1.5 Factoring Polynomials

Factoring Polynomials

Factoring is the process of breaking a polynomial into simpler expressions that multiply together to give the original. It's one of the most important skills in algebra because it's how you'll solve equations, simplify rational expressions, and work with higher-level math throughout this course.

This section covers several factoring techniques, from pulling out a greatest common factor to recognizing special patterns like difference of squares and sum/difference of cubes.

Introduction to Polynomials

A polynomial is an expression made up of variables and coefficients, combined using addition, subtraction, and multiplication. Polynomials are classified by their number of terms:

• Monomial: one term (e.g., $7x^3$)
• Binomial: two terms (e.g., $x^2 - 9$)
• Trinomial: three terms (e.g., $2x^2 + 5x + 3$)

Factoring means writing a polynomial as a product of simpler expressions. Think of it as reversing the multiplication (distribution) you already know how to do.

Greatest Common Factor in Polynomials

Always start here. Before trying any other technique, check whether all the terms share a common factor.

Steps to factor out the GCF:

1. List the factors of each term (both the numerical coefficients and the variables with their exponents).
2. Identify the largest number that divides every coefficient and the lowest power of each variable that appears in every term.
3. Divide each term by the GCF.
4. Write the result as the GCF times the remaining expression in parentheses.

Example: Factor $12x^3 - 18x^2 + 6x$.

• The coefficients are 12, 18, and 6. The GCF of those is 6.
• The variable parts are $x^3$, $x^2$, and $x$. The lowest power is $x$.
• GCF = $6x$
• Divide each term: $12x^3 \div 6x = 2x^2$, $-18x^2 \div 6x = -3x$, $6x \div 6x = 1$
• Result: $6x(2x^2 - 3x + 1)$

Methods for Quadratic Trinomials

A quadratic trinomial has the form $ax^2 + bx + c$, where $a \neq 0$. The goal is to write it as a product of two binomials.

When $a = 1$ (simple case):

Find two numbers that multiply to $c$ and add to $b$. Those numbers go directly into the binomial factors.

Example: Factor $x^2 + 5x + 6$.

You need two numbers that multiply to 6 and add to 5. That's 2 and 3.

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

When $a \neq 1$ (AC method):

1. Multiply $a \cdot c$ to get the product $ac$.
2. Find two numbers that multiply to $ac$ and add to $b$.
3. Rewrite the middle term $bx$ as two terms using those numbers.
4. Factor by grouping (split into two pairs, factor each pair, then factor out the common binomial).

Example: Factor $6x^2 + 7x - 3$.

1. $a \cdot c = 6 \times (-3) = -18$

2. Find two numbers that multiply to $-18$ and add to $7$. That's $9$ and $-2$.

3. Rewrite: $6x^2 + 9x - 2x - 3$

4. Group: $(6x^2 + 9x) + (-2x - 3) = 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3)$

Using the quadratic formula as a backup:

If you can't spot the factors, the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ gives you the roots $r_1$ and $r_2$. Then the factored form is $a(x - r_1)(x - r_2)$.

Grouping Method for Longer Polynomials

For polynomials with four or more terms, factoring by grouping is your main tool.

1. Group the terms into pairs that share a common factor.
2. Factor the GCF out of each pair.
3. If both groups produce the same binomial factor, factor that binomial out.

Example: Factor $x^3 + 2x^2 - 3x - 6$.

1. Group: $(x^3 + 2x^2) + (-3x - 6)$

2. Factor each group: $x^2(x + 2) - 3(x + 2)$

3. Factor out $(x + 2)$: $(x^2 - 3)(x + 2)$

If the grouping doesn't produce a common binomial, try rearranging the terms before grouping again.

Perfect Square Trinomials

These follow a specific pattern worth memorizing:

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

How to recognize one: The first and last terms must be perfect squares, and the middle term must equal $2ab$ (twice the product of their square roots).

Example: Is $x^2 + 6x + 9$ a perfect square trinomial?

• $\sqrt{x^2} = x$ and $\sqrt{9} = 3$
• Check the middle term: $2(x)(3) = 6x$. Yes, it matches.
• Factored form: $(x + 3)^2$

Example: $y^2 - 14y + 49 = (y - 7)^2$ because $2(y)(7) = 14y$ and the middle term is negative.

Difference of Squares

This pattern applies to binomials where two perfect squares are subtracted:

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

Example: $x^2 - 25 = (x + 5)(x - 5)$

Example: $9y^2 - 16 = (3y + 4)(3y - 4)$

A common mistake: there is no "sum of squares" factoring pattern. $a^2 + b^2$ does not factor over the real numbers.

Sum and Difference of Cubes

Two formulas to know here:

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

A helpful memory trick: the sign in the binomial factor matches the original sign, and the first sign in the trinomial factor is the opposite.

Example: Factor $x^3 + 8$.

• Rewrite as $x^3 + 2^3$, so $a = x$ and $b = 2$.
• $(x + 2)(x^2 - 2x + 4)$

Example: Factor $27y^3 - 125$.

• Rewrite as $(3y)^3 - 5^3$, so $a = 3y$ and $b = 5$.
• $(3y - 5)(9y^2 + 15y + 25)$

Factoring with Non-Integer Exponents

Sometimes you'll encounter expressions with fractional or negative exponents. The same factoring patterns apply; you just need to rewrite the expression first.

Fractional exponents: Rewrite terms so you can recognize a familiar pattern.

Example: Factor $x^{3/2} - 8$.

• Rewrite as $(x^{1/2})^3 - 2^3$. This is a difference of cubes with $a = x^{1/2}$ and $b = 2$.
• $(x^{1/2} - 2)(x + 2x^{1/2} + 4)$

Negative exponents: You can use substitution to make the expression look like a standard polynomial. Let $u = x^{-1}$, factor in terms of $u$, then substitute back.

Example: Factor $x^{-2} - 9x^{-1} + 8$.

• Let $u = x^{-1}$: the expression becomes $u^2 - 9u + 8$.
• Factor: $(u - 1)(u - 8)$
• Substitute back: $(x^{-1} - 1)(x^{-1} - 8) = \frac{(1 - x)(1 - 8x)}{x^2}$