Advanced Algebraic Structures and Equations

Polynomial Operations and Closure is an important idea in algebra because it helps us understand what happens when we combine polynomials in different ways. In Grade 12, you are expected to add, subtract, and multiply polynomials accurately, and also recognize that these operations keep us inside the set of polynomials.

This idea is called the closure property. Informally, a set is closed under an operation if performing that operation on members of the set always gives another member of the same set.

For polynomials, this means that if you add, subtract, or multiply two polynomials, the result is still a polynomial. This is one reason polynomials are so useful in algebra.

First, let us recall what a polynomial is. A polynomial is an expression made of variables, coefficients, and whole-number exponents. A polynomial in one variable usually looks like

$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0$$

where the exponents are non-negative integers: \(0,1,2,3,\dots\).

Examples of polynomials include:

• \(3x^2-5x+1\)
• \(-7x^4+2x-9\)
• \(6\)
• \(\frac{1}{2}x^3+4x\)

Examples that are not polynomials include:

• \(x^{-1}+2\) because of the negative exponent
• \(\sqrt{x}+1\) because \(\sqrt{x}=x^{1/2}\), and the exponent is not a whole number
• \(\frac{1}{x}+3\) because this is the same as \(x^{-1}+3\)

Categorizing polynomials helps you describe them clearly. We usually classify polynomials in two common ways: by number of terms and by degree.

By number of terms:

• Monomial: one term, such as \(5x^3\)
• Binomial: two terms, such as \(x^2-4\)
• Trinomial: three terms, such as \(2x^2+3x-1\)
• Polynomial: any expression with one or more polynomial terms

By degree:

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

The degree of a polynomial is the highest exponent of the variable after the polynomial has been simplified.

Now let us study the main operations.

1. Adding polynomials

To add polynomials, combine like terms. Like terms have the same variable raised to the same power.

For example, \(3x^2\) and \(-5x^2\) are like terms, but \(3x^2\) and \(3x\) are not.

When adding polynomials, it helps to:

1. Remove brackets if needed.
2. Group like terms together.
3. Add the coefficients of like terms.

2. Subtracting polynomials

Subtracting polynomials is very similar to adding, but you must first distribute the negative sign across the polynomial being subtracted.

For example,

$$\left(4x^2+3x-1\right)-\left(x^2-5x+2\right)$$

becomes

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

Then combine like terms.

3. Multiplying polynomials

To multiply polynomials, use the distributive property. Every term in one polynomial must be multiplied by every term in the other polynomial.

For a monomial times a polynomial, distribute the monomial to each term.

For a binomial times a binomial, you may use expansion methods such as distribution or FOIL, but remember FOIL is just a special case of distribution.

When multiplying powers of the same variable, add the exponents:

$$x^a\cdot x^b=x^{a+b}$$

Why closure matters

The set of polynomials is closed under addition, subtraction, and multiplication. This means:

• If \(p(x)\) and \(q(x)\) are polynomials, then \(p(x)+q(x)\) is a polynomial.
• If \(p(x)\) and \(q(x)\) are polynomials, then \(p(x)-q(x)\) is a polynomial.
• If \(p(x)\) and \(q(x)\) are polynomials, then \(p(x)q(x)\) is a polynomial.

This works because:

• Adding or subtracting coefficients still gives valid coefficients.
• Multiplying terms keeps exponents as whole numbers, since whole numbers added together are still whole numbers.

However, polynomials are not always closed under division. For example,

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

and \(\frac{1}{x}\) is not a polynomial term. So division can take us outside the set of polynomials.

Worked Example 1: Adding polynomials

Simplify:

$$\left(2x^3-5x^2+4x-7\right)+\left(3x^3+x^2-6x+9\right)$$

Step 1: Group like terms.

$$2x^3+3x^3-5x^2+x^2+4x-6x-7+9$$

Step 2: Add coefficients.

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

Answer: \(5x^3-4x^2-2x+2\)

This result is still a polynomial, so this shows closure under addition.

Worked Example 2: Subtracting polynomials

Simplify:

$$\left(6x^4-2x^2+x-8\right)-\left(3x^4+5x^2-4x+1\right)$$

Step 1: Distribute the minus sign.

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

Step 2: Combine like terms.

$$\left(6x^4-3x^4\right)+\left(-2x^2-5x^2\right)+\left(x+4x\right)+\left(-8-1\right)$$

$$3x^4-7x^2+5x-9$$

Answer: \(3x^4-7x^2+5x-9\)

This is also a polynomial, so subtraction preserves closure.

Worked Example 3: Multiplying a monomial by a polynomial

Simplify and categorize the result:

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

Step 1: Distribute \(-2x^2\) to each term.

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

Step 2: Multiply coefficients and add exponents where needed.

$$-6x^5+2x^3-10x^2$$

Step 3: Categorize the result.

• It has three terms, so it is a trinomial.
• The highest power is \(5\), so it is a degree 5 polynomial.

Answer: \(-6x^5+2x^3-10x^2\)

Again, the product is a polynomial, showing closure under multiplication.

Worked Example 4: Multiplying two binomials

Simplify and classify:

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

Step 1: Distribute each term in the first polynomial across the second.

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

Step 2: Multiply each term.

$$2x^3-5x^2+6x^2-15x$$

Step 3: Combine like terms.

$$2x^3+x^2-15x$$

Step 4: Classify the result.

• It has three terms, so it is a trinomial.
• The highest exponent is \(3\), so it is a cubic polynomial.

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

Important patterns to remember

• Like terms can only be combined if the variable part is exactly the same.
• When subtracting, be careful with negative signs.
• When multiplying, every term must be included.
• The degree of a sum or difference is at most the highest degree present.
• The degree of a product is found by adding degrees of factors, as long as the product is not zero.

Common mistakes

• Combining unlike terms: \(3x^2+2x\neq 5x^3\)
• Forgetting to distribute a negative sign: \(-(x^2-3x+1)=-x^2+3x-1\)
• Missing terms in multiplication: every term in one polynomial must multiply every term in the other
• Incorrect exponent rules: \(x^2\cdot x^3=x^5\), not \(x^6\)

Checking for closure in a quick way

If you start with two valid polynomials, ask:

• Are you adding them? Result stays polynomial.
• Are you subtracting them? Result stays polynomial.
• Are you multiplying them? Result stays polynomial.
• Are you dividing them? Be careful, because the result may not be a polynomial.

Brief Summary

Polynomials are algebraic expressions with variables raised only to whole-number exponents. You can add and subtract polynomials by combining like terms, and multiply them using the distributive property. The set of polynomials is closed under addition, subtraction, and multiplication, which means these operations always produce another polynomial. This closure property is a key reason polynomials form an important algebraic structure.

Advanced Factoring Strategies is about breaking complicated algebraic expressions into simpler factors. Factoring is useful because it helps us solve equations, simplify rational expressions, and understand the structure of polynomials.

In Grade 12, factoring goes beyond simple trinomials. You may need to recognize patterns, rearrange terms, use substitution, or combine several methods in one problem. The key skill is learning which strategy fits the expression.

In this lesson, we will focus on four important strategies:

• Factoring by grouping
• Using substitution
• Difference of squares
• Sum and difference of cubes

Before using any advanced method, always check for a greatest common factor (GCF) first. Factoring out a common factor can make the remaining expression much easier to work with.

For example, in \(6x^3 - 12x^2\), both terms share \(6x^2\). Factoring it out gives

$$6x^3 - 12x^2 = 6x^2(x - 2).$$

Once the GCF is removed, look for deeper patterns.

1. Factoring by Grouping

Factoring by grouping is useful when an expression has four terms. The idea is to group terms in pairs, factor each pair, and then look for a common binomial factor.

The general process is:

1. Group the terms into two pairs.
2. Factor out the GCF from each pair.
3. If the same binomial appears, factor it out.

For example, suppose we have

$$ax + ay + bx + by.$$

Group the first two terms and the last two terms:

$$a(x+y) + b(x+y).$$

Now factor out the common binomial \((x+y)\):

$$a(x+y) + b(x+y) = (a+b)(x+y).$$

This method depends on spotting the repeated factor after grouping.

2. Difference of Squares

A difference of squares has the form

$$A^2 - B^2.$$

It factors as

$$A^2 - B^2 = (A-B)(A+B).$$

This pattern only works for a difference, not a sum. So \(A^2 + B^2\) does not factor in the same way over the real numbers.

Examples of difference of squares include:

• \(x^2 - 25 = (x-5)(x+5)\)
• \(9y^2 - 16 = (3y-4)(3y+4)\)
• \(4a^4 - b^2 = (2a^2-b)(2a^2+b)\)

Notice that the squared parts may be more complicated than just a single variable.

3. Sum and Difference of Cubes

A sum of cubes has the form

$$A^3 + B^3 = (A+B)(A^2 - AB + B^2).$$

A difference of cubes has the form

$$A^3 - B^3 = (A-B)(A^2 + AB + B^2).$$

A useful memory pattern is:

• Same sign, opposite sign, always positive

This means:

• For \(A^3 + B^3\), the first bracket uses \(+\), the middle term in the second bracket uses \(-\), and the last term is positive.
• For \(A^3 - B^3\), the first bracket uses \(-\), the middle term in the second bracket uses \(+\), and the last term is positive.

Examples:

• \(x^3 + 8 = x^3 + 2^3 = (x+2)(x^2 - 2x + 4)\)
• \(27y^3 - 1 = (3y)^3 - 1^3 = (3y-1)(9y^2 + 3y + 1)\)

4. Using Substitution

Some high-degree polynomials look complicated, but they become easier if you treat part of the expression as a single variable.

For example, consider

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

This is not a regular quadratic in \(x\), but it is a quadratic in \(x^2\). Let

$$u = x^2.$$

Then the expression becomes

$$u^2 - 5u + 4.$$

Now factor as usual:

$$u^2 - 5u + 4 = (u-1)(u-4).$$

Substitute back \(u=x^2\):

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

Then continue factoring each difference of squares:

$$x^2-1 = (x-1)(x+1), \qquad x^2-4 = (x-2)(x+2).$$

So the complete factorization is

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

Substitution is especially useful when the powers follow a pattern such as \(x^4, x^2,\) constant or \(x^6, x^3,\) constant.

How to Choose a Factoring Strategy

When you see a polynomial, ask yourself these questions in order:

1. Is there a GCF I can factor out first?
2. Are there 4 terms that suggest grouping?
3. Is it a difference of squares?
4. Is it a sum or difference of cubes?
5. Does it look like a quadratic in disguise, so substitution would help?
6. After factoring once, can any factor be factored further?

This step-by-step thinking helps avoid missing part of the factorization.

Worked Example 1: Grouping

Factor:

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

Step 1: Group the terms.

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

Step 2: Factor each group.

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

Step 3: Factor out the common binomial.

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

So the factorization is

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

Notice that \(x^2+2\) does not factor further over the real numbers.

Worked Example 2: Difference of Squares with More Than One Step

Factor:

$$16x^4 - 81$$

Step 1: Recognize a difference of squares.

$$16x^4 - 81 = (4x^2)^2 - 9^2$$

So

$$16x^4 - 81 = (4x^2-9)(4x^2+9).$$

Step 2: Check whether any factor can be factored further.

The factor \(4x^2-9\) is also a difference of squares:

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

The factor \(4x^2+9\) is a sum of squares, so it does not factor over the real numbers.

Therefore, the complete factorization is

$$16x^4 - 81 = (2x-3)(2x+3)(4x^2+9).$$

Worked Example 3: Sum and Difference of Cubes

Factor:

$$8x^3 - 27$$

Step 1: Identify the cube terms.

$$8x^3 = (2x)^3, \,\, 27 = 3^3$$

So this is a difference of cubes:

$$A^3 - B^3 = (A-B)(A^2+AB+B^2).$$

Step 2: Apply the formula.

$$8x^3 - 27 = (2x-3)(4x^2 + 6x + 9).$$

Step 3: Check for further factoring.

The quadratic \(4x^2 + 6x + 9\) does not factor over the real numbers, so we are done.

Final answer:

$$8x^3 - 27 = (2x-3)(4x^2+6x+9).$$

Worked Example 4: Substitution and Then More Factoring

Factor:

$$x^6 - 9x^3 + 20$$

Step 1: Notice the power pattern.

The exponents are \(6, 3, 0\), so this is a quadratic in \(x^3\).

Let

$$u = x^3.$$

Then the expression becomes

$$u^2 - 9u + 20.$$

Step 2: Factor the quadratic.

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

Step 3: Substitute back.

$$x^6 - 9x^3 + 20 = (x^3-5)(x^3-4).$$

Step 4: Check whether the factors can be factored further.

Neither \(x^3-5\) nor \(x^3-4\) is a difference of cubes, because 5 and 4 are not perfect cubes. So the factorization is complete.

Final answer:

$$x^6 - 9x^3 + 20 = (x^3-5)(x^3-4).$$

Common Mistakes to Avoid

• Forgetting the GCF first. Always check for a common factor before using advanced methods.
• Using the difference of squares on a sum. \(a^2+b^2\) does not factor as \((a+b)(a-b)\).
• Mixing up the cube formulas. In the second factor, the last term is always positive.
• Stopping too early. After factoring once, always check whether any factor can be factored again.
• Incorrect substitution. If you let \(u=x^2\) or \(u=x^3\), make sure every term matches that pattern correctly.

Quick Pattern Review

• Difference of squares: $$A^2-B^2=(A-B)(A+B)$$
• Sum of cubes: $$A^3+B^3=(A+B)(A^2-AB+B^2)$$
• Difference of cubes: $$A^3-B^3=(A-B)(A^2+AB+B^2)$$

Final Summary

Advanced factoring means recognizing structure. Some expressions factor by grouping, some by special patterns like difference of squares or cubes, and others become simpler through substitution.

The most successful approach is systematic: factor out any GCF, look for patterns, and keep factoring until no factor can be broken down further. With practice, you will become much faster at identifying the right method.

Rational Expressions and Domain Restrictions

A rational expression is a fraction where the numerator, denominator, or both are polynomials. Examples include \(\frac{x+3}{x-2}\), \(\frac{2x^2-8}{x+1}\), and \(\frac{3}{x^2-9}\).

In 12th Grade algebra, it is very important not only to simplify, multiply, and divide rational expressions, but also to state the domain restrictions. These are the values of the variable that are not allowed because they make a denominator equal to zero.

Why does this matter? Division by zero is undefined. So even if an expression looks simpler after factoring and canceling, the original restrictions still remain. A value that made the original denominator zero is excluded from the domain forever.

This lesson will show you how to:

• identify rational expressions,
• find excluded values from denominators,
• simplify rational expressions correctly,
• multiply and divide rational expressions, and
• keep track of domain restrictions throughout the process.

1. What is a domain restriction?

The domain of an expression is the set of all input values that make the expression defined. For rational expressions, the only values we usually need to exclude are those that make a denominator equal to zero.

For example, consider

$$\frac{5}{x-4}$$

The denominator is \(x-4\). If \(x=4\), then the denominator becomes \(0\), and the expression is undefined. So the domain restriction is

$$x \ne 4$$

This is often read as “\(x\) cannot equal 4.”

2. How to find excluded values

To find domain restrictions:

1. Look at every denominator in the expression.
2. Set each denominator equal to zero.
3. Solve for the variable.
4. Exclude those values from the domain.

Example:

$$\frac{2x+1}{x^2-5x+6}$$

Factor the denominator:

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

Now set each factor equal to zero:

$$x-2=0 \Rightarrow x=2$$ $$x-3=0 \Rightarrow x=3$$

So the domain restrictions are

$$x \ne 2, \quad x \ne 3$$

3. Simplifying rational expressions

To simplify a rational expression, factor the numerator and denominator completely. Then cancel only common factors, not terms.

This distinction is very important. You may cancel \((x-3)\) with \((x-3)\), but you may not cancel the \(x\) terms inside \(x+3\) and \(x+5\), because those are terms, not factors.

For example, this is correct:

$$\frac{(x-3)(x+2)}{(x-3)(x+5)}=\frac{x+2}{x+5}$$

But this is not correct:

$$\frac{x+2}{x+5} \ne \frac{2}{5}$$

You cannot cancel parts of terms that are connected by addition or subtraction.

Very important: even after canceling, you must keep the restrictions from the original denominator.

Worked Example 1: Simplifying and stating restrictions

Simplify:

$$\frac{x^2-9}{x^2-3x}$$

Step 1: Factor numerator and denominator.

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

So the expression becomes

$$\frac{(x-3)(x+3)}{x(x-3)}$$

Step 2: Find domain restrictions from the original denominator.

The original denominator is \(x(x-3)\). Set each factor equal to zero:

$$x=0 \quad \text{or} \quad x-3=0 \Rightarrow x=3$$

So the excluded values are

$$x \ne 0, \quad x \ne 3$$

Step 3: Cancel common factors.

$$\frac{(x-3)(x+3)}{x(x-3)}=\frac{x+3}{x}$$

Final answer:

$$\frac{x+3}{x}, \quad x \ne 0, 3$$

Notice that even though \((x-3)\) canceled, \(x=3\) is still excluded because it made the original denominator zero.

4. Multiplying rational expressions

To multiply rational expressions:

1. Factor all numerators and denominators completely.
2. State domain restrictions from every original denominator.
3. Cancel common factors.
4. Multiply what remains.

Worked carefully, this method makes problems much easier.

Worked Example 2: Multiplying rational expressions

Simplify:

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

Step 1: Factor everything.

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

Substitute these into the expression:

$$\frac{(x-2)(x+2)}{(x-3)(x+2)} \cdot \frac{x-3}{x+2}$$

Step 2: Find domain restrictions from the original denominators.

The original denominators are \(x^2-x-6\) and \(x+2\).

From \(x^2-x-6=(x-3)(x+2)\), we get

$$x \ne 3, \quad x \ne -2$$

From \(x+2\), we also get

$$x \ne -2$$

So the restrictions are

$$x \ne 3, \quad x \ne -2$$

Step 3: Cancel common factors.

There is an \((x+2)\) factor in the numerator and denominator, and an \((x-3)\) factor in the numerator and denominator:

$$\frac{(x-2)(x+2)}{(x-3)(x+2)} \cdot \frac{x-3}{x+2} =\frac{x-2}{x+2}$$

Final answer:

$$\frac{x-2}{x+2}, \quad x \ne 3, -2$$

5. Dividing rational expressions

Dividing by a rational expression is the same as multiplying by its reciprocal.

In symbols,

$$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \cdot \frac{d}{c}$$

But be careful: when dividing rational expressions, domain restrictions come from:

• any original denominator, and
• the divisor itself cannot equal zero.

If the divisor is zero, division is undefined.

Worked Example 3: Dividing rational expressions

Simplify:

$$\frac{x^2-16}{x^2+3x-4} \div \frac{x-4}{x+4}$$

Step 1: Rewrite division as multiplication by the reciprocal.

$$\frac{x^2-16}{x^2+3x-4} \cdot \frac{x+4}{x-4}$$

Step 2: Factor everything.

$$x^2-16=(x-4)(x+4)$$ $$x^2+3x-4=(x+4)(x-1)$$

Now substitute:

$$\frac{(x-4)(x+4)}{(x+4)(x-1)} \cdot \frac{x+4}{x-4}$$

Step 3: Find domain restrictions.

From the original denominator \(x^2+3x-4=(x+4)(x-1)\):

$$x \ne -4, \quad x \ne 1$$

Also, the divisor is

$$\frac{x-4}{x+4}$$

This divisor must itself be defined, so \(x \ne -4\).

In addition, the divisor cannot equal zero. A rational expression is zero when its numerator is zero and denominator is not zero. So if

$$x-4=0 \Rightarrow x=4$$

then the divisor equals zero, which means division would be undefined. So we must also exclude \(x=4\).

Therefore the full restrictions are

$$x \ne -4, \quad x \ne 1, \quad x \ne 4$$

Step 4: Cancel common factors.

$$\frac{(x-4)(x+4)}{(x+4)(x-1)} \cdot \frac{x+4}{x-4} = \frac{x+4}{x-1}$$

Final answer:

$$\frac{x+4}{x-1}, \quad x \ne -4, 1, 4$$

6. Why canceled values still matter

Sometimes students think that if a factor cancels, then its value is no longer restricted. That is not true.

Look at this expression:

$$\frac{x-5}{x^2-10x+25}$$

Factor the denominator:

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

So

$$\frac{x-5}{(x-5)^2}=\frac{1}{x-5}$$

Even though one \((x-5)\) factor canceled, \(x=5\) is still excluded because the original denominator was zero there.

This is one of the most important ideas in this topic.

7. Common mistakes to avoid

• Forgetting to factor first. You can only cancel factors, so expressions must usually be factored before simplifying.
• Canceling terms instead of factors. For example, in \(\frac{x+2}{x+5}\), nothing cancels.
• Ignoring excluded values. Always state what values make any original denominator zero.
• For division, forgetting the divisor cannot be zero. This can create an extra excluded value.
• Using the simplified form to find restrictions. Restrictions must come from the original expression, not just the final simplified one.

Worked Example 4: A more complete mixed example

Simplify:

$$\frac{x^2+7x+10}{x^2-25} \div \frac{x+2}{x-5}$$

Step 1: Rewrite as multiplication by the reciprocal.

$$\frac{x^2+7x+10}{x^2-25} \cdot \frac{x-5}{x+2}$$

Step 2: Factor all polynomials.

$$x^2+7x+10=(x+5)(x+2)$$ $$x^2-25=(x-5)(x+5)$$

Now substitute:

$$\frac{(x+5)(x+2)}{(x-5)(x+5)} \cdot \frac{x-5}{x+2}$$

Step 3: Find domain restrictions.

From the original denominator \(x^2-25=(x-5)(x+5)\):

$$x \ne 5, \quad x \ne -5$$

The divisor is

$$\frac{x+2}{x-5}$$

It must be defined, so \(x \ne 5\).

Also, the divisor cannot be zero. That happens when its numerator is zero:

$$x+2=0 \Rightarrow x=-2$$

So the full restrictions are

$$x \ne 5, \quad x \ne -5, \quad x \ne -2$$

Step 4: Cancel common factors.

$$\frac{(x+5)(x+2)}{(x-5)(x+5)} \cdot \frac{x-5}{x+2}=1$$

Final answer:

$$1, \quad x \ne 5, -5, -2$$

This example is a great reminder: even if the final answer is just \(1\), the original expression is not defined for certain values. Those restrictions must still be stated.

8. A step-by-step strategy you can always use

When working with rational expressions, follow this checklist:

1. Factor every polynomial completely.
2. Find all values that make any original denominator zero.
3. If dividing, also make sure the divisor is not zero.
4. Rewrite division as multiplication by the reciprocal if needed.
5. Cancel only common factors.
6. Multiply or simplify what remains.
7. Write the final simplified expression together with all excluded values.

9. Final summary

Rational expressions are quotients of polynomials, and they are undefined whenever a denominator equals zero. That is why domain restrictions are essential.

To simplify, multiply, or divide rational expressions, first factor completely. Then find excluded values from the original denominators, cancel only common factors, and keep all restrictions in the final answer.

If you remember one key idea, make it this: canceling a factor does not remove the original restriction. Excluded values come from where the expression started, not from what it looks like after simplification.

Partial Fraction Decomposition is a method for rewriting a complicated rational expression as a sum of simpler fractions. A rational expression is a fraction where the numerator and denominator are polynomials, such as \(\frac{3x+5}{x^2-1}\).

This skill is important in algebra because it makes rational expressions easier to simplify, solve, and work with. It is also very useful later in calculus, especially when integrating rational functions.

In this lesson, you will learn:

• when partial fraction decomposition can be used,
• how to set up the correct form,
• how to solve for the unknown constants,
• and how to handle different types of factors in the denominator.

1. When can we use partial fractions?

Partial fraction decomposition is used on a proper rational expression. This means the degree of the numerator must be less than the degree of the denominator.

For example:

• \(\frac{2x+1}{(x-1)(x+3)}\) is proper because the numerator has degree 1 and the denominator has degree 2.
• \(\frac{x^2+1}{x+4}\) is not proper because the numerator has degree 2 and the denominator has degree 1.

If the expression is not proper, you must first use polynomial division to rewrite it as a polynomial plus a proper rational expression.

2. The main idea

If the denominator can be factored, then we can write the expression as a sum of simpler fractions based on those factors.

For example, if

$$\frac{5x+1}{(x-2)(x+3)}$$

is decomposed, we write

$$\frac{5x+1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}$$

where \(A\) and \(B\) are constants we need to find.

3. Forms to use for different denominators

The form of the decomposition depends on the factorization of the denominator.

Case A: Distinct linear factors

If the denominator has different linear factors, such as \((x-1)(x+4)\), then write

$$\frac{P(x)}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}$$

Case B: Repeated linear factors

If a linear factor repeats, you need a term for each power.

For example, if the denominator is \((x-2)^3\), then use

$$\frac{P(x)}{(x-2)^3}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}$$

Case C: Irreducible quadratic factors

Sometimes the denominator includes a quadratic that cannot be factored over the real numbers, such as \(x^2+1\).

In that case, the numerator over that factor must be linear.

So if the denominator contains \((x^2+1)\), use

$$\frac{P(x)}{x^2+1}=\frac{Ax+B}{x^2+1}$$

If the denominator is \((x-1)(x^2+1)\), then the decomposition has the form

$$\frac{P(x)}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$

4. General steps for partial fraction decomposition

1. Check that the fraction is proper. If not, use polynomial division first.
2. Factor the denominator completely.
3. Write the decomposition form based on the factors.
4. Multiply both sides by the common denominator to clear fractions.
5. Solve for the unknown constants by substituting convenient values or by comparing coefficients.
6. Write the final decomposed expression.

Worked Example 1: Distinct linear factors

Decompose

$$\frac{7}{(x-1)(x+2)}$$

Step 1: Set up the form.

$$\frac{7}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$$

Step 2: Clear the denominators.

Multiply both sides by \((x-1)(x+2)\):

$$7=A(x+2)+B(x-1)$$

Step 3: Solve for \(A\) and \(B\).

Let \(x=1\):

$$7=A(3)+B(0)$$ $$7=3A$$ $$A=\frac{7}{3}$$

Let \(x=-2\):

$$7=A(0)+B(-3)$$ $$7=-3B$$ $$B=-\frac{7}{3}$$

Final answer:

$$\frac{7}{(x-1)(x+2)}=\frac{7/3}{x-1}-\frac{7/3}{x+2}$$

You may also write it as

$$\frac{7}{(x-1)(x+2)}=\frac{7}{3(x-1)}-\frac{7}{3(x+2)}$$

Worked Example 2: Distinct linear factors with a variable numerator

Decompose

$$\frac{3x+5}{(x-1)(x+4)}$$

Step 1: Set up the form.

$$\frac{3x+5}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}$$

Step 2: Clear the denominators.

$$3x+5=A(x+4)+B(x-1)$$

Step 3: Expand the right side.

$$3x+5=Ax+4A+Bx-B$$ $$3x+5=(A+B)x+(4A-B)$$

Step 4: Compare coefficients.

Match the coefficients of corresponding powers of \(x\):

• Coefficient of \(x\): \(A+B=3\)
• Constant term: \(4A-B=5\)

Now solve the system:

$$A+B=3$$ $$4A-B=5$$

Add the equations:

$$5A=8 \Rightarrow A=\frac{8}{5}$$

Then

$$B=3-\frac{8}{5}=\frac{7}{5}$$

Final answer:

$$\frac{3x+5}{(x-1)(x+4)}=\frac{8}{5(x-1)}+\frac{7}{5(x+4)}$$

Worked Example 3: Repeated linear factor

Decompose

$$\frac{2x+3}{(x-1)^2}$$

Step 1: Use the correct form for a repeated factor.

$$\frac{2x+3}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$$

Step 2: Clear the denominators.

$$2x+3=A(x-1)+B$$

Step 3: Expand.

$$2x+3=Ax-A+B$$

Step 4: Compare coefficients.

• Coefficient of \(x\): \(A=2\)
• Constant term: \(-A+B=3\)

Substitute \(A=2\):

$$-2+B=3 \Rightarrow B=5$$

Final answer:

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

Worked Example 4: One linear factor and one irreducible quadratic factor

Decompose

$$\frac{x^2+2}{(x-1)(x^2+1)}$$

Step 1: Set up the correct form.

Because \(x^2+1\) is an irreducible quadratic, its numerator must be linear:

$$\frac{x^2+2}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$$

Step 2: Clear denominators.

$$x^2+2=A(x^2+1)+(Bx+C)(x-1)$$

Step 3: Expand carefully.

$$x^2+2=Ax^2+A+Bx^2-Bx+Cx-C$$ $$x^2+2=(A+B)x^2+(-B+C)x+(A-C)$$

Step 4: Compare coefficients.

On the left side,

$$x^2+2=1x^2+0x+2$$

So we get the system:

• \(A+B=1\)
• \(-B+C=0\)
• \(A-C=2\)

From \(-B+C=0\), we have \(C=B\).

Substitute into the other equations:

$$A+B=1$$ $$A-B=2$$

Subtract the second equation from the first:

$$2B=-1 \Rightarrow B=-\frac{1}{2}$$

Then

$$C=-\frac{1}{2}$$

And

$$A=1-\left(-\frac{1}{2}\right)=\frac{3}{2}$$

Final answer:

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

A cleaner way to write it is

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

5. What if the fraction is improper?

If the degree of the numerator is greater than or equal to the degree of the denominator, do polynomial division first.

For example, consider

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

Divide \(x^2+3x+2\) by \(x+1\):

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

Since the division leaves no remainder, there is no partial fraction part needed.

Here is another example:

$$\frac{x^2+1}{x-1}$$

Divide:

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

Now the rational part \(\frac{2}{x-1}\) is already simple.

6. Common mistakes to avoid

• Forgetting to check if the fraction is proper. Always compare degrees first.
• Not factoring the denominator completely. Partial fractions depend on the factors.
• Using the wrong numerator form. Linear factors get constant numerators, but irreducible quadratic factors need linear numerators.
• For repeated factors, missing terms. If the denominator has \((x-2)^3\), you need terms for \((x-2)\), \((x-2)^2\), and \((x-2)^3\).
• Algebra errors when expanding. Be careful with signs and distribution.

7. Quick pattern guide

• For \((x-a)(x-b)\), use \(\frac{A}{x-a}+\frac{B}{x-b}\)
• For \((x-a)^2\), use \(\frac{A}{x-a}+\frac{B}{(x-a)^2}\)
• For \((x-a)^3\), use \(\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}\)
• For \(x^2+1\), use \(\frac{Ax+B}{x^2+1}\)
• For \((x-a)(x^2+1)\), use \(\frac{A}{x-a}+\frac{Bx+C}{x^2+1}\)

8. Why this matters

Partial fraction decomposition turns a complicated rational expression into simpler pieces. These simpler pieces are easier to:

• simplify,
• evaluate,
• solve in equations,
• and use in later topics such as integration.

Think of it as breaking one hard fraction into smaller fractions that are easier to manage.

Brief Summary

To use partial fraction decomposition, first make sure the rational expression is proper, then factor the denominator completely. Next, write the correct fraction form based on the factors, clear denominators, and solve for the unknown constants. The key is choosing the correct setup: constants over linear factors, extra terms for repeated factors, and linear numerators over irreducible quadratic factors.

Radical Expressions and Fractional Exponents are two ways of writing the same idea. In this lesson, you will learn how to move between radical notation and rational exponent notation, simplify expressions, and rationalize numerators and denominators.

This topic is important because it appears in algebra, equation solving, function work, and later courses. If you understand how roots and fractional powers are connected, many expressions become easier to simplify and compare.

Let’s begin with the key idea: a radical is another way to write a fractional exponent.

For example, the square root of a number can be written as a power of \(\frac{1}{2}\):

$$\sqrt{x} = x^{1/2}$$

Similarly, the cube root can be written as a power of \(\frac{1}{3}\):

$$\sqrt[3]{x} = x^{1/3}$$

More generally, for a positive integer \(n\),

$$\sqrt[n]{x} = x^{1/n}$$

If the exponent has the form \(\frac{m}{n}\), then it means both a power and a root:

$$x^{m/n} = \left(\sqrt[n]{x}\right)^m = \sqrt[n]{x^m}$$

This means:

• the denominator of the fraction tells you the root,
• the numerator tells you the power.

For example,

$$x^{3/4} = \sqrt[4]{x^3} = \left(\sqrt[4]{x}\right)^3$$

Both forms are equivalent. You can use whichever one makes the problem easier.

Common translations between forms are:

• \(\sqrt{x} = x^{1/2}\)
• \(\sqrt[3]{x} = x^{1/3}\)
• \(\sqrt[5]{x^2} = x^{2/5}\)
• \(x^{7/3} = \sqrt[3]{x^7}\)
• \(a^{3/2} = \left(\sqrt{a}\right)^3 = \sqrt{a^3}\)

When simplifying, it helps to remember the basic exponent laws still work with fractional exponents, as long as the expressions are defined.

Exponent rules to use:

• \(a^m \cdot a^n = a^{m+n}\)
• \(\dfrac{a^m}{a^n} = a^{m-n}\), for \(a \ne 0\)
• \((a^m)^n = a^{mn}\)
• \((ab)^m = a^m b^m\)
• \(a^{-m} = \dfrac{1}{a^m}\), for \(a \ne 0\)

These rules combine naturally with radicals. For instance,

$$x^{1/2} \cdot x^{3/2} = x^{4/2} = x^2$$

and

$$\frac{x^{5/3}}{x^{2/3}} = x^{3/3} = x$$

Now let’s look at simplifying radical expressions.

To simplify a radical, factor out perfect powers when possible. For square roots, look for perfect squares. For cube roots, look for perfect cubes.

For example,

$$\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$$

In fractional exponent form, this is

$$72^{1/2} = (36 \cdot 2)^{1/2} = 36^{1/2} \cdot 2^{1/2} = 6\sqrt{2}$$

For a cube root,

$$\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$$

Worked Example 1: Translate between radical and fractional exponent notation

Write \(\sqrt[4]{x^3}\) using a fractional exponent.

The index of the radical is \(4\), so the denominator of the exponent is \(4\). The power on \(x\) is \(3\), so the numerator is \(3\).

$$\sqrt[4]{x^3} = x^{3/4}$$

Now go the other way: write \(a^{5/2}\) in radical form.

The denominator \(2\) means square root, and the numerator \(5\) means fifth power:

$$a^{5/2} = \sqrt{a^5} = \left(\sqrt{a}\right)^5$$

Both forms are correct.

Worked Example 2: Simplify an expression with fractional exponents

Simplify

$$x^{1/2} \cdot x^{3/2}$$

Because the bases are the same, add the exponents:

$$x^{1/2} \cdot x^{3/2} = x^{(1/2 + 3/2)} = x^{4/2} = x^2$$

In radical form, this would be

$$\sqrt{x} \cdot \sqrt{x^3} = x^2$$

This example shows that fractional exponents follow the same exponent laws as whole-number exponents.

Worked Example 3: Rationalize a denominator

Simplify

$$\frac{5}{\sqrt{3}}$$

A denominator is rationalized when it has no radical in it. To remove \(\sqrt{3}\) from the denominator, multiply the fraction by \(\frac{\sqrt{3}}{\sqrt{3}}\), which equals 1.

$$\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$$

The denominator is now rational, so the simplified result is

$$\frac{5\sqrt{3}}{3}$$

If the denominator has two terms and contains a square root, you usually rationalize using the conjugate.

The conjugate of \(a+b\) is \(a-b\), and the conjugate of \(a-b\) is \(a+b\). This works because

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

For example, to rationalize

$$\frac{1}{2+\sqrt{5}}$$

multiply by the conjugate:

$$\frac{1}{2+\sqrt{5}} \cdot \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{2-\sqrt{5}}{4-5} = \frac{2-\sqrt{5}}{-1} = \sqrt{5}-2$$

Worked Example 4: Rationalize a numerator and simplify

Simplify

$$\frac{\sqrt{6}}{3}$$

This expression is already acceptable because the denominator is rational. But if a problem asks you to rationalize the numerator, the idea is to remove the radical from the top.

Multiply by \(\frac{\sqrt{6}}{\sqrt{6}}\):

$$\frac{\sqrt{6}}{3} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{6}{3\sqrt{6}} = \frac{2}{\sqrt{6}}$$

If desired, you can then rationalize the denominator too:

$$\frac{2}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$$

This shows that rationalizing the numerator is less common than rationalizing the denominator, but it uses the same idea: multiply by a form of 1 that removes the radical.

Important notes about domain:

• For even roots, such as square roots, the expression under the radical must be nonnegative in real-number work.
• For odd roots, such as cube roots, negative values are allowed.
• Be careful when simplifying expressions involving variables, especially with even roots.

For example,

$$\sqrt{x^2} = |x|$$

not just \(x\), because square root means the principal nonnegative root.

Tips for success:

• Read fractional exponents as “power over root.”
• If you see \(x^{m/n}\), think: root \(n\), then power \(m\).
• Use exponent laws to combine like bases.
• When simplifying radicals, factor out perfect powers.
• To rationalize a denominator with one radical term, multiply by that radical.
• To rationalize a denominator with two terms, multiply by the conjugate.

Brief Summary

Radical notation and fractional exponent notation represent the same mathematical idea. You can translate between them using

$$x^{m/n} = \sqrt[n]{x^m}$$

You also simplify these expressions using exponent rules, and you rationalize denominators or numerators by multiplying by a suitable form of 1. With practice, these skills make complicated algebraic expressions much easier to handle.

Solving Radical and Rational Equations is an important algebra skill because many equations are not solved by simple inverse operations alone. When an equation contains a radical such as a square root, or a rational expression such as a fraction with a variable in the denominator, we often need special methods to solve it correctly.

The most important idea in this lesson is that not every answer you get from your algebra steps is actually valid. Some steps, especially squaring both sides or multiplying by expressions involving variables, can create answers that do not truly satisfy the original equation. These are called extraneous solutions.

In this lesson, you will learn how to solve radical and rational equations carefully, how to recognize restrictions, and how to check your answers in the original equation.

1. What is a radical equation?

A radical equation is an equation in which the variable appears inside a radical, such as:

$$\sqrt{x+5}=3$$

To solve radical equations, we usually try to isolate the radical first, then remove it by raising both sides to a power. For square roots, we square both sides. For cube roots, we cube both sides.

Basic process for solving radical equations:

1. Isolate the radical on one side of the equation.
2. Raise both sides to the appropriate power to remove the radical.
3. Solve the resulting equation.
4. Check every solution in the original equation.

Why checking matters: Squaring both sides can introduce solutions that were not originally true. So checking is not optional for radical equations.

Example 1: Solve a simple radical equation

Solve:

$$\sqrt{x+7}=5$$

Step 1: Isolate the radical.

The radical is already isolated.

Step 2: Square both sides.

$$\left(\sqrt{x+7}\right)^2=5^2$$ $$x+7=25$$

Step 3: Solve.

$$x=18$$

Step 4: Check in the original equation.

$$\sqrt{18+7}=\sqrt{25}=5$$

This is true, so the solution is:

$$x=18$$

2. Radical equations with more than one step

Sometimes the radical is not isolated at first. In that case, move other terms before squaring.

Example 2: Solve a radical equation carefully

Solve:

$$\sqrt{2x+3}=x-1$$

Step 1: Isolate the radical.

The radical is already isolated.

Step 2: Square both sides.

$$\left(\sqrt{2x+3}\right)^2=(x-1)^2$$ $$2x+3=x^2-2x+1$$

Step 3: Rearrange into a standard quadratic equation.

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

So:

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

Step 4: Solve using the quadratic formula.

$$x=\frac{4\pm\sqrt{(-4)^2-4(1)(-2)}}{2(1)}$$ $$x=\frac{4\pm\sqrt{16+8}}{2}$$ $$x=\frac{4\pm\sqrt{24}}{2}$$ $$x=\frac{4\pm 2\sqrt{6}}{2}$$ $$x=2\pm\sqrt{6}$$

Step 5: Check both answers in the original equation.

First, test \(x=2+\sqrt{6}\). This makes the right side \(x-1=1+\sqrt{6}\), which is positive, so it may work. Substituting shows it does satisfy the equation.

Now test \(x=2-\sqrt{6}\). Since \(\sqrt{6}\approx 2.45\), this value is less than 0. Then \(x-1\) is negative. But a square root is always nonnegative, so this cannot satisfy the original equation.

Therefore, the only valid solution is:

$$x=2+\sqrt{6}$$

Important note: In equations like \(\sqrt{2x+3}=x-1\), the right side must also be nonnegative, because it equals a square root. This can help you predict which answers might be impossible.

3. What is a rational equation?

A rational equation is an equation that contains one or more rational expressions. A rational expression is a fraction involving a variable, such as:

$$\frac{3}{x-2}=5$$

To solve rational equations, we usually eliminate the denominators by multiplying every term by the least common denominator (LCD).

Basic process for solving rational equations:

1. Find all values that would make any denominator equal to 0. These values are not allowed.
2. Find the LCD of all denominators.
3. Multiply every term in the equation by the LCD.
4. Solve the resulting equation.
5. Check that the solution does not make a denominator 0, and verify it in the original equation.

Why restrictions matter: A denominator can never be 0. So if your algebra gives a value that makes a denominator 0, that value must be rejected.

Example 3: Solve a rational equation

Solve:

$$\frac{2}{x-1}+\frac{3}{x+2}=\frac{7}{(x-1)(x+2)}$$

Step 1: State restrictions.

The denominators are 0 when:

$$x-1=0 \Rightarrow x=1$$ $$x+2=0 \Rightarrow x=-2$$

So \(x\neq 1\) and \(x\neq -2\).

Step 2: Find the LCD.

The LCD is:

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

Step 3: Multiply every term by the LCD.

$$ (x-1)(x+2)\left(\frac{2}{x-1}\right)+(x-1)(x+2)\left(\frac{3}{x+2}\right)=(x-1)(x+2)\left(\frac{7}{(x-1)(x+2)}\right) $$

Simplify:

$$2(x+2)+3(x-1)=7$$

Step 4: Solve.

$$2x+4+3x-3=7$$ $$5x+1=7$$ $$5x=6$$ $$x=\frac{6}{5}$$

Step 5: Check restrictions and original equation.

\(\frac{6}{5}\) is not 1 or \(-2\), so it does not make any denominator 0. Substituting into the original equation confirms it works.

Therefore:

$$x=\frac{6}{5}$$

4. Rational equations that become quadratic

Some rational equations simplify into quadratic equations. This means you may get more than one possible answer, and you must test each one against the denominator restrictions.

Example 4: Solve and reject extraneous values

Solve:

$$\frac{1}{x}+\frac{1}{x-2}=\frac{4}{x(x-2)}$$

Step 1: State restrictions.

$$x\neq 0,\quad x\neq 2$$

Step 2: Find the LCD.

The LCD is:

$$x(x-2)$$

Step 3: Multiply every term by the LCD.

$$x(x-2)\left(\frac{1}{x}\right)+x(x-2)\left(\frac{1}{x-2}\right)=x(x-2)\left(\frac{4}{x(x-2)}\right)$$

Simplify:

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

Step 4: Check.

Since \(3\neq 0\) and \(3\neq 2\), it does not violate the restrictions. Substitute into the original equation:

$$\frac{1}{3}+\frac{1}{1}=\frac{1}{3}+1=\frac{4}{3}$$

Right side:

$$\frac{4}{3(1)}=\frac{4}{3}$$

Both sides are equal, so:

$$x=3$$

5. How extraneous solutions happen

There are two common reasons invalid solutions appear:

• When solving radical equations, squaring both sides can create extra answers.
• When solving rational equations, algebra may produce a value that makes a denominator 0.

That is why checking in the original equation is one of the most important habits in this topic.

6. Tips for solving correctly

• Always isolate the radical before removing it.
• Always write denominator restrictions first in rational equations.
• Multiply every term by the LCD, not just part of the equation.
• Be careful when squaring binomials: \((a-b)^2=a^2-2ab+b^2\).
• Check every solution in the original equation, especially after squaring.

7. Common mistakes to avoid

• Forgetting to check solutions after squaring both sides.
• Accepting a value that makes a denominator 0.
• Using the wrong LCD.
• Not distributing correctly after multiplying by the LCD.
• Squaring incorrectly, such as thinking \((x-1)^2=x^2-1\), which is false.

8. Quick comparison

Here is the main difference between the two types:

• Radical equations: remove radicals by raising both sides to a power, then check for extraneous solutions.
• Rational equations: remove denominators using the LCD, then reject any values that make denominators 0.

Brief Summary

To solve radical equations, isolate the radical, remove it by raising both sides to a power, solve, and then check your answers in the original equation. To solve rational equations, first identify values that make denominators 0, then multiply every term by the LCD, solve, and reject any invalid answers.

The key skill in both cases is being careful about extraneous solutions. A correct algebra process is important, but a final answer is only valid if it satisfies the original equation.

Absolute value measures distance from zero on a number line. Because distance is never negative, absolute value is always nonnegative.

For any real number \(x\), the absolute value of \(x\) is written \(|x|\) and defined by the piecewise rule

$$ |x|= \begin{cases} x & \text{if } x\ge 0 \\ -x & \text{if } x<0 \end{cases} $$

This piecewise definition is the key to solving absolute value equations and inequalities. When the expression inside the absolute value could be positive or negative, we must consider cases carefully.

In this lesson, you will learn how to solve:

• absolute value equations such as \(|2x-5|=7\)
• absolute value inequalities such as \(|3x+1|<10\)
• multi-step problems with more than one absolute value
• compound inequalities by using piecewise cases and interval reasoning

1. Understanding what absolute value means

Think of \(|x-a|\) as the distance between \(x\) and \(a\) on the number line.

• \(|x-a|=b\) means the distance from \(x\) to \(a\) is exactly \(b\).
• \(|x-a|<b\) means the distance from \(x\) to \(a\) is less than \(b\).
• \(|x-a|>b\) means the distance from \(x\) to \(a\) is greater than \(b\).

This idea helps explain why equations usually split into two cases, and inequalities often become compound inequalities.

2. Solving absolute value equations

Suppose we want to solve

$$|A|=k$$

where \(A\) is an algebraic expression.

If \(k<0\), there is no solution, because an absolute value cannot be negative.

If \(k\ge 0\), then

$$|A|=k \iff A=k \text{ or } A=-k$$

This works because both \(k\) and \(-k\) are exactly \(k\) units from zero.

Steps for solving an absolute value equation

1. Isolate the absolute value expression if needed.
2. Check whether the number on the other side is negative.
3. Write two equations: one positive case and one negative case.
4. Solve each equation.
5. Check your answers in the original equation.

Worked Example 1: Basic equation

Solve \(|2x-5|=7\).

Since the absolute value equals a positive number, split into two cases:

$$2x-5=7 \quad \text{or} \quad 2x-5=-7$$

Solve each one:

$$2x=12 \Rightarrow x=6$$ $$2x=-2 \Rightarrow x=-1$$

Solution: \(x=6\) or \(x=-1\).

Check:

• If \(x=6\), then \(|2(6)-5|=|12-5|=|7|=7\).
• If \(x=-1\), then \(|2(-1)-5|=|-2-5|=|-7|=7\).

Both are correct.

3. Solving absolute value inequalities

The rules for inequalities depend on whether the sign is less than or greater than.

Case A: Less than

If \(b>0\), then

$$|A|<b \iff -b<A<b$$

and

$$|A|\le b \iff -b\le A\le b$$

This means the expression must stay within a distance of \(b\) from zero.

Case B: Greater than

If \(b>0\), then

$$|A|>b \iff A<-b \text{ or } A>b$$

and

$$|A|\ge b \iff A\le -b \text{ or } A\ge b$$

This means the expression is farther than \(b\) from zero.

Important special cases

• \(|A|<0\): no solution
• \(|A|\le 0\): only possible when \(A=0\)
• \(|A|>0\): all real numbers except where \(A=0\)
• \(|A|\ge 0\): always true for all real numbers

Worked Example 2: Absolute value inequality

Solve \(|3x+1|<10\).

Use the rule for a “less than” inequality:

$$-10<3x+1<10$$

Now solve all three parts together:

$$-11<3x<9$$ $$-\frac{11}{3}<x<3$$

Solution:

$$\left(-\frac{11}{3},\,3\right)$$

This means every \(x\) between \(-\frac{11}{3}\) and \(3\), not including the endpoints.

Worked Example 3: “Greater than” inequality

Solve \(|2x-4|\ge 6\).

Use the rule for a “greater than or equal to” inequality:

$$2x-4\le -6 \quad \text{or} \quad 2x-4\ge 6$$

Solve each inequality:

$$2x\le -2 \Rightarrow x\le -1$$ $$2x\ge 10 \Rightarrow x\ge 5$$

Solution:

$$(-\infty,-1] \cup [5,\infty)$$

This matches the distance idea: numbers at least 6 units away from 4.

4. Solving by piecewise cases

Sometimes a problem has more than one absolute value, or it does not simplify easily with the standard rules. In these situations, use the piecewise definition directly.

The main idea is:

• Find the values that make each inside expression equal to zero.
• These values split the number line into intervals.
• On each interval, remove the absolute values by deciding whether each inside expression is positive or negative.
• Solve the resulting equation or inequality on that interval.
• Keep only answers that actually belong to the interval you used.

Worked Example 4: Multi-step equation with two absolute values

Solve

$$|x-1|+|x+2|=7$$

The expressions inside the absolute values are zero at \(x=1\) and \(x=-2\). These points split the number line into three intervals:

• \(x<-2\)
• \(-2\le x<1\)
• \(x\ge 1\)

Case 1: \(x<-2\)

Both \(x-1\) and \(x+2\) are negative, so

$$|x-1|=-(x-1)=-x+1, \quad |x+2|=-(x+2)=-x-2$$

Substitute:

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

This answer fits the interval \(x<-2\), so \(x=-4\) is valid.

Case 2: \(-2\le x<1\)

Here \(x+2\ge 0\), but \(x-1<0\). So

$$|x-1|=-x+1, \quad |x+2|=x+2$$

Substitute:

$$(-x+1)+(x+2)=7$$ $$3=7$$

This is impossible, so there is no solution in this interval.

Case 3: \(x\ge 1\)

Both expressions are nonnegative, so

$$|x-1|=x-1, \quad |x+2|=x+2$$

Substitute:

$$(x-1)+(x+2)=7$$ $$2x+1=7$$ $$2x=6 \Rightarrow x=3$$

This fits the interval \(x\ge 1\), so \(x=3\) is valid.

Solution: \(x=-4\) or \(x=3\).

5. Piecewise method for inequalities

The same case-by-case strategy works for inequalities. This is especially useful when several absolute values appear in one problem.

For example, in an inequality like

$$|x-2|-|x+1|>1$$

you would split the number line at \(x=2\) and \(x=-1\), then solve on each interval.

The process is:

1. Find all critical points where the inside of an absolute value is zero.
2. Break the number line into intervals.
3. Rewrite each absolute value appropriately on each interval.
4. Solve the simpler inequality.
5. Intersect the result with the interval.
6. Combine all valid pieces.

6. Common mistakes to avoid

• Forgetting the negative case when solving \(|A|=k\). You need both \(A=k\) and \(A=-k\).
• Using “and” instead of “or” for greater-than inequalities. For example, \(|A|>5\) becomes \(A<-5\) or \(A>5\).
• Using “or” instead of “and” for less-than inequalities. For example, \(|A|<5\) becomes \(-5<A<5\).
• Not checking interval restrictions in piecewise problems.
• Ignoring impossible cases, such as \(|A|=-3\).

7. Quick guide: which rule should I use?

• If you see \(|A|=k\), write two equations.
• If you see \(|A|<k\) or \(|A|\le k\), write a compound inequality with and.
• If you see \(|A|>k\) or \(|A|\ge k\), write two inequalities joined by or.
• If there are multiple absolute values, use piecewise cases.

8. Final summary

Absolute value tells you distance, and that distance idea explains why solving these problems often involves two cases.

For equations, isolate the absolute value and split into positive and negative cases. For inequalities, remember the pattern:

• \(|A|<b\) means between \(-b\) and \(b\)
• \(|A|>b\) means outside that interval

When the problem is more complicated, especially with multiple absolute values, use the piecewise definition and solve separately on each interval. Always check that each solution fits the case you used and satisfies the original equation or inequality.

Systems of Non-Linear Equations are systems in which at least one equation is not linear. Instead of two straight lines, you may be finding where a line meets a parabola, where two parabolas meet, or where a circle meets another curve.

When you solve a system of non-linear equations, you are finding the intersection points of the graphs. Each solution is an ordered pair \\( (x,y) \\) that makes every equation in the system true at the same time.

In Grade 12, the most common non-linear systems involve equations such as quadratic equations, circles, rational equations, or radical equations. The key idea is still the same as with linear systems: use algebra to reduce the system to one variable, solve, and then substitute back.

Main idea: the two most useful methods are substitution and elimination. In non-linear systems, substitution is often the fastest because one equation may already be solved for one variable.

Common types of systems:

• a line and a parabola
• two quadratic equations
• a line and a circle
• systems involving radicals or rational expressions

Why non-linear systems can have several solutions

A pair of lines usually intersects once, is parallel, or is the same line. But curves can cross in more than one place. That means a non-linear system can have:

• no solution if the graphs do not meet,
• one solution if they touch at one point,
• two or more solutions if they cross multiple times.

Every algebraic solution should make sense on the graphs. It is helpful to think visually: if a line cuts a parabola, there may be two intersection points, one intersection point, or none.

General strategy for solving a non-linear system

1. Choose one equation and solve for one variable if needed.
2. Substitute that expression into the other equation.
3. Solve the resulting one-variable equation.
4. Substitute each value back to find the matching second variable.
5. Check each solution in the original system.

Important note about checking: in systems with radicals or rational expressions, algebraic steps can sometimes create extraneous solutions. These are values that appear during the work but do not satisfy the original equations. Always check at the end.

Worked Example 1: Line and parabola

Solve the system

$$ \begin{cases} y = x^2 - 1 \\ y = 2x + 3 \end{cases} $$

Since both equations equal \\(y\\), set them equal to each other:

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

Move all terms to one side:

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

Now solve using the quadratic formula:

$$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5} $$

Find the matching \\(y\\)-values using \\(y = 2x + 3\\).

If \\(x = 1 + \sqrt{5}\\), then

$$ y = 2(1 + \sqrt{5}) + 3 = 5 + 2\sqrt{5} $$

If \\(x = 1 - \sqrt{5}\\), then

$$ y = 2(1 - \sqrt{5}) + 3 = 5 - 2\sqrt{5} $$

So the solutions are

$$ (1+\sqrt{5},\; 5+2\sqrt{5}) \quad \text{and} \quad (1-\sqrt{5},\; 5-2\sqrt{5}) $$

This means the line and parabola intersect at two points.

Worked Example 2: Circle and line

Solve the system

$$ \begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases} $$

Substitute \\(y = x+1\\) into the circle equation:

$$ x^2 + (x+1)^2 = 25 $$

Expand:

$$ x^2 + x^2 + 2x + 1 = 25 $$

$$ 2x^2 + 2x - 24 = 0 $$

Divide by 2:

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

Factor:

$$ (x+4)(x-3)=0 $$

So

$$ x=-4 \quad \text{or} \quad x=3 $$

Now find \\(y\\):

• If \\(x=-4\\), then \\(y=x+1=-3\\).
• If \\(x=3\\), then \\(y=x+1=4\\).

The solutions are

$$ (-4,-3) \quad \text{and} \quad (3,4) $$

These are the two points where the line intersects the circle.

Worked Example 3: Using elimination with two non-linear equations

Solve the system

$$ \begin{cases} y = x^2 \\ y = 6 - x^2 \end{cases} $$

Set the two expressions for \\(y\\) equal:

$$ x^2 = 6 - x^2 $$

Add \\(x^2\\) to both sides:

$$ 2x^2 = 6 $$

$$ x^2 = 3 $$

$$ x = \pm \sqrt{3} $$

Now use \\(y = x^2\\):

$$ y = 3 $$

So the solutions are

$$ (\sqrt{3},3) \quad \text{and} \quad (-\sqrt{3},3) $$

This is a good example of two parabolas intersecting in two symmetric points.

Worked Example 4: A system with a radical

Solve the system

$$ \begin{cases} y = \sqrt{x+1} \\ y = x - 1 \end{cases} $$

Set the equations equal:

$$ \sqrt{x+1} = x - 1 $$

Because a square root is always non-negative, we must have

$$ x - 1 \ge 0 \Rightarrow x \ge 1 $$

Now square both sides:

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

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

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

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

So the possible values are \\(x=0\\) or \\(x=3\\).

Now check them in the original equation \\(\sqrt{x+1} = x-1\\).

• If \\(x=0\\), then \\(\sqrt{1}=1\\) but \\(x-1=-1\\). Since \\(1 \ne -1\\), this does not work.
• If \\(x=3\\), then \\(\sqrt{4}=2\\) and \\(x-1=2\\). This works.

Now find \\(y\\):

$$ y = 3-1 = 2 $$

The only solution is

$$ (3,2) $$

This example shows why checking is essential. Squaring both sides created an extra solution that was not actually valid.

How to choose a method

• Use substitution when one equation is already solved for \\(x\\) or \\(y\\).
• Use elimination when matching terms can cancel easily.
• Use factoring whenever the resulting equation is factorable.
• Use the quadratic formula when factoring is difficult or impossible.

Things to watch out for

• Forgetting all solutions: a quadratic equation may give two \\(x\\)-values.
• Not checking domain restrictions: radicals need non-negative values inside the root, and rational expressions cannot have zero in the denominator.
• Losing or adding solutions: operations like squaring both sides can change the solution set.
• Not using the original equations to check: always test final answers in the original system.

Geometric meaning of the answers

Each solution represents a point where the graphs intersect. If you find two ordered pairs, the curves cross twice. If you find one ordered pair, they meet once. If there is no real solution, the graphs do not intersect in the real coordinate plane.

Quick problem-solving checklist

1. Identify the type of curves involved.
2. Choose substitution or elimination.
3. Reduce to one equation in one variable.
4. Solve carefully.
5. Substitute back to get the second variable.
6. Check all answers in the original system.
7. State solutions as ordered pairs.

Summary

Systems of non-linear equations are solved by finding the points that satisfy both equations at once. Most often, you substitute one equation into the other and solve the resulting quadratic or other non-linear equation. Because curves can intersect more than once, these systems may have multiple solutions. Always check your answers, especially when radicals or rational expressions are involved.