$2x(x-2)=2x^2-4x$$

Subtract:

$$\left(2x^2-4x+1\right)-\left(2x^2-4x\right)=1$$

The remainder \(1\) has degree less than the divisor, so we stop.

Final answer:

$$x^3-4x+1=(x-2)(x^2+2x)+1$$

Worked Example 4: Dividing by a quadratic polynomial

Divide \(2x^3 + 3x^2 - x + 5\) by \(x^2 + 1\).

Since the divisor is quadratic, the remainder must have degree less than 2. So the remainder will be of the form \(ax+b\) or a constant.

Start with the leading terms:

$$\frac{2x^3}{x^2}=2x$$

Put \(2x\) in the quotient.

Multiply:

$$2x(x^2+1)=2x^3+2x$$

Subtract:

$$\left(2x^3+3x^2-x+5\right)-\left(2x^3+2x\right)=3x^2-3x+5$$

Next step:

$$\frac{3x^2}{x^2}=3$$

Put \(3\) in the quotient.

Multiply:

$$3(x^2+1)=3x^2+3$$

Subtract:

$$\left(3x^2-3x+5\right)-\left(3x^2+3\right)=-3x+2$$

Now the remainder is \(-3x+2\), which has degree 1. Since this is less than the degree of the divisor, we stop.

Final answer:

$$2x^3+3x^2-x+5=(x^2+1)(2x+3)+(-3x+2)$$

What the Division Algorithm tells us after dividing

After any polynomial division, we should be able to write:

$$\text{Dividend}=(\text{Divisor})(\text{Quotient})+(\text{Remainder})$$

This is not just a final answer format. It is the actual meaning of the Division Algorithm.

For example, if dividing \(f(x)\) by \(x-3\) gives quotient \(q(x)\) and remainder \(5\), then:

$$f(x)=(x-3)q(x)+5$$

This form is useful because it connects to the remainder theorem and factor theorem later in polynomial study.

Common mistakes to avoid

• Not arranging terms in descending order. Always write highest power to lowest power.
• Forgetting missing terms. Use terms like \(0x^2\) if needed.
• Subtracting incorrectly. Be careful with negative signs.
• Stopping too early. Keep going until the remainder has lower degree than the divisor.
• Writing a remainder with degree too large. If dividing by \(x^2+1\), the remainder cannot still contain an \(x^2\) term.

Quick check questions

• If you divide by \(x+4\), what kind of remainder should you expect? A constant.
• If you divide by \(x^2-9\), what is the highest possible degree of the remainder? Degree 1.
• What form must every division answer have? Dividend = Divisor × Quotient + Remainder

Summary

The Division Algorithm for Polynomials says that when a polynomial \(f(x)\) is divided by a nonzero polynomial \(d(x)\), we can write:

$$f(x)=d(x)q(x)+r(x)$$

where the degree of \(r(x)\) is less than the degree of \(d(x)\).

To divide polynomials, use long division: divide the leading terms, multiply, subtract, and repeat. Always check that the remainder has lower degree than the divisor. With practice, this process becomes much easier and helps with many other polynomial topics.

The Remainder Theorem

When dividing polynomials, long division can take time. The Remainder Theorem gives a faster way to find the remainder when a polynomial is divided by a linear binomial of the form \(x-a\).

This theorem is very useful because it connects division and substitution. Instead of dividing, you can simply plug a number into the polynomial.

What the theorem says:

If a polynomial \(P(x)\) is divided by \(x-a\), then the remainder is \(P(a)\).

In symbols:

$$\text{Remainder when } P(x) \text{ is divided by } (x-a) = P(a)$$

For example, if you want the remainder when \(P(x)\) is divided by \(x-3\), you do not need long division. You just find \(P(3)\).

Why this works

When a polynomial is divided by \(x-a\), the result can always be written like this:

$$P(x)=(x-a)Q(x)+R$$

Here:

• \(Q(x)\) is the quotient polynomial
• \(R\) is the remainder

Because you are dividing by a linear expression \((x-a)\), the remainder must be a constant number.

Now substitute \(x=a\):

$$P(a)=(a-a)Q(a)+R$$ $$P(a)=0+R$$ $$P(a)=R$$

So the remainder is exactly the value of the polynomial at \(x=a\).

Important idea to remember

• Divide by \(x-a\)  use \(P(a)\)
• Divide by \(x+a\)  use \(P(-a)\), because \(x+a=x-(-a)\)

This sign change is a very common place for mistakes, so be careful.

How to use the Remainder Theorem

1. Identify the divisor.
2. Write it in the form \(x-a\).
3. Substitute \(a\) into the polynomial.
4. Simplify the result.
5. The answer is the remainder.

Worked Example 1: A basic example

Find the remainder when \(P(x)=2x^2+5x-1\) is divided by \(x-2\).

Step 1: The divisor is \(x-2\), so \(a=2\).

Step 2: Find \(P(2)\).

$$P(2)=2(2^2)+5(2)-1$$ $$=2(4)+10-1$$ $$=8+10-1$$ $$=17$$

Answer: The remainder is \(17\).

Worked Example 2: Watch the sign carefully

Find the remainder when \(P(x)=x^3-4x+7\) is divided by \(x+1\).

Step 1: Rewrite the divisor in the form \(x-a\).

Since \(x+1=x-(-1)\), we have \(a=-1\).

Step 2: Find \(P(-1)\).

$$P(-1)=(-1)^3-4(-1)+7$$ $$=-1+4+7$$ $$=10$$

Answer: The remainder is \(10\).

Worked Example 3: A higher-degree polynomial

Find the remainder when \(P(x)=3x^4-x^3+2x^2-6\) is divided by \(x-1\).

Step 1: The divisor is \(x-1\), so \(a=1\).

Step 2: Evaluate \(P(1)\).

$$P(1)=3(1)^4-(1)^3+2(1)^2-6$$ $$=3-1+2-6$$ $$=-2$$

Answer: The remainder is \(-2\).

Worked Example 4: Finding an unknown value

Suppose \(P(x)=2x^3+kx^2-5x+4\). If the remainder when \(P(x)\) is divided by \(x-2\) is \(10\), find \(k\).

By the Remainder Theorem, the remainder when dividing by \(x-2\) is \(P(2)\).

So:

$$P(2)=10$$

Now calculate \(P(2)\):

$$P(2)=2(2^3)+k(2^2)-5(2)+4$$ $$=2(8)+4k-10+4$$ $$=16+4k-10+4$$ $$=10+4k$$

Set this equal to the remainder:

$$10+4k=10$$ $$4k=0$$ $$k=0$$

Answer: \(k=0\).

Connection to the Factor Theorem

The Remainder Theorem also helps explain the Factor Theorem.

If the remainder is \(0\), then the divisor divides the polynomial exactly. That means the divisor is a factor.

So:

• If \(P(a)=0\), then \(x-a\) is a factor of \(P(x)\).
• If \(P(a)\neq 0\), then \(x-a\) is not a factor.

Example: Is \(x-3\) a factor of \(P(x)=x^2-5x+6\)?

Check \(P(3)\):

$$P(3)=3^2-5(3)+6=9-15+6=0$$

Because the remainder is \(0\), \(x-3\) is a factor.

Common mistakes to avoid

• Using the wrong sign: For \(x+4\), substitute \(-4\), not \(4\).
• Forgetting brackets: When substituting negative numbers, use brackets. For example, write \((-2)^3\), not \(-2^3\).
• Confusing quotient and remainder: The theorem only gives the remainder, not the full result of the division.
• Arithmetic errors: Simplify carefully, especially with powers and negative signs.

Quick practice ideas

Try these on your own:

• Find the remainder when \(x^2+3x+2\) is divided by \(x-1\).
• Find the remainder when \(2x^3-x+5\) is divided by \(x+2\).
• Determine whether \(x+1\) is a factor of \(x^3+2x^2-x-2\).

Final summary

The Remainder Theorem says that the remainder when dividing \(P(x)\) by \(x-a\) is \(P(a)\). This lets you find a remainder quickly by substitution instead of long division.

Always write the divisor in the form \(x-a\), substitute carefully, and watch the sign. If the result is \(0\), then the divisor is a factor of the polynomial.

Lesson: The Factor Theorem

When working with polynomials, we often want to know whether a smaller expression, such as \(x-3\) or \(x+2\), divides evenly into a polynomial. The Factor Theorem gives us a quick way to test this without doing long division every time.

This theorem is closely connected to the idea of a remainder. If dividing a polynomial by a linear expression leaves remainder 0, then that linear expression is a factor. The Factor Theorem tells us exactly how to check that using substitution.

What is the Factor Theorem?

For a polynomial \(P(x)\):

If \(P(a)=0\), then \((x-a)\) is a factor of \(P(x)\).

And the reverse is also true:

If \((x-a)\) is a factor of \(P(x)\), then \(P(a)=0\).

This means there is a direct connection between:

• the expression \((x-a)\),
• the value \(a\), and
• whether the polynomial equals 0 when \(x=a\).

Important matching idea:

• If the factor is \((x-5)\), test \(x=5\).
• If the factor is \((x+4)\), test \(x=-4\).
• If the factor is \((x-a)\), test \(x=a\).
• If the factor is \((x+a)\), test \(x=-a\).

Many students make sign mistakes here, so always match the factor carefully.

Why does this work?

The Factor Theorem comes from the Remainder Theorem. The Remainder Theorem says that when a polynomial \(P(x)\) is divided by \((x-a)\), the remainder is \(P(a)\).

So if \(P(a)=0\), then the remainder is 0. A remainder of 0 means the division is exact, so \((x-a)\) is a factor.

How to use the Factor Theorem

1. Identify the possible factor, such as \((x-a)\) or \((x+a)\).
2. Find the value you must substitute into the polynomial.
3. Calculate \(P(a)\).
4. If the result is 0, the expression is a factor.
5. If the result is not 0, the expression is not a factor.

Worked Example 1: Simple check

Determine whether \((x-2)\) is a factor of

$$P(x)=x^2-5x+6$$

Since the factor is \((x-2)\), we test \(x=2\).

$$P(2)=2^2-5(2)+6$$

$$=4-10+6$$

$$=0$$

Because \(P(2)=0\), \((x-2)\) is a factor of \(x^2-5x+6\).

Worked Example 2: Be careful with signs

Determine whether \((x+3)\) is a factor of

$$P(x)=x^3-4x^2-3x+18$$

The factor is \((x+3)\), so we test \(x=-3\), not \(x=3\).

$$P(-3)=(-3)^3-4(-3)^2-3(-3)+18$$

$$=-27-4(9)+9+18$$

$$=-27-36+9+18$$

$$=-36$$

Because \(P(-3)\neq 0\), \((x+3)\) is not a factor.

Worked Example 3: A cubic that does have the factor

Determine whether \((x-1)\) is a factor of

$$P(x)=2x^3-3x^2-8x+9$$

Since the factor is \((x-1)\), test \(x=1\).

$$P(1)=2(1)^3-3(1)^2-8(1)+9$$

$$=2-3-8+9$$

$$=0$$

So \((x-1)\) is a factor of the polynomial.

If we wanted, we could then divide the polynomial by \((x-1)\) to find the other factor, but the Factor Theorem only asks us to test whether it is a factor.

Worked Example 4: Using the theorem with an unknown number

Find the value of \(k\) if \((x-2)\) is a factor of

$$P(x)=x^2+kx-10$$

If \((x-2)\) is a factor, then by the Factor Theorem:

$$P(2)=0$$

Substitute \(x=2\):

$$P(2)=2^2+k(2)-10$$

$$=4+2k-10$$

$$=2k-6$$

Now set this equal to 0:

$$2k-6=0$$

$$2k=6$$

$$k=3$$

So the value of \(k\) is 3.

What the theorem helps us understand

• If \((x-a)\) is a factor, then \(a\) is a root or zero of the polynomial.
• If \(a\) is a root, then the graph of the polynomial crosses or touches the \(x\)-axis at \(x=a\).
• Testing factors becomes much faster than using full polynomial division.

Common mistakes to avoid

• Sign errors: For \((x+2)\), substitute \(-2\), not \(2\).
• Arithmetic mistakes: Use brackets carefully, especially with powers and negative numbers.
• Forgetting the conclusion: After calculating \(P(a)\), clearly state whether the expression is a factor or not.
• Mixing up factor and root: \((x-a)\) is the factor, while \(a\) is the root.

Quick practice thinking

If you are asked whether \((x-4)\) is a factor of a polynomial, ask yourself:

• What value should I substitute? Answer: \(4\).
• What result am I hoping for? Answer: \(0\).

If you are asked whether \((x+1)\) is a factor, then substitute \(-1\) and check whether the result is 0.

Summary

The Factor Theorem says that \((x-a)\) is a factor of \(P(x)\) exactly when \(P(a)=0\). To test a linear factor, substitute the matching value into the polynomial. If the result is 0, the factor works; if not, it does not. This theorem is a fast and powerful way to connect factors, roots, and the behavior of polynomials.

Factorizing Quadratic Trinomials

A quadratic trinomial is an expression with three terms and highest power 2. Some common examples are \(x^2+5x+6\), \(2x^2+7x+3\), and \(3x^2-11x-4\).

To factorize a quadratic trinomial means to rewrite it as a product of two simpler expressions, usually two binomials. For example, factorizing

$$x^2+5x+6$$

gives

$$x^2+5x+6=(x+2)(x+3).$$

Factorizing is useful because it helps us solve equations, find roots, and understand the behavior of a polynomial.

In this lesson, you will learn how to factor quadratic trinomials by:

• recognizing the form of a quadratic trinomial,
• using algebraic identities when possible,
• splitting the middle term,
• and checking that your factorization is correct.

1. The basic form of a quadratic trinomial

A quadratic trinomial usually has the form

$$ax^2+bx+c$$

where:

• \(a\), \(b\), and \(c\) are numbers,
• \(a\neq 0\),
• \(ax^2\) is the quadratic term,
• \(bx\) is the middle term,
• \(c\) is the constant term.

Our goal is to rewrite it in a form like

$$ax^2+bx+c=(px+q)(rx+s).$$

2. Factorizing when the coefficient of \(x^2\) is 1

First, let us look at the simpler case where \(a=1\). Then the trinomial looks like

$$x^2+bx+c.$$

We want two numbers that:

• multiply to give \(c\),
• add to give \(b\).

If we find those two numbers, we can write

$$x^2+bx+c=(x+m)(x+n)$$

where \(m+n=b\) and \(mn=c\).

Worked Example 1

Factorize:

$$x^2+7x+12$$

We need two numbers that multiply to \(12\) and add to \(7\).

• \(1\) and \(12\) add to \(13\)
• \(2\) and \(6\) add to \(8\)
• \(3\) and \(4\) add to \(7\)

So the correct pair is \(3\) and \(4\).

$$x^2+7x+12=(x+3)(x+4)$$

Check:

$$ (x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12 $$

The factorization is correct.

Worked Example 2

Factorize:

$$x^2-5x+6$$

We need two numbers that multiply to \(6\) and add to \(-5\).

Since the product is positive and the sum is negative, both numbers must be negative.

• \(-2\) and \(-3\) multiply to \(6\)
• \(-2)+(-3)=-5\)

So,

$$x^2-5x+6=(x-2)(x-3).$$

3. Special algebraic identities

Some quadratic trinomials match well-known patterns. These are important because they make factorizing faster.

Perfect square trinomials

$$x^2+2ax+a^2=(x+a)^2$$ $$x^2-2ax+a^2=(x-a)^2$$

Examples:

• \(x^2+6x+9=(x+3)^2\)
• \(x^2-10x+25=(x-5)^2\)

Difference of squares

$$x^2-a^2=(x-a)(x+a)$$

Example:

• \(x^2-16=(x-4)(x+4)\)

Notice that \(x^2-16\) is not a trinomial, but this identity is still very useful when factorizing expressions.

4. Factorizing when the coefficient of \(x^2\) is not 1

Now consider a trinomial like

$$ax^2+bx+c$$

where \(a\neq 1\). A very useful method is splitting the middle term.

Here is the idea:

1. Multiply \(a\) and \(c\).
2. Find two numbers that multiply to \(ac\) and add to \(b\).
3. Split the middle term using those two numbers.
4. Factor by grouping.

Worked Example 3

Factorize:

$$2x^2+7x+3$$

Step 1: Multiply \(a\) and \(c\).

$$a\cdot c=2\cdot 3=6$$

Step 2: Find two numbers that multiply to \(6\) and add to \(7\).

Those numbers are \(6\) and \(1\).

Step 3: Split the middle term \(7x\) as \(6x+x\).

$$2x^2+7x+3=2x^2+6x+x+3$$

Step 4: Factor by grouping.

$$2x^2+6x+x+3=2x(x+3)+1(x+3)$$

Now both groups contain \((x+3)\).

$$2x(x+3)+1(x+3)=(x+3)(2x+1)$$

So the factorization is

$$2x^2+7x+3=(x+3)(2x+1).$$

Check:

$$ (x+3)(2x+1)=2x^2+x+6x+3=2x^2+7x+3 $$

Worked Example 4

Factorize:

$$3x^2-11x-4$$

Step 1: Multiply \(a\) and \(c\).

$$a\cdot c=3\cdot(-4)=-12$$

Step 2: Find two numbers that multiply to \(-12\) and add to \(-11\).

Those numbers are \(-12\) and \(1\).

Step 3: Split the middle term.

$$3x^2-11x-4=3x^2-12x+x-4$$

Step 4: Factor by grouping.

$$3x^2-12x+x-4=3x(x-4)+1(x-4)$$

Now factor out the common binomial:

$$3x(x-4)+1(x-4)=(x-4)(3x+1)$$

So,

$$3x^2-11x-4=(x-4)(3x+1).$$

5. How to know the signs of the factors

The signs in the factors depend on the middle term and constant term.

• If the constant term is positive, the two numbers have the same sign.
• If the constant term is negative, the two numbers have different signs.
• If the middle term is positive, the larger positive value usually appears in the sum.
• If the middle term is negative, the negative value must be stronger in the sum.

For example:

• \(x^2+8x+15=(x+3)(x+5)\)
• \(x^2-8x+15=(x-3)(x-5)\)
• \(x^2+2x-15=(x+5)(x-3)\)

6. Common mistakes to avoid

• Using numbers that multiply correctly but do not add correctly. Always check both conditions.
• Forgetting the signs. Negative signs matter a lot in factorization.
• Not factoring out a common factor first. For example, before factorizing \(2x^2+4x+2\), first take out \(2\): $$2x^2+4x+2=2(x^2+2x+1)=2(x+1)^2$$
• Stopping too early. Make sure the expression is fully factorized.

7. A step-by-step strategy

When you see a quadratic trinomial, use this checklist:

1. Look for a greatest common factor first.
2. Check whether it matches a special identity, like a perfect square.
3. If the trinomial is \(x^2+bx+c\), find two numbers that multiply to \(c\) and add to \(b\).
4. If the trinomial is \(ax^2+bx+c\) with \(a\neq 1\), use splitting the middle term.
5. Expand your answer to check that it matches the original expression.

8. Why this works

When we multiply two binomials, we use expansion. For example,

$$ (x+m)(x+n)=x^2+(m+n)x+mn $$

This shows exactly why we look for numbers that add to the middle coefficient and multiply to the constant term.

For expressions like \(ax^2+bx+c\), splitting the middle term helps us rewrite the trinomial into four terms, so that grouping becomes possible.

Brief Summary

To factorize a quadratic trinomial, rewrite it as a product of two binomials. If the coefficient of \(x^2\) is 1, find two numbers that multiply to the constant term and add to the middle coefficient. If the coefficient of \(x^2\) is not 1, multiply \(a\) and \(c\), split the middle term, and factor by grouping. Always check for common factors first and confirm your answer by expanding.

Factorizing Cubic Polynomials means rewriting a polynomial of degree 3 as a product of simpler factors. A cubic polynomial usually looks like \(ax^3+bx^2+cx+d\), where \(a \neq 0\).

In this lesson, you will learn how to factorize cubic polynomials by:

• finding a rational root,
• using the factor theorem, and
• dividing the cubic by the factor you found to get a quadratic.

This is an important skill because once a cubic is factorized, it becomes much easier to solve equations, find roots, and understand the graph of the polynomial.

Step 1: Understand what a root means

If a number \(r\) is a root of a polynomial, then substituting \(x=r\) makes the polynomial equal to 0.

For example, if \(f(x)=x^3-6x^2+11x-6\), then \(x=1\) is a root if:

$$f(1)=1^3-6(1)^2+11(1)-6=1-6+11-6=0$$

Since the value is 0, \(1\) is a root.

Step 2: Use the factor theorem

The factor theorem says:

If \(f(r)=0\), then \((x-r)\) is a factor of \(f(x)\).

So if you test a value and get 0, you have found a factor immediately.

For example:

• If \(f(2)=0\), then \((x-2)\) is a factor.
• If \(f(-3)=0\), then \((x+3)\) is a factor.

Step 3: Choose possible rational roots

To factorize a cubic, we usually start by testing simple integer values. The best place to start is with the factors of the constant term.

For a polynomial like

$$f(x)=x^3+2x^2-5x-6$$

the constant term is \(-6\), so the possible integer roots are:

$$\pm 1,\ \pm 2,\ \pm 3,\ \pm 6$$

Why these numbers? Because if a rational root is an integer, it often comes from the factors of the constant term.

Step 4: Test the possible roots

Substitute each possible value into the polynomial until you get 0.

Suppose we test \(x=1\):

$$1^3+2(1)^2-5(1)-6=1+2-5-6=-8$$

Not 0, so \(x=1\) is not a root.

Now test \(x=2\):

$$2^3+2(2)^2-5(2)-6=8+8-10-6=0$$

So \(x=2\) is a root, which means \((x-2)\) is a factor.

Step 5: Divide the cubic by the factor

Once you know one factor, divide the cubic by that factor. The result will be a quadratic polynomial.

You can use polynomial division. Then factorize the quadratic if possible.

For example, divide

$$x^3+2x^2-5x-6$$

by \((x-2)\). The quotient is:

$$x^2+4x+3$$

So:

$$x^3+2x^2-5x-6=(x-2)(x^2+4x+3)$$

Now factor the quadratic:

$$x^2+4x+3=(x+1)(x+3)$$

So the full factorization is:

$$x^3+2x^2-5x-6=(x-2)(x+1)(x+3)$$

A useful strategy

1. Write down the possible integer roots from the constant term.
2. Test them one by one using substitution.
3. When you find a root, write the corresponding linear factor.
4. Divide the cubic by that factor.
5. Factor the quadratic if possible.

This method works especially well when the cubic has integer coefficients and at least one rational root.

Worked Example 1

Factorize:

$$x^3-6x^2+11x-6$$

Step 1: List possible roots

The constant term is \(-6\), so possible integer roots are:

$$\pm 1,\ \pm 2,\ \pm 3,\ \pm 6$$

Step 2: Test a value

Try \(x=1\):

$$1^3-6(1)^2+11(1)-6=1-6+11-6=0$$

So \((x-1)\) is a factor.

Step 3: Divide by \((x-1)\)

Dividing \(x^3-6x^2+11x-6\) by \((x-1)\) gives:

$$x^2-5x+6$$

So:

$$x^3-6x^2+11x-6=(x-1)(x^2-5x+6)$$

Step 4: Factor the quadratic

$$x^2-5x+6=(x-2)(x-3)$$

Final answer:

$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$$

Worked Example 2

Factorize:

$$x^3+2x^2-5x-6$$

Step 1: Possible roots

$$\pm 1,\ \pm 2,\ \pm 3,\ \pm 6$$

Step 2: Test values

Test \(x=1\):

$$1+2-5-6=-8$$

Not a root.

Test \(x=2\):

$$8+8-10-6=0$$

So \((x-2)\) is a factor.

Step 3: Divide

$$x^3+2x^2-5x-6 \div (x-2)=x^2+4x+3$$

Step 4: Factor the quadratic

$$x^2+4x+3=(x+1)(x+3)$$

Final answer:

$$x^3+2x^2-5x-6=(x-2)(x+1)(x+3)$$

Worked Example 3

Factorize:

$$2x^3-x^2-8x+4$$

This example is a little harder because the coefficient of \(x^3\) is 2, not 1.

Step 1: Look for a rational root

Start by testing simple values. Try \(x=2\):

$$2(2^3)-(2^2)-8(2)+4=16-4-16+4=0$$

So \(x=2\) is a root, and \((x-2)\) is a factor.

Step 2: Divide by \((x-2)\)

$$2x^3-x^2-8x+4 \div (x-2)=2x^2+3x-2$$

So:

$$2x^3-x^2-8x+4=(x-2)(2x^2+3x-2)$$

Step 3: Factor the quadratic

We need two numbers that multiply to \(2 \cdot (-2)=-4\) and add to \(3\). Those numbers are \(4\) and \(-1\).

So:

$$2x^2+3x-2=(2x-1)(x+2)$$

Final answer:

$$2x^3-x^2-8x+4=(x-2)(2x-1)(x+2)$$

Worked Example 4

Factorize:

$$x^3-4x^2-7x+10$$

Step 1: Possible roots

The constant term is 10, so test:

$$\pm 1,\ \pm 2,\ \pm 5,\ \pm 10$$

Step 2: Test values

Try \(x=1\):

$$1-4-7+10=0$$

So \((x-1)\) is a factor.

Step 3: Divide

$$x^3-4x^2-7x+10 \div (x-1)=x^2-3x-10$$

So:

$$x^3-4x^2-7x+10=(x-1)(x^2-3x-10)$$

Step 4: Factor the quadratic

$$x^2-3x-10=(x-5)(x+2)$$

Final answer:

$$x^3-4x^2-7x+10=(x-1)(x-5)(x+2)$$

Common mistakes to avoid

• Forgetting negative roots: If the constant term is 6, test both positive and negative factors: \(\pm 1, \pm 2, \pm 3, \pm 6\).
• Using the wrong factor: If \(x=3\) is a root, the factor is \((x-3)\), not \((x+3)\).
• Stopping too early: After finding one factor, you still need to factor the quadratic quotient.
• Arithmetic errors when testing roots: Substitute carefully and use brackets when needed.

How this connects to solving equations

If you are asked to solve a cubic equation such as

$$x^3+2x^2-5x-6=0$$

first factorize it:

$$x^3+2x^2-5x-6=(x-2)(x+1)(x+3)$$

Then set each factor equal to 0:

$$x-2=0 \quad \text{or} \quad x+1=0 \quad \text{or} \quad x+3=0$$

So the solutions are:

$$x=2,\ -1,\ -3$$

Quick checklist for factorizing cubic polynomials

• Is the expression written in descending powers of \(x\)?
• Did you list the possible roots from the constant term?
• Did you test values using substitution?
• Did you use the factor theorem correctly?
• Did you divide by the factor you found?
• Did you fully factor the quadratic quotient?

Summary

To factorize a cubic polynomial, first look for a rational root by testing likely values, usually the factors of the constant term. When a value makes the polynomial equal to 0, the factor theorem tells you the corresponding linear factor.

Next, divide the cubic by that linear factor to get a quadratic. Then factor the quadratic if possible. This turns one complicated cubic expression into a product of simpler factors.

Relationship Between Zeros and Coefficients of a Quadratic Polynomial

When we work with a quadratic polynomial, we often want to know about its zeros, also called its roots. These are the values of the variable that make the polynomial equal to zero.

For example, if a polynomial is \(x^2-5x+6\), then its zeros are the values of \(x\) that satisfy

$$x^2-5x+6=0$$

In this lesson, you will learn a very useful shortcut: instead of solving the equation completely every time, we can connect the zeros directly to the coefficients of the quadratic. This relationship is called the relationship between zeros and coefficients.

It helps us:

• find the sum of the zeros quickly,
• find the product of the zeros quickly,
• form a quadratic polynomial when the zeros are known,
• check whether our answers are correct.

1. Standard form of a quadratic polynomial

A quadratic polynomial is written in the form

$$ax^2+bx+c$$

where \(a\), \(b\), and \(c\) are real numbers and \(a \ne 0\).

If \(\alpha\) and \(\beta\) are the two zeros of the polynomial, then

$$a(x-\alpha)(x-\beta)$$

is another way to write the same quadratic.

Let us expand this expression:

$$a(x-\alpha)(x-\beta)=a\left(x^2-(\alpha+\beta)x+\alpha\beta\right)$$ $$=ax^2-a(\alpha+\beta)x+a\alpha\beta$$

Now compare this with the standard form

$$ax^2+bx+c$$

Matching the coefficients of both expressions gives us:

$$-a(\alpha+\beta)=b$$ $$a\alpha\beta=c$$

Now divide by \(a\):

$$\alpha+\beta=-\frac{b}{a}$$ $$\alpha\beta=\frac{c}{a}$$

These are the formulas we need.

2. Main formulas

If \(\alpha\) and \(\beta\) are the zeros of the quadratic polynomial \(ax^2+bx+c\), then:

$$\alpha+\beta=-\frac{b}{a}$$ $$\alpha\beta=\frac{c}{a}$$

So:

• Sum of zeros = negative of coefficient of \(x\) divided by coefficient of \(x^2\)
• Product of zeros = constant term divided by coefficient of \(x^2\)

This is often called Vieta's formulas for quadratics.

3. Understanding the idea with factor form

If the zeros are \(\alpha\) and \(\beta\), then the polynomial can be written as

$$a(x-\alpha)(x-\beta)$$

The factor \((x-\alpha)\) becomes zero when \(x=\alpha\), and the factor \((x-\beta)\) becomes zero when \(x=\beta\). That is why \(\alpha\) and \(\beta\) are the zeros.

When we expand the factors, the middle term contains \((\alpha+\beta)\), and the constant term contains \(\alpha\beta\). That is why the sum and product of the zeros are connected to the coefficients.

4. Worked Example 1: Find sum and product of zeros

Find the sum and product of the zeros of

$$x^2-7x+12$$

Here,

• \(a=1\)
• \(b=-7\)
• \(c=12\)

Use the formulas:

$$\alpha+\beta=-\frac{b}{a}=-\frac{-7}{1}=7$$ $$\alpha\beta=\frac{c}{a}=\frac{12}{1}=12$$

Answer:

• Sum of zeros = \(7\)
• Product of zeros = \(12\)

We can check by factoring:

$$x^2-7x+12=(x-3)(x-4)$$

The zeros are \(3\) and \(4\).

Now check:

$$3+4=7, \quad 3\times 4=12$$

The formulas work.

5. Worked Example 2: Quadratic with leading coefficient not equal to 1

Find the sum and product of the zeros of

$$2x^2-5x-3$$

Here,

• \(a=2\)
• \(b=-5\)
• \(c=-3\)

Apply the formulas:

$$\alpha+\beta=-\frac{b}{a}=-\frac{-5}{2}=\frac{5}{2}$$ $$\alpha\beta=\frac{c}{a}=\frac{-3}{2}$$

Answer:

• Sum of zeros = \(\frac{5}{2}\)
• Product of zeros = \(-\frac{3}{2}\)

Let us check by factoring:

$$2x^2-5x-3=(2x+1)(x-3)$$

The zeros are:

$$2x+1=0 \Rightarrow x=-\frac{1}{2}$$ $$x-3=0 \Rightarrow x=3$$

Now check:

$$-\frac{1}{2}+3=\frac{5}{2}$$ $$-\frac{1}{2}\times 3=-\frac{3}{2}$$

Again, the formulas are correct.

6. Worked Example 3: Form a quadratic polynomial from its zeros

Form a quadratic polynomial whose zeros are \(2\) and \(-5\).

If the zeros are \(\alpha\) and \(\beta\), then a quadratic polynomial having those zeros is

$$k(x-\alpha)(x-\beta)$$

For the simplest polynomial, take \(k=1\).

So,

$$p(x)=(x-2)(x-(-5))=(x-2)(x+5)$$

Expand:

$$p(x)=x^2+5x-2x-10$$ $$p(x)=x^2+3x-10$$

Answer: A quadratic polynomial is

$$x^2+3x-10$$

Now check the relationship:

• Sum of zeros = \(2+(-5)=-3\)
• Product of zeros = \(2\times(-5)=-10\)

For the polynomial \(x^2+3x-10\), we have \(a=1\), \(b=3\), \(c=-10\).

Using the formulas:

$$-\frac{b}{a}=-\frac{3}{1}=-3$$ $$\frac{c}{a}=\frac{-10}{1}=-10$$

Everything matches.

7. Worked Example 4: Find the quadratic polynomial using sum and product

The sum of the zeros of a quadratic polynomial is \(8\), and their product is \(15\). Find a quadratic polynomial with these zeros.

Let the zeros be \(\alpha\) and \(\beta\).

We know:

$$\alpha+\beta=8$$ $$\alpha\beta=15$$

A monic quadratic polynomial with these zeros is

$$x^2-(\alpha+\beta)x+\alpha\beta$$

Substitute the given values:

$$x^2-8x+15$$

Answer: One such polynomial is

$$x^2-8x+15$$

We can factor it:

$$x^2-8x+15=(x-3)(x-5)$$

The zeros are \(3\) and \(5\), and indeed:

$$3+5=8, \quad 3\times 5=15$$

8. Important result: polynomial from sum and product

If a quadratic polynomial has zeros \(\alpha\) and \(\beta\), then one polynomial with those zeros is

$$x^2-(\alpha+\beta)x+\alpha\beta$$

This is very useful when the sum and product are given directly.

For example, if sum = \(S\) and product = \(P\), then the polynomial is

$$x^2-Sx+P$$

9. Common mistakes to avoid

• Forgetting the negative sign in the sum formula. The sum of zeros is \(-\frac{b}{a}\), not \(\frac{b}{a}\).
• Using the wrong coefficients. Always compare the polynomial with \(ax^2+bx+c\) carefully.
• Ignoring the value of \(a\). If \(a\neq 1\), you must divide by \(a\).
• Mixing up sum and product. Sum corresponds to the coefficient of \(x\), while product corresponds to the constant term.

10. Quick practice ideas

Try applying the formulas to each of these:

• \(x^2+6x+8\)
• \(3x^2-x-2\)
• Quadratic with zeros \(4\) and \(\frac{1}{2}\)
• Quadratic whose sum of zeros is \(-2\) and product is \(-15\)

11. Final summary

For a quadratic polynomial \(ax^2+bx+c\), if the zeros are \(\alpha\) and \(\beta\), then:

$$\alpha+\beta=-\frac{b}{a}$$ $$\alpha\beta=\frac{c}{a}$$

These formulas let us connect the zeros directly to the coefficients.

They also help us write a quadratic polynomial from its zeros:

$$x^2-(\alpha+\beta)x+\alpha\beta$$

This relationship is one of the most useful tools in solving quadratic polynomial problems quickly and accurately.

Relationship Between Zeros and Coefficients of a Cubic Polynomial

When a polynomial is written in factored form, its zeros are easy to see. But many times, a cubic polynomial is given in expanded form, and the zeros are not obvious. In those cases, the relationship between zeros and coefficients helps us connect the roots directly to the numbers in the polynomial.

For quadratic polynomials, students often learn that the sum and product of the zeros can be found from the coefficients. The same idea also works for cubic polynomials, but now there are three zeros instead of two.

In this lesson, we will learn how to connect the zeros of a cubic polynomial to its coefficients, how to write a cubic polynomial when the zeros are known, and how to use these relationships to solve problems.

1. Standard form of a cubic polynomial

A cubic polynomial is a polynomial of degree 3. Its standard form is

$$p(x)=ax^3+bx^2+cx+d$$

where:

• \(a, b, c, d\) are constants,

• \(a \neq 0\), because otherwise it would not be cubic.

Suppose the three zeros of this polynomial are \(\alpha\), \(\beta\), and \(\gamma\). Then the polynomial can be written as

$$p(x)=a(x-\alpha)(x-\beta)(x-\gamma)$$

This factored form is the starting point for finding the relationship between zeros and coefficients.

2. Expanding the factored form

Let us expand

$$a(x-\alpha)(x-\beta)(x-\gamma)$$

First, multiply two factors:

$$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta$$

Now multiply by \((x-\gamma)\):

$$\big(x^2-(\alpha+\beta)x+\alpha\beta\big)(x-\gamma)$$

After expanding and simplifying, we get

$$x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$$

Now multiply the whole expression by \(a\):

$$p(x)=a\left[x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma\right]$$

So,

$$p(x)=ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\beta\gamma+\gamma\alpha)x-a\alpha\beta\gamma$$

Compare this with the standard form

$$p(x)=ax^3+bx^2+cx+d$$

By matching the coefficients, we get the important formulas.

3. Vieta's formulas for a cubic polynomial

If \(\alpha\), \(\beta\), and \(\gamma\) are the zeros of

$$ax^3+bx^2+cx+d$$

then:

$$\alpha+\beta+\gamma=-\frac{b}{a}$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}$$ $$\alpha\beta\gamma=-\frac{d}{a}$$

These three equations are the relationship between zeros and coefficients for a cubic polynomial.

It is important to notice the pattern of signs:

• The sum of the zeros is negative: \(-\frac{b}{a}\)

• The sum of the products of the zeros taken two at a time is positive: \(\frac{c}{a}\)

• The product of the three zeros is negative: \(-\frac{d}{a}\)

4. What does “products taken two at a time” mean?

If the zeros are \(\alpha\), \(\beta\), and \(\gamma\), then the products taken two at a time are:

$$\alpha\beta,\ \beta\gamma,\ \gamma\alpha$$

Their sum is

$$\alpha\beta+\beta\gamma+\gamma\alpha$$

This middle relationship is very important in cubic polynomials, so be careful not to forget any term.

5. How to use these formulas

You can use the relationships in two main ways:

1. If the polynomial is given, you can find the sum of zeros, the sum of pairwise products, and the product of zeros without solving the polynomial.

2. If the zeros are given, you can build the cubic polynomial.

Let us now see worked examples.

Worked Example 1: Find the relationships from a given polynomial

Find the sum of zeros, the sum of products of zeros taken two at a time, and the product of zeros for

$$p(x)=2x^3-5x^2+3x-7$$

Here,

$$a=2,\quad b=-5,\quad c=3,\quad d=-7$$

Using the formulas:

$$\alpha+\beta+\gamma=-\frac{b}{a}=-\frac{-5}{2}=\frac{5}{2}$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=\frac{3}{2}$$ $$\alpha\beta\gamma=-\frac{d}{a}=-\frac{-7}{2}=\frac{7}{2}$$

Answer:

• Sum of zeros = \(\frac{5}{2}\)

• Sum of pairwise products = \(\frac{3}{2}\)

• Product of zeros = \(\frac{7}{2}\)

Notice that we did not need to actually find the zeros.

Worked Example 2: Form a cubic polynomial when the zeros are known

Form a cubic polynomial whose zeros are \(1\), \(2\), and \(3\).

If the zeros are \(1, 2, 3\), then a polynomial with these zeros is

$$p(x)=(x-1)(x-2)(x-3)$$

First multiply two factors:

$$(x-1)(x-2)=x^2-3x+2$$

Now multiply by \((x-3)\):

$$ (x^2-3x+2)(x-3) $$ $$ =x^3-3x^2-3x^2+9x+2x-6 $$ $$ =x^3-6x^2+11x-6 $$

So the required polynomial is

$$p(x)=x^3-6x^2+11x-6$$

Let us check using the relationships:

• Sum of zeros = \(1+2+3=6\)

• Coefficient formula gives \(-\frac{b}{a}=-\frac{-6}{1}=6\)

• Sum of pairwise products = \((1)(2)+(2)(3)+(3)(1)=2+6+3=11\)

• Coefficient formula gives \(\frac{c}{a}=\frac{11}{1}=11\)

• Product of zeros = \((1)(2)(3)=6\)

• Coefficient formula gives \(-\frac{d}{a}=-\frac{-6}{1}=6\)

Everything matches correctly.

Worked Example 3: Form a polynomial using the relationships directly

Find a cubic polynomial whose zeros are \(2\), \(-1\), and \(4\).

Let the polynomial be monic, which means the coefficient of \(x^3\) is 1. Then:

$$p(x)=x^3-b_1x^2+b_2x-b_3$$

Instead of using new formulas, we will use the relationships directly.

First find the needed values:

$$\alpha+\beta+\gamma=2+(-1)+4=5$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=(2)(-1)+(-1)(4)+(4)(2)=-2-4+8=2$$ $$\alpha\beta\gamma=(2)(-1)(4)=-8$$

Now compare with

$$x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$$

Substitute the values:

$$p(x)=x^3-5x^2+2x-(-8)$$ $$p(x)=x^3-5x^2+2x+8$$

So the required polynomial is

$$p(x)=x^3-5x^2+2x+8$$

Worked Example 4: Find an unknown number using the relationships

One zero of the cubic polynomial \(x^3-9x^2+23x-k\) is \(2\). Find \(k\) if the other two zeros are also real numbers.

Let the three zeros be \(2\), \(\beta\), and \(\gamma\).

From the polynomial,

$$a=1,\quad b=-9,\quad c=23,\quad d=-k$$

Using the sum of zeros:

$$2+\beta+\gamma=-\frac{b}{a}=9$$ $$\beta+\gamma=7$$

Using the sum of products of zeros taken two at a time:

$$2\beta+2\gamma+\beta\gamma=\frac{c}{a}=23$$

Since \(\beta+\gamma=7\),

$$2(\beta+\gamma)+\beta\gamma=23$$ $$2(7)+\beta\gamma=23$$ $$14+\beta\gamma=23$$ $$\beta\gamma=9$$

Now use the product of zeros:

$$2\cdot\beta\cdot\gamma=-\frac{d}{a}=k$$

Since \(\beta\gamma=9\),

$$2\cdot 9=k$$ $$k=18$$

So the value of \(k\) is

$$k=18$$

6. Important observations

• The formulas work whether the zeros are positive, negative, integers, fractions, or repeated zeros.

• If a zero is repeated, it is counted again. For example, if the zeros are \(2,2,-1\), then all three are used in the formulas.

• You do not need to factor the polynomial to use these relationships.

• Be careful with signs, especially in \(-\frac{b}{a}\) and \(-\frac{d}{a}\).

7. Common mistakes to avoid

• Forgetting the middle expression: The sum of products taken two at a time is \(\alpha\beta+\beta\gamma+\gamma\alpha\), not just one product.

• Sign errors: For a cubic, the product of the zeros is \(-\frac{d}{a}\), not \(\frac{d}{a}\).

• Ignoring the leading coefficient: If the polynomial is not monic, always divide by \(a\) in the formulas.

• Mixing up coefficients: In \(ax^3+bx^2+cx+d\), the coefficient of \(x^2\) is \(b\), of \(x\) is \(c\), and the constant term is \(d\).

8. Quick review table

For the cubic polynomial

$$ax^3+bx^2+cx+d$$

with zeros \(\alpha, \beta, \gamma\):

• \(\alpha+\beta+\gamma=-\frac{b}{a}\)

• \(\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}\)

• \(\alpha\beta\gamma=-\frac{d}{a}\)

If the polynomial is monic, meaning \(a=1\), then

$$x^3+bx^2+cx+d$$

has:

• Sum of zeros = \(-b\)

• Sum of pairwise products = \(c\)

• Product of zeros = \(-d\)

9. Summary

The zeros of a cubic polynomial and its coefficients are closely connected. If \(\alpha\), \(\beta\), and \(\gamma\) are the zeros of \(ax^3+bx^2+cx+d\), then:

$$\alpha+\beta+\gamma=-\frac{b}{a},\quad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},\quad \alpha\beta\gamma=-\frac{d}{a}$$

These formulas let you find information about the zeros without solving the polynomial, and they also help you build a polynomial when the zeros are known. With careful attention to signs and coefficients, these relationships become a powerful tool for working with cubic polynomials.

Algebraic Identities and Expansions are short, reliable formulas that help us expand expressions quickly and accurately. Instead of multiplying everything from scratch every time, we can use these patterns to save time and reduce mistakes.

In this lesson, you will learn the most important identities for squares and cubes of binomials, how to expand expressions carefully, and how to simplify the result. You will also see how these identities connect to polynomials, since expansions often produce polynomial expressions.

What is an algebraic identity?

An algebraic identity is an equation that is true for all values of the variables involved. For example,

$$ (a+b)^2=a^2+2ab+b^2 $$

is an identity because no matter what numbers we choose for \(a\) and \(b\), both sides are equal.

These identities are especially useful when working with expressions like \((x+3)^2\), \((2x-5)^2\), or \((x+y)^3\).

Main identities to know

You should memorize these standard identities:

• Square of a sum: \((a+b)^2=a^2+2ab+b^2\)
• Square of a difference: \((a-b)^2=a^2-2ab+b^2\)
• Product of sum and difference: \((a+b)(a-b)=a^2-b^2\)
• Cube of a sum: \((a+b)^3=a^3+3a^2b+3ab^2+b^3\)
• Cube of a difference: \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)

Notice the patterns carefully. For squares, there are three terms. For cubes, there are four terms.

Understanding square expansions

Let us start with \((a+b)^2\). This means

$$ (a+b)^2=(a+b)(a+b) $$

Now multiply each term:

$$ (a+b)(a+b)=a^2+ab+ab+b^2=a^2+2ab+b^2 $$

That is why the middle term is \(2ab\), not just \(ab\).

Similarly,

$$ (a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ab+b^2 $$

A very common mistake is writing

$$ (a-b)^2=a^2-b^2 $$

This is wrong. The correct expansion includes the middle term:

$$ (a-b)^2=a^2-2ab+b^2 $$

The difference of squares

Now look at

$$ (a+b)(a-b) $$

Expanding gives:

$$ (a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2 $$

The middle terms cancel out. This identity is called the difference of squares.

Understanding cube expansions

The cube identities are:

$$ (a+b)^3=a^3+3a^2b+3ab^2+b^3 $$

$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3 $$

These are important patterns. Notice the coefficients \(1, 3, 3, 1\). Also notice that in \((a-b)^3\), the signs alternate:

$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3 $$

How to expand step by step

When expanding, follow these steps:

1. Identify whether the expression matches a known identity.
2. Substitute the correct parts into the formula.
3. Expand carefully, especially the middle term.
4. Simplify the final expression.

If the expression does not directly match an identity, you can still expand by multiplication and then simplify.

Worked Example 1: Expanding a square of a binomial

Expand \((x+5)^2\).

Use the identity

$$ (a+b)^2=a^2+2ab+b^2 $$

Here, \(a=x\) and \(b=5\).

Substitute into the formula:

$$ (x+5)^2=x^2+2(x)(5)+5^2 $$

Now simplify:

$$ (x+5)^2=x^2+10x+25 $$

Answer: \(x^2+10x+25\)

Worked Example 2: Expanding a square with a negative sign

Expand \((2x-3)^2\).

Use the identity

$$ (a-b)^2=a^2-2ab+b^2 $$

Here, \(a=2x\) and \(b=3\).

Substitute:

$$ (2x-3)^2=(2x)^2-2(2x)(3)+3^2 $$

Simplify each part:

$$ (2x)^2=4x^2,\, -2(2x)(3)=-12x,\, 3^2=9 $$

So,

$$ (2x-3)^2=4x^2-12x+9 $$

Answer: \(4x^2-12x+9\)

Worked Example 3: Using the difference of squares

Expand \((3x+4)(3x-4)\).

This matches the identity

$$ (a+b)(a-b)=a^2-b^2 $$

Here, \(a=3x\) and \(b=4\).

Substitute:

$$ (3x+4)(3x-4)=(3x)^2-4^2 $$

Simplify:

$$ (3x+4)(3x-4)=9x^2-16 $$

Answer: \(9x^2-16\)

Worked Example 4: Expanding a cube

Expand \((x+2)^3\).

Use the identity

$$ (a+b)^3=a^3+3a^2b+3ab^2+b^3 $$

Here, \(a=x\) and \(b=2\).

Substitute:

$$ (x+2)^3=x^3+3x^2(2)+3x(2^2)+2^3 $$

Simplify step by step:

$$ (x+2)^3=x^3+6x^2+12x+8 $$

Answer: \(x^3+6x^2+12x+8\)

Expanding expressions with more than one identity

Sometimes an expression may contain a square and then need simplification with other terms. For example, if you need to simplify

$$ (x+1)^2+(x-1)^2 $$

expand each part first:

$$ (x+1)^2=x^2+2x+1 $$

$$ (x-1)^2=x^2-2x+1 $$

Now add them:

$$ (x^2+2x+1)+(x^2-2x+1)=2x^2+2 $$

The \(+2x\) and \(-2x\) cancel out.

What about trinomials?

A trinomial has three terms, such as \((a+b+c)\). You may sometimes need to square a trinomial.

A useful method is to treat it as a binomial first:

$$ (a+b+c)^2=((a+b)+c)^2 $$

Then use the square identity:

$$ ((a+b)+c)^2=(a+b)^2+2(a+b)c+c^2 $$

Now expand \((a+b)^2\):

$$ (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc $$

This is the standard expansion for the square of a trinomial.

For example,

$$ (x+y+2)^2=x^2+y^2+4+2xy+4x+4y $$

Important tips for avoiding mistakes

• Do not forget the middle term in square identities.
• Be very careful with negative signs.
• Square the entire term, not just the number. For example, \((2x)^2=4x^2\), not \(2x^2\).
• For cubes, remember the pattern \(1,3,3,1\).
• Always simplify completely at the end.

Common errors

• Writing \((a+b)^2=a^2+b^2\) instead of \(a^2+2ab+b^2\)
• Writing \((a-b)^2=a^2-b^2\) instead of \(a^2-2ab+b^2\)
• Forgetting that \((-b)^2=b^2\)
• Mixing up \((a-b)^3\) and \((a+b)^3\)

Quick practice to think about

• Expand \((x-7)^2\)
• Expand \((2x+1)^3\)
• Expand \((5y+3)(5y-3)\)
• Expand \((a+b+c)^2\)

Try using the identities before doing direct multiplication. With practice, you will begin to recognize the patterns immediately.

Summary

Algebraic identities are standard formulas that make expansion faster and easier. The most important ones for this topic are the square and cube identities for sums and differences, and the difference of squares identity.

When expanding, identify the pattern, substitute carefully, and simplify. If you watch the signs and remember the middle terms, you can expand polynomial expressions accurately and with confidence.