Polynomial Operations and Factoring

Adding, subtracting, and multiplying polynomials means working with algebra expressions that have one or more terms, such as \(3x^2 + 2x - 5\).

A polynomial is made of terms. Each term has a coefficient and may include a variable with a whole-number exponent. Examples of polynomials are \(4x+7\), \(2x^2-3x+1\), and \(5a^3-2a+9\).

In this lesson, you will learn how to:

• Add polynomials by combining like terms
• Subtract polynomials by distributing the minus sign and then combining like terms
• Multiply polynomials by using the distributive property carefully

These skills are important because polynomial expressions show up often in algebra, equations, graphs, and later factoring work.

1. Understanding terms and like terms

A term is a part of an expression separated by a plus or minus sign. For example, in \(5x^2 - 3x + 8\), the terms are \(5x^2\), \(-3x\), and \(8\).

Like terms have the same variable part. That means the variables and their exponents must match exactly.

• \(3x\) and \(-7x\) are like terms
• \(4x^2\) and \(9x^2\) are like terms
• \(5ab\) and \(-2ab\) are like terms
• \(x\) and \(x^2\) are not like terms
• \(2a\) and \(2b\) are not like terms

You can only combine like terms.

2. Adding polynomials

To add polynomials, remove parentheses if needed and then combine like terms.

It helps to line up terms with the same degree, but this is not required. What matters most is grouping like terms correctly.

Steps for adding polynomials:

1. Write the expressions next to each other with a plus sign
2. Remove parentheses
3. Group like terms
4. Add the coefficients
5. Write the answer in standard form, from greatest exponent to least

Worked Example 1: Adding polynomials

Simplify:

$$ (3x^2 + 5x - 4) + (2x^2 - 7x + 6) $$

Combine like terms:

$$ 3x^2 + 2x^2 + 5x - 7x - 4 + 6 $$ $$ = 5x^2 - 2x + 2 $$

Answer: \(5x^2 - 2x + 2\)

Notice that:

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

3. Subtracting polynomials

Subtracting polynomials is very similar to adding, but there is one important extra step: distribute the subtraction sign to every term in the second polynomial.

This means each sign inside the second set of parentheses changes.

Steps for subtracting polynomials:

1. Write the subtraction problem clearly
2. Change the signs of every term in the polynomial being subtracted
3. Remove parentheses
4. Combine like terms

Worked Example 2: Subtracting polynomials

Simplify:

$$ (4x^2 - 3x + 1) - (x^2 + 5x - 6) $$

Distribute the minus sign to the second polynomial:

$$ 4x^2 - 3x + 1 - x^2 - 5x + 6 $$

Now combine like terms:

$$ 4x^2 - x^2 - 3x - 5x + 1 + 6 $$ $$ = 3x^2 - 8x + 7 $$

Answer: \(3x^2 - 8x + 7\)

A common mistake is forgetting to change every sign in the second polynomial. Be careful with this step.

4. Multiplying polynomials

To multiply polynomials, use the distributive property. This means each term in one polynomial must be multiplied by each term in the other polynomial.

When multiplying terms:

• Multiply the coefficients
• Multiply variables using exponent rules

For example:

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

because \(3 \cdot 2 = 6\) and \(x^2 \cdot x^3 = x^5\).

Multiplying a monomial by a polynomial

If one expression has only one term, multiply that term by each term in the polynomial.

Worked Example 3: Monomial times polynomial

Simplify:

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

Distribute \(3x\) to each term:

$$ (3x)(2x^2) + (3x)(-5x) + (3x)(4) $$ $$ = 6x^3 - 15x^2 + 12x $$

Answer: \(6x^3 - 15x^2 + 12x\)

Multiplying a binomial by a binomial

A binomial has two terms. When multiplying two binomials, each term in the first binomial must multiply each term in the second binomial.

One organized way is:

• Multiply the first term by both terms in the second binomial
• Multiply the second term by both terms in the second binomial
• Combine like terms

Worked Example 4: Binomial times binomial

Simplify:

$$ (x + 3)(x + 5) $$

Distribute:

$$ x(x+5) + 3(x+5) $$

Now multiply each term:

$$ x^2 + 5x + 3x + 15 $$

Combine like terms:

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

Answer: \(x^2 + 8x + 15\)

Here is the same multiplication shown term by term:

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

5. Important rules to remember

• Only like terms can be added or subtracted.
• When subtracting, change every sign in the polynomial being subtracted.
• When multiplying, every term must be distributed to every term.
• Write answers in standard form, starting with the highest power.

6. Common mistakes

• Combining unlike terms: \(2x + 3x^2\) cannot become \(5x^2\) or \(5x\)
• Forgetting a negative sign when subtracting
• Not multiplying every term when multiplying polynomials
• Forgetting to combine like terms at the end

7. Quick check examples

Simplify:

\((2x + 7) + (5x - 3)\)

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

Simplify:

\((6a^2 + 2a - 1) - (4a^2 - 3a + 5)\)

$$ 6a^2 + 2a - 1 - 4a^2 + 3a - 5 = 2a^2 + 5a - 6 $$

Simplify:

\((2x - 1)(x + 4)\)

$$ (2x)(x) + (2x)(4) + (-1)(x) + (-1)(4) $$ $$ = 2x^2 + 8x - x - 4 $$ $$ = 2x^2 + 7x - 4 $$

Brief summary

Adding and subtracting polynomials is mostly about combining like terms. For subtraction, be especially careful to change the signs in the second polynomial.

Multiplying polynomials uses the distributive property. Multiply each term carefully, then combine like terms and write the final answer in standard form.

With practice, these steps become much easier and more natural.

Special Binomial Products are shortcuts for multiplying certain binomials. A binomial is an expression with two terms, such as \(x+3\) or \(2a-b\).

Instead of multiplying term by term every time, we can recognize patterns and expand more quickly. In 9th Grade Maths, the most important special binomial products are:

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

Learning these patterns helps you expand expressions faster and also helps later when you factor polynomials.

1. Square of a Sum

When a binomial is multiplied by itself, we call it a square. For example, \((a+b)^2\) means \((a+b)(a+b)\).

Let us expand it:

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

So the pattern is:

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

This is called a perfect square trinomial because the result has three terms and comes from squaring a binomial.

Important: The middle term is always \(2ab\), not just \(ab\).

2. Square of a Difference

Now look at \((a-b)^2\). This means \((a-b)(a-b)\).

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

So the pattern is:

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

This is also a perfect square trinomial.

Notice: The last term is still positive, because \((-b)(-b)=+b^2\). Only the middle term is negative.

3. Product of a Sum and Difference

Now consider \((a+b)(a-b)\).

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

The middle terms cancel out, so we get the pattern:

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

This is called a difference of squares.

It has only two terms, and there is a minus sign between them.

Why These Patterns Matter

These products appear often in algebra. Recognizing them saves time and helps avoid mistakes.

• They make expansion faster.
• They help you check whether an answer is correct.
• They connect expansion and factoring.

How to Recognize Each Pattern

• If you see \((a+b)^2\), use \(a^2+2ab+b^2\).
• If you see \((a-b)^2\), use \(a^2-2ab+b^2\).
• If you see \((a+b)(a-b)\), use \(a^2-b^2\).

Worked Example 1: Square of a Sum

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

Use the pattern \((a+b)^2=a^2+2ab+b^2\).

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

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

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

Worked Example 2: Square of a Difference

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

Use the pattern \((a-b)^2=a^2-2ab+b^2\).

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

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

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

Worked Example 3: Difference of Squares

Expand \((y+4)(y-4)\).

Use the pattern \((a+b)(a-b)=a^2-b^2\).

Here, \(a=y\) and \(b=4\).

$$ (y+4)(y-4)=y^2-4^2 $$$$ =y^2-16 $$

Answer: \(y^2-16\)

Worked Example 4: A More Challenging Example

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

Use the pattern \((a+b)^2=a^2+2ab+b^2\).

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

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

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

Common Mistakes to Avoid

1. Forgetting the middle term
   For example, \((x+3)^2\) is not \(x^2+9\).
   It is $$x^2+6x+9$$
2. Using the wrong sign in \((a-b)^2\)
   \((a-b)^2\) is not \(a^2-b^2\).
   It is $$a^2-2ab+b^2$$
3. Squaring only one term
   Every part of the binomial must be included in the pattern.
4. Mixing up square of a difference and difference of squares
   These are different:
   $$ (a-b)^2=a^2-2ab+b^2 $$$$ (a+b)(a-b)=a^2-b^2 $$

Quick Check

Try these on your own using the patterns:

• \((x+7)^2\)
• \((p-6)^2\)
• \((n+9)(n-9)\)

The answers are:

• \(x^2+14x+49\)
• \(p^2-12p+36\)
• \(n^2-81\)

Summary

Special binomial products are useful algebra patterns that help you expand expressions quickly. The three main patterns are:

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

If you learn to recognize these forms, you can work faster and more accurately with polynomials.

Factoring Trinomials When \(a=1\)

In this lesson, you will learn how to factor trinomials like \(x^2+5x+6\). A trinomial is a polynomial with three terms. Factoring means rewriting the trinomial as a product of two binomials.

We will focus on trinomials of the form

$$x^2+bx+c$$

because the coefficient of \(x^2\) is \(1\). This is what “\(a=1\)” means.

The main idea is to find two numbers that:

• multiply to make the constant term \(c\), and
• add to make the middle coefficient \(b\).

If we find those two numbers, we can write

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

where \(m\) and \(n\) are the two numbers.

Why does this work? If we multiply the binomials, we get

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

So the middle coefficient comes from the sum \(m+n\), and the last term comes from the product \(mn\).

Step-by-step method

1. Write the trinomial in the form \(x^2+bx+c\).
2. Find two integers whose product is \(c\).
3. Check whether those two integers add to \(b\).
4. Write the factors as \((x+m)(x+n)\).
5. Check by multiplying the binomials back out.

Worked Example 1: Positive constant and positive middle term

Factor:

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

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

Let’s list factor pairs of \(12\):

• \(1\) and \(12\) → sum is \(13\)
• \(2\) and \(6\) → sum is \(8\)
• \(3\) and \(4\) → sum is \(7\)

The correct pair is \(3\) and \(4\).

So,

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

Check:

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

Worked Example 2: Positive constant and negative middle term

Factor:

$$x^2-9x+20$$

We need two numbers that multiply to \(20\) and add to \(-9\).

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

Factor pairs of \(20\):

• \(-1\) and \(-20\) → sum is \(-21\)
• \(-2\) and \(-10\) → sum is \(-12\)
• \(-4\) and \(-5\) → sum is \(-9\)

The correct pair is \(-4\) and \(-5\).

So,

$$x^2-9x+20=(x-4)(x-5)$$

Check:

$$ (x-4)(x-5)=x^2-5x-4x+20=x^2-9x+20 $$

Worked Example 3: Negative constant

Factor:

$$x^2+x-12$$

We need two numbers that multiply to \(-12\) and add to \(1\).

Since the product is negative, one number must be positive and the other must be negative.

Factor pairs of \(12\):

• \(1\) and \(12\)
• \(2\) and \(6\)
• \(3\) and \(4\)

Now test signs:

• \(12+(-1)=11\)
• \(6+(-2)=4\)
• \(4+(-3)=1\)

The correct pair is \(4\) and \(-3\).

So,

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

Check:

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

Worked Example 4: Larger numbers

Factor:

$$x^2+13x+40$$

We need two numbers that multiply to \(40\) and add to \(13\).

Factor pairs of \(40\):

• \(1\) and \(40\) → sum is \(41\)
• \(2\) and \(20\) → sum is \(22\)
• \(4\) and \(10\) → sum is \(14\)
• \(5\) and \(8\) → sum is \(13\)

The correct pair is \(5\) and \(8\).

So,

$$x^2+13x+40=(x+5)(x+8)$$

Check:

$$ (x+5)(x+8)=x^2+8x+5x+40=x^2+13x+40 $$

How to think about signs

• If \(c\) is positive and \(b\) is positive, both numbers are positive.
• If \(c\) is positive and \(b\) is negative, both numbers are negative.
• If \(c\) is negative, the numbers have different signs.

Common mistakes to avoid

• Using numbers that multiply correctly but do not add correctly. You must check both.
• Forgetting negative signs. The signs matter in both the product and the sum.
• Not checking your answer. Multiply the binomials to make sure you get the original trinomial.

Quick practice ideas

Try using the same method on these:

• \(x^2+8x+15\)
• \(x^2-11x+24\)
• \(x^2-2x-15\)

Brief Summary

To factor a trinomial of the form \(x^2+bx+c\), find two numbers that multiply to \(c\) and add to \(b\). Then write the trinomial as \((x+m)(x+n)\). Always check your answer by multiplying the factors back together.

Factoring Trinomials When the Leading Coefficient Is Greater Than 1

In this lesson, you will learn how to factor trinomials of the form \(ax^2+bx+c\) when \(a>1\).

A trinomial has three terms. For example, \(6x^2+11x+3\) is a trinomial. Factoring means rewriting it as a product of two binomials.

When the coefficient of \(x^2\) is 1, factoring is usually quicker. But when the coefficient is bigger than 1, we need a more careful method. One of the most useful methods is called the AC method.

The goal is to change a trinomial like

$$ax^2+bx+c$$

into something like

$$mx+n)(px+q)$$

so that multiplying the binomials gives the original trinomial.

Why factoring matters

• It helps solve quadratic equations.
• It helps simplify algebra expressions.
• It helps you see structure in a polynomial.

The AC Method

To factor \(ax^2+bx+c\), follow these steps:

1. Multiply \(a\) and \(c\).
2. Find two integers whose product is \(ac\) and whose sum is \(b\).
3. Rewrite the middle term \(bx\) using those two numbers.
4. Factor by grouping.
5. Check your answer by multiplying.

This method works because splitting the middle term lets us make matching groups.

Important sign patterns

• If \(c\) is positive and \(b\) is positive, both numbers are positive.
• If \(c\) is positive and \(b\) is negative, both numbers are negative.
• If \(c\) is negative, the numbers have different signs.

Keeping track of signs helps you choose the correct pair faster.

Worked Example 1

Factor:

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

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

$$6\cdot 3=18$$

Step 2: Find two numbers that multiply to 18 and add to 11.

Those numbers are 9 and 2, because

$$9\cdot 2=18 \qquad 9+2=11$$

Step 3: Rewrite the middle term.

$$6x^2+11x+3=6x^2+9x+2x+3$$

Step 4: Factor by grouping.

Group the first two terms and the last two terms:

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

Factor out the greatest common factor from each group:

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

Now both groups contain \((2x+3)\):

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

Check:

$$(3x+1)(2x+3)=6x^2+9x+2x+3=6x^2+11x+3$$

So the factorization is correct.

Worked Example 2

Factor:

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

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

$$8\cdot (-3)=-24$$

Step 2: Find two numbers that multiply to \(-24\) and add to \(-2\).

Those numbers are \(-6\) and 4, because

$$(-6)(4)=-24 \qquad -6+4=-2$$

Step 3: Rewrite the middle term.

$$8x^2-2x-3=8x^2-6x+4x-3$$

Step 4: Factor by grouping.

$$=(8x^2-6x)+(4x-3)$$

Factor each group:

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

Factor out the common binomial:

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

Check:

$$(2x+1)(4x-3)=8x^2-6x+4x-3=8x^2-2x-3$$

Worked Example 3

Factor:

$$12x^2+7x-10$$

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

$$12\cdot (-10)=-120$$

Step 2: Find two numbers that multiply to \(-120\) and add to 7.

We want opposite signs because the product is negative. The pair is 15 and \(-8\), because

$$15(-8)=-120 \qquad 15+(-8)=7$$

Step 3: Rewrite the middle term.

$$12x^2+7x-10=12x^2+15x-8x-10$$

Step 4: Factor by grouping.

$$=(12x^2+15x)+(-8x-10)$$

Factor each group:

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

Now factor out \((4x+5)\):

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

Check:

$$(3x-2)(4x+5)=12x^2+15x-8x-10=12x^2+7x-10$$

What if there is a greatest common factor first?

Always check for a greatest common factor (GCF) before using the AC method.

For example, factor:

$$4x^2+12x+8$$

All three terms have a GCF of 4, so factor that out first:

$$4x^2+12x+8=4(x^2+3x+2)$$

Now factor the trinomial inside:

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

So the full factorization is

$$4(x+1)(x+2)$$

Worked Example 4

Factor:

$$15x^2-14x-8$$

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

$$15\cdot (-8)=-120$$

Step 2: Find two numbers that multiply to \(-120\) and add to \(-14\).

Those numbers are \(-20\) and 6, because

$$(-20)(6)=-120 \qquad -20+6=-14$$

Step 3: Rewrite the middle term.

$$15x^2-14x-8=15x^2-20x+6x-8$$

Step 4: Factor by grouping.

$$=(15x^2-20x)+(6x-8)$$

Factor each group:

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

Now factor the common binomial:

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

Common mistakes to avoid

• Forgetting the GCF first: Always look for a common factor before anything else.
• Using the wrong sign pair: Check whether the product \(ac\) is positive or negative.
• Stopping too early: After splitting the middle term, you must still factor by grouping.
• Not checking: Multiply your factors back together to make sure you get the original trinomial.

Quick strategy checklist

1. Check for a GCF.
2. Identify \(a\), \(b\), and \(c\).
3. Multiply \(a\cdot c\).
4. Find two integers with product \(ac\) and sum \(b\).
5. Split the middle term.
6. Factor by grouping.
7. Check by multiplying.

Brief Summary

To factor a trinomial with \(a>1\), the most reliable method is the AC method. Multiply the first and last coefficients, find two numbers that multiply to that product and add to the middle coefficient, split the middle term, and factor by grouping. Always check for a GCF first, and always multiply your answer to make sure it works.

Factoring Special Cases means recognizing patterns in polynomials that let you factor quickly and accurately. Instead of guessing factors, you look for a familiar structure. In this lesson, you will learn three important special cases: difference of squares, perfect square trinomials, and sum or difference of cubes.

Factoring is the reverse of multiplying. For example, if you know that $$ (x+3)(x+3)=x^2+6x+9, $$ then factoring means taking the polynomial \(x^2+6x+9\) and writing it as \((x+3)^2\).

These special patterns appear often in algebra, so learning to spot them will save time and help you avoid mistakes.

Before factoring any polynomial, always check for a greatest common factor (GCF) first. If all terms share a common number or variable, factor that out before looking for a special case.

1. Difference of Squares

A difference of squares has two terms, both of which are perfect squares, separated by subtraction.

The pattern is:

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

To use this pattern, ask yourself:

• Are there exactly two terms?
• Is there a subtraction sign?
• Is each term a perfect square?

Some common perfect squares are:

• \(1=x^0\), \(4=2^2\), \(9=3^2\), \(16=4^2\), \(25=5^2\)
• \(x^2=(x)^2\), \(y^2=(y)^2\), \(25x^2=(5x)^2\)

For example, in \(x^2-16\), both \(x^2\) and \(16\) are perfect squares. So:

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

Important: A sum of squares does not factor the same way. For example, \(x^2+16\) is not \((x+4)(x+4)\).

2. Perfect Square Trinomials

A trinomial has three terms. A perfect square trinomial comes from squaring a binomial.

There are two main patterns:

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

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

To recognize a perfect square trinomial, check:

• Is the first term a perfect square?
• Is the last term a perfect square?
• Is the middle term equal to \(2ab\), where \(a\) and \(b\) are the square roots of the first and last terms?

For example, consider \(x^2+10x+25\).

• \(x^2=(x)^2\)
• \(25=(5)^2\)
• The middle term is \(10x=2(x)(5)\)

So this is a perfect square trinomial:

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

If the middle term is negative, the binomial will have a minus sign:

$$ x^2-8x+16=(x-4)^2 $$

3. Sum and Difference of Cubes

A cube is a number or variable raised to the third power. For example, \(x^3\), \(8=2^3\), and \(27y^3=(3y)^3\).

The patterns are:

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

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

Notice something important:

• For a sum of cubes, the second factor has a minus in the middle.
• For a difference of cubes, the second factor has a plus in the middle.

A helpful memory trick is: same, opposite, always positive.

• The first sign in the factor matches the original sign.
• The second sign is the opposite.
• The last term is always positive.

Example: \(x^3-27\)

• \(x^3=(x)^3\)
• \(27=(3)^3\)

So:

$$ x^3-27=(x-3)(x^2+3x+9) $$

And for a sum:

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

How to Decide Which Special Case to Use

When you see a polynomial, go through these steps:

1. Factor out the GCF, if there is one.
2. Count the terms: are there 2 terms or 3 terms?
3. If there are 2 terms, check for difference of squares or sum/difference of cubes.
4. If there are 3 terms, check for a perfect square trinomial.
5. After factoring, check whether any factor can be factored more.

Worked Example 1: Difference of Squares

Factor \(9x^2-49\).

Step 1: Check whether each term is a perfect square.

• \(9x^2=(3x)^2\)
• \(49=(7)^2\)

Step 2: Notice that this is subtraction, so use the difference of squares formula.

$$ 9x^2-49=(3x-7)(3x+7) $$

Answer: \((3x-7)(3x+7)\)

Worked Example 2: Perfect Square Trinomial

Factor \(x^2-12x+36\).

Step 1: Check the first and last terms.

• \(x^2=(x)^2\)
• \(36=(6)^2\)

Step 2: Check the middle term.

$$ 2(x)(6)=12x $$

Since the middle term is \(-12x\), the binomial must have a minus sign.

$$ x^2-12x+36=(x-6)^2 $$

Answer: \((x-6)^2\)

Worked Example 3: Difference of Cubes

Factor \(8x^3-125\).

Step 1: Identify each cube.

• \(8x^3=(2x)^3\)
• \(125=(5)^3\)

Step 2: Use the difference of cubes formula.

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

Here, \(a=2x\) and \(b=5\).

$$ 8x^3-125=(2x-5)(4x^2+10x+25) $$

Answer: \((2x-5)(4x^2+10x+25)\)

Worked Example 4: GCF First, Then Special Case

Factor \(2x^3-18x\).

Step 1: Factor out the GCF.

Both terms have a common factor of \(2x\).

$$ 2x^3-18x=2x(x^2-9) $$

Step 2: Look at what remains.

\(x^2-9\) is a difference of squares because \(x^2=(x)^2\) and \(9=(3)^2\).

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

Step 3: Write the full factorization.

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

Answer: \(2x(x-3)(x+3)\)

Common Mistakes to Avoid

• Forgetting the GCF first. Always check before using a special pattern.
• Factoring a sum of squares like a difference of squares. \(x^2+9\) does not factor using special cases in this lesson.
• Mixing up cube formulas. Remember: same, opposite, always positive.
• Missing the square root or cube root. For example, the square root of \(25x^2\) is \(5x\), not just \(5\).
• Stopping too early. After factoring once, check whether any factor can be factored again.

Quick Pattern Review

• Difference of squares: $$ a^2-b^2=(a-b)(a+b) $$
• Perfect square trinomial: $$ a^2+2ab+b^2=(a+b)^2 $$
• Perfect square trinomial: $$ a^2-2ab+b^2=(a-b)^2 $$
• Sum of cubes: $$ a^3+b^3=(a+b)(a^2-ab+b^2) $$
• Difference of cubes: $$ a^3-b^3=(a-b)(a^2+ab+b^2) $$

Summary

Factoring special cases is about spotting patterns. If you see two perfect squares with subtraction, use difference of squares. If you see a trinomial with perfect square first and last terms and the correct middle term, use a perfect square trinomial. If you see two perfect cubes, use the sum or difference of cubes formulas. With practice, these patterns become much easier to recognize.

Zero Product Property is one of the most useful ideas in algebra for solving equations that involve multiplication.

It says this:

If two or more factors are multiplied together and the product is 0, then at least one of the factors must be 0.

In symbols, if

$$ab=0$$

then

$$a=0 \quad \text{or} \quad b=0$$

This idea works because the only way multiplication can give 0 is if one of the numbers being multiplied is 0.

For example,

$$5\cdot 0=0 \qquad \text{and} \qquad 0\cdot (-3)=0$$

But if neither factor is 0, the product will not be 0.

This property becomes very important when solving quadratic equations and other polynomial equations that can be written in factored form.

For instance, if you have

$$ (x-4)(x+2)=0 $$

then the Zero Product Property tells us that either

$$x-4=0 \quad \text{or} \quad x+2=0$$

Then we solve each simple equation:

$$x=4 \quad \text{or} \quad x=-2$$

So the solutions are 4 and -2.

Important: The Zero Product Property only works when one side of the equation is 0.

For example, if you have

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

you cannot say that

$$x-1=5 \quad \text{or} \quad x+3=5$$

That is not how the property works. You must first get the equation into a form where the product equals 0.

Here is the basic process for using the Zero Product Property:

1. Write the equation so one side is 0.

2. Factor the polynomial completely.

3. Set each factor equal to 0.

4. Solve each resulting equation.

5. Check your solutions if needed.

Let’s go through some examples.

Example 1: Solve a simple factored equation

Solve:

$$x(x-7)=0$$

Since the product is 0, set each factor equal to 0:

$$x=0 \quad \text{or} \quad x-7=0$$

Solve each one:

$$x=0 \quad \text{or} \quad x=7$$

Answer: The solutions are

$$x=0 \text{ and } x=7$$

Example 2: Solve a quadratic by factoring first

Solve:

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

This equation is not yet written as a product, so we factor it:

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

Now the equation becomes

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

Apply the Zero Product Property:

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

Solve:

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

Answer: The solutions are

$$x=-2 \text{ and } x=-3$$

Example 3: Move everything to one side first

Solve:

$$x^2=9x$$

The equation is not equal to 0 yet, so subtract 9x from both sides:

$$x^2-9x=0$$

Now factor:

$$x(x-9)=0$$

Apply the Zero Product Property:

$$x=0 \quad \text{or} \quad x-9=0$$

Solve:

$$x=0 \quad \text{or} \quad x=9$$

Answer: The solutions are

$$x=0 \text{ and } x=9$$

Example 4: More than two factors

The Zero Product Property also works with three or more factors.

Solve:

$$ (x-1)(x+4)(x-6)=0 $$

If the product of three factors is 0, then at least one factor must be 0. So set each factor equal to 0:

$$x-1=0 \quad \text{or} \quad x+4=0 \quad \text{or} \quad x-6=0$$

Solve each equation:

$$x=1, \quad x=-4, \quad x=6$$

Answer: The solutions are

$$x=1, -4, 6$$

Why factoring matters

The Zero Product Property is powerful because many polynomial equations are hard to solve in expanded form, but become much easier once they are factored.

For example, solving

$$x^2-4x-12=0$$

is easier if we factor:

$$x^2-4x-12=(x-6)(x+2)$$

Then:

$$ (x-6)(x+2)=0 $$

So:

$$x-6=0 \quad \text{or} \quad x+2=0$$

which gives

$$x=6 \quad \text{or} \quad x=-2$$

Common mistakes to avoid

• Using the property before the equation equals 0. You must have an equation like $$\text{something}=0$$ before setting factors equal to 0.

• Forgetting to factor completely. If the expression is not fully factored, you might miss solutions.

• Solving only one factor. Every factor must be set equal to 0.

• Sign errors. Be careful when solving equations like $$x+5=0$$, which gives $$x=-5$$.

Quick check

Try these on your own:

• $$ (x+1)(x-8)=0 $$

• $$ x^2-11x+24=0 $$

• $$ x^2+2x=0 $$

The answers are:

• $$x=-1, 8$$

• $$x=3, 8$$

• $$x=0, -2$$

Summary

The Zero Product Property says that if a product equals 0, then at least one factor must equal 0.

To use it, first make sure the equation equals 0, then factor, set each factor equal to 0, and solve.

This is one of the main tools for solving quadratic and polynomial equations in factored form.

Polynomial Long Division is a method for dividing one polynomial by another, much like regular long division with numbers.

You use it when a polynomial does not divide in one quick step, or when you want to rewrite a fraction involving polynomials in a simpler form.

For example, if you want to divide \(x^2+5x+6\) by \(x+2\), polynomial long division helps you find the result step by step.

Goal: Write the division in the form

$$\text{dividend} = (\text{divisor})(\text{quotient}) + \text{remainder}$$

So if you divide \(P(x)\) by \(D(x)\), you get

$$P(x)=D(x)\cdot Q(x)+R(x)$$

The remainder must have lower degree than the divisor. In simpler words, the remainder must be “smaller” than the divisor.

Important idea: Polynomial long division works a lot like number long division:

• Divide the first term.
• Put the answer on top.
• Multiply back.
• Subtract.
• Bring down the next term.
• Repeat.

Before starting, make sure the polynomials are written in descending powers. That means the terms go from highest exponent to lowest exponent.

If a power is missing, use a placeholder with coefficient 0. For example,

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

This helps keep all the terms lined up correctly.

Steps for Polynomial Long Division

1. Write the dividend inside the division symbol and the divisor outside.
2. Divide the first term of the dividend by the first term of the divisor.
3. Write that result above the division bar.
4. Multiply the entire divisor by that term.
5. Subtract the result from the current polynomial.
6. Bring down the next term.
7. Repeat until the remainder has lower degree than the divisor.

Let’s practice with worked examples.

Example 1: Divide \(x^2+5x+6\) by \(x+2\)

We want to find

$$\frac{x^2+5x+6}{x+2}$$

Step 1: Divide the first terms.

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

Put \(x\) on top.

Step 2: Multiply back.

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

Write that under \(x^2+5x\), then subtract:

$$ \begin{aligned} (x^2+5x+6)-(x^2+2x) &= 3x+6 \end{aligned} $$

Step 3: Divide again.

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

Put \(+3\) on top.

Step 4: Multiply back.

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

Subtract:

$$(3x+6)-(3x+6)=0$$

So the quotient is

$$x+3$$

and the remainder is \(0\).

Final answer:

$$\frac{x^2+5x+6}{x+2}=x+3$$

This means

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

Example 2: Divide \(2x^2+7x+3\) by \(x+3\)

We start with

$$\frac{2x^2+7x+3}{x+3}$$

Step 1: Divide the first terms.

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

Put \(2x\) on top.

Step 2: Multiply back.

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

Subtract:

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

Step 3: Divide again.

$$\frac{x}{x}=1$$

Put \(+1\) on top.

Step 4: Multiply back.

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

Subtract:

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

Final answer:

$$\frac{2x^2+7x+3}{x+3}=2x+1$$

Example 3: Divide \(x^3+4x^2+x-6\) by \(x+2\)

This example has a cubic polynomial, so there will be more steps.

Step 1: Divide the first terms.

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

Put \(x^2\) on top.

Step 2: Multiply back.

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

Subtract:

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

Step 3: Divide again.

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

Put \(+2x\) on top.

Step 4: Multiply back.

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

Subtract:

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

Step 5: Divide again.

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

Put \(-3\) on top.

Step 6: Multiply back.

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

Subtract:

$$(-3x-6)-(-3x-6)=0$$

Final answer:

$$\frac{x^3+4x^2+x-6}{x+2}=x^2+2x-3$$

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

This example is important because there is a missing term. Notice that \(x^3+2x-5\) has no \(x^2\) term.

So we rewrite it as

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

Now divide:

$$\frac{x^3+0x^2+2x-5}{x-1}$$

Step 1: Divide the first terms.

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

Put \(x^2\) on top.

Step 2: Multiply back.

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

Subtract:

$$ (x^3+0x^2+2x-5)-(x^3-x^2)=x^2+2x-5 $$

Step 3: Divide again.

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

Put \(+x\) on top.

Step 4: Multiply back.

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

Subtract:

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

Step 5: Divide again.

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

Put \(+3\) on top.

Step 6: Multiply back.

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

Subtract:

$$ (3x-5)-(3x-3)=-2 $$

Now the remainder is \(-2\). Since this remainder has degree 0, which is lower than the degree of \(x-1\), we stop.

Final answer:

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

or, using quotient and remainder form,

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

How to Check Your Answer

After dividing, multiply the divisor by the quotient and then add the remainder. You should get the original dividend.

For Example 4:

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

Multiply:

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

Now subtract 2:

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

So the answer is correct.

Common Mistakes to Avoid

• Forgetting missing terms. Always include terms like \(0x^2\) if needed.
• Only dividing part of the divisor. When you multiply back, multiply the top term by the entire divisor.
• Sign errors when subtracting. Be very careful with negative signs.
• Stopping too early. Keep going until the remainder has lower degree than the divisor.
• Terms not lined up correctly. Match powers of \(x\) in columns.

Helpful Tips

• Always arrange polynomials in descending powers.
• Work one term at a time.
• After each subtraction, simplify before moving on.
• Check your answer by multiplying back.

Summary

Polynomial long division is a step-by-step way to divide polynomials, just like long division with numbers. You divide the leading terms, write the result on top, multiply back, subtract, and repeat.

If there is a remainder, write the answer as quotient with a remainder, or as a fraction over the divisor. With practice, the process becomes organized and predictable.