Quadratic factoring

Understanding Quadratic Expressions to Factor

In this chapter, we focus on factoring quadratic expressions, that is, algebraic expressions whose highest power of $x$ is $2$.

A quadratic expression in one variable usually looks like
$$
ax^2 + bx + c,
$$
where $a$, $b$, and $c$ are constants and $a \neq 0$.

To “factor a quadratic” means to rewrite it as a product of two simpler expressions, typically two linear factors:
$$
ax^2 + bx + c = (px + q)(rx + s),
$$
for some numbers $p, q, r, s$.

Factoring quadratics is especially important because it lets you solve quadratic equations by setting each factor equal to zero.

We will look at the most common patterns and techniques:

• Factoring out a greatest common factor first
• Quadratics with leading coefficient $1$: $x^2 + bx + c$
• Quadratics with leading coefficient not $1$: $ax^2 + bx + c$
• Special patterns you can recognize quickly
• Checking your factorization

(General factoring ideas and common factors have been discussed in the parent “Factoring” chapter; here we apply them specifically to quadratics.)

Always Start: Look for a Common Factor

Before using any specific quadratic technique, check if all terms share a common factor.

For example:
$$
6x^2 + 9x = 3x(2x + 3).
$$

Once you factor out the greatest common factor (GCF), you then focus on factoring the quadratic that remains, if possible.

Example:

• Factor $4x^2 + 8x + 4$.

First, notice each term is divisible by $4$:
$$
4x^2 + 8x + 4 = 4(x^2 + 2x + 1).
$$

Now the quadratic inside the parentheses might be factorable further (we will see how soon). So the overall process is often:

1. Factor out a GCF.
2. Factor the remaining quadratic (if possible).

Quadratics with Leading Coefficient 1: $x^2 + bx + c$

The simplest type of quadratic to factor has the form
$$
x^2 + bx + c,
$$
with coefficient of $x^2$ equal to $1$.

The “Product and Sum” Idea

To factor $x^2 + bx + c$, you look for two numbers:

• whose product is $c$, and
• whose sum is $b$.

If you can find such numbers $m$ and $n$, then
$$
x^2 + bx + c = (x + m)(x + n).
$$

The two numbers $m$ and $n$ are called the roots or zeros of the factored form, but here we just think of them as the numbers that make the pattern work.

Example: Positive $c$

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

We need two numbers that multiply to $6$ and add to $5$.

Possible factor pairs of $6$:

• $1$ and $6$ (sum $7$),
• $2$ and $3$ (sum $5$).

We want $5$, so use $2$ and $3$.

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

Example: Negative $c$

Factor $x^2 + x - 6$.

We need two numbers that multiply to $-6$ and add to $1$.

Factor pairs of $-6$:

• $-1$ and $6$ (sum $5$),
• $1$ and $-6$ (sum $-5$),
• $-2$ and $3$ (sum $1$),
• $2$ and $-3$ (sum $-1$).

We want $1$, so use $-2$ and $3$.

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

Example: Both negative numbers

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

We need two numbers that multiply to $12$ and add to $-7$.

Factor pairs of $12$:

• $1$ and $12$ (sum $13$),
• $2$ and $6$ (sum $8$),
• $3$ and $4$ (sum $7$).

We need $-7$ with a positive product $12$. That means both numbers are negative:

• $-3$ and $-4$ multiply to $12$ and add to $-7$.

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

When It Does Not Factor Nicely

Not every quadratic with leading coefficient $1$ factors neatly with integer numbers.

For example, try $x^2 + x + 1$:

• Factor pairs of $1$ are just $1$ and $1$.
• Sums: $1+1 = 2$ or $-1-1 = -2$.

You cannot get a sum of $1$, so $x^2 + x + 1$ does not factor over the integers. In this course level, we usually stop and say it is “prime” (or “irreducible”) with respect to factoring into integer-coefficient linear factors.

Quadratics with Leading Coefficient Not 1: $ax^2 + bx + c$

When the coefficient of $x^2$ is not $1$, such as in
$$
2x^2 + 7x + 3
\quad\text{or}\quad
3x^2 - 5x - 2,
$$
factoring can be a bit more involved.

There are several methods. Here we focus on one that is systematic and widely used: the $ac$ method (sometimes called the “split the middle term” method).

The $ac$ Method

Consider
$$
ax^2 + bx + c
$$
with $a \neq 1$.

1. Compute the product $ac$.
2. Find two numbers $m$ and $n$ such that:
• $m \cdot n = ac$,
• $m + n = b$.
3. Rewrite the middle term $bx$ as $mx + nx$.
4. Factor by grouping in two pairs.
5. If done correctly, you will get a product of two binomials.

Example: $2x^2 + 7x + 3$

1. Here $a = 2$, $b = 7$, $c = 3$.

Compute $ac = 2 \cdot 3 = 6$.

2. Find $m$ and $n$ such that:
• $m \cdot n = 6$,
• $m + n = 7$.

The pair $1$ and $6$ works: $1 \cdot 6 = 6$ and $1 + 6 = 7$.

3. Rewrite $7x$ as $1x + 6x$:
   $$
   2x^2 + 7x + 3 = 2x^2 + x + 6x + 3.
   $$
4. Now factor by grouping:

Group the first two and last two terms:
$$
(2x^2 + x) + (6x + 3).
$$

Factor each group:

• From $2x^2 + x$ factor out $x$: $x(2x + 1)$.
• From $6x + 3$ factor out $3$: $3(2x + 1)$.

So we have
$$
2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1).
$$

Now $(2x + 1)$ is common:
$$
= (2x + 1)(x + 3).
$$

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

Example with Negative $c$: $3x^2 - 5x - 2$

1. $a = 3$, $b = -5$, $c = -2$.

Compute $ac = 3 \cdot (-2) = -6$.

2. Find $m$ and $n$ such that:
• $m \cdot n = -6$,
• $m + n = -5$.

Factor pairs of $-6$:

• $-1$ and $6$ (sum $5$),
• $1$ and $-6$ (sum $-5$),
• $-2$ and $3$ (sum $1$),
• $2$ and $-3$ (sum $-1$).

We want $-5$, so use $1$ and $-6$.

3. Rewrite $-5x$ as $1x - 6x$:
   $$
   3x^2 - 5x - 2 = 3x^2 + x - 6x - 2.
   $$
4. Factor by grouping:
   $$
   (3x^2 + x) + (-6x - 2).
   $$
   Factor each group:
• From $3x^2 + x$ factor out $x$: $x(3x + 1)$.
• From $-6x - 2$ factor out $-2$: $-2(3x + 1)$.

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

Factor out $(3x + 1)$:
$$
= (3x + 1)(x - 2).
$$

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

Example: Common Factor plus $ac$ Method

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

1. First factor out the GCF $4$:
   $$
   4x^2 + 12x + 8 = 4(x^2 + 3x + 2).
   $$
2. Now factor $x^2 + 3x + 2$ as a simple quadratic ($a=1$).

Find two numbers with product $2$ and sum $3$: $1$ and $2$.

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

3. Combine with the GCF:
   $$
   4x^2 + 12x + 8 = 4(x + 1)(x + 2).
   $$

Special Quadratic Patterns

Some quadratics follow common patterns you can recognize and factor very quickly. Here we focus on those most often seen in quadratic factoring.

Difference of Squares

A difference of squares is any expression of the form
$$
A^2 - B^2,
$$
which always factors as
$$
A^2 - B^2 = (A - B)(A + B).
$$

In a quadratic in $x$, this often appears as
$$
x^2 - a^2 = (x - a)(x + a).
$$

Examples:

1. $x^2 - 9$:
   = 3^2$, so
   $$
   x^2 - 9 = x^2 - 3^2 = (x - 3)(x + 3).
   $$
2. $4x^2 - 25$:
   x^2 = (2x)^2$ and = 5^2$, so
   $$
   4x^2 - 25 = (2x)^2 - 5^2 = (2x - 5)(2x + 5).
   $$
3. $9x^2 - 16y^2$ (a binomial quadratic in $x$ and $y$):
   $$
   9x^2 = (3x)^2,\quad 16y^2 = (4y)^2,
   $$
   so
   $$
   9x^2 - 16y^2 = (3x - 4y)(3x + 4y).
   $$

The key clues for a difference of squares are:

• exactly two terms (a binomial),
• a subtraction sign,
• each term is a perfect square.

Perfect Square Trinomials

A perfect square trinomial comes from squaring a binomial, for example:
$$
(x + a)^2 = x^2 + 2ax + a^2,
$$
$$
(x - a)^2 = x^2 - 2ax + a^2.
$$

Recognizing these in reverse lets you factor quickly.

The patterns in $x$ are:

1. $x^2 + 2ax + a^2 = (x + a)^2$,
2. $x^2 - 2ax + a^2 = (x - a)^2$.

Key features:

• first term is a perfect square ($x^2$ or something like $4x^2$),
• last term is a perfect square ($a^2$),
• the middle term’s coefficient is twice the product of the square roots of first and last terms.

Examples

1. Factor $x^2 + 6x + 9$.
• First term: $x^2 = x^2$.
• Last term: $9 = 3^2$.
• Middle term: $6x = 2 \cdot x \cdot 3$.

So it matches $x^2 + 2ax + a^2$ with $a = 3$:
$$
x^2 + 6x + 9 = (x + 3)^2.
$$

2. Factor $4x^2 - 12x + 9$.
• First term: $4x^2 = (2x)^2$.
• Last term: $9 = 3^2$.
• Middle term: $-12x = 2 \cdot (2x) \cdot (-3)$.

So it matches $(2x - 3)^2$:
$$
4x^2 - 12x + 9 = (2x - 3)^2.
$$

3. Not a perfect square: $x^2 + 5x + 4$.
• First term: $x^2$.
• Last term: $4 = 2^2$.
• Twice the product of $x$ and $2$ is $2\cdot x \cdot 2 = 4x$, not $5x$.

So this is not a perfect square trinomial, though it still factors:
$$
x^2 + 5x + 4 = (x + 1)(x + 4).
$$

Deciding Which Factoring Method to Use

When you see a quadratic expression to factor, you can follow a short checklist:

1. Is there a common factor?
   Factor out the GCF first if there is one.
2. Is it a binomial (two terms)?
• Check if it is a difference of squares, $A^2 - B^2$.
• If so, factor as $(A - B)(A + B)$.
3. Is it a trinomial of the form $x^2 + bx + c$?
• Use the product and sum method to find $m$ and $n$ such that $m \cdot n = c$ and $m + n = b$.
4. Is it a trinomial $ax^2 + bx + c$ with $a \neq 1$?
• Use the $ac$ method (split the middle term and factor by grouping).
5. Does it look like a perfect square trinomial?
• Recognize $(x + a)^2$ or $(px + q)^2$ patterns.
6. If none of these work with integers, the quadratic might be irreducible over the integers at this level.

Checking Your Factoring

It is very important to verify your factored form. The easiest way is to multiply the factors back together and see whether you recover the original quadratic.

For example, if you factored
$$
3x^2 - 5x - 2
$$
as
$$
(3x + 1)(x - 2),
$$
check:

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

Since the multiplication gives the original quadratic, the factoring is correct.

If the multiplication does not match, go back and look for:

• a sign error (very common),
• a mistake in choosing numbers with the correct product and sum,
• a forgotten GCF.

Factoring Quadratic Equations vs. Expressions

In this chapter, the focus is on factoring expressions like $2x^2 + 7x + 3$.

To solve a quadratic equation such as
$$
2x^2 + 7x + 3 = 0,
$$
you would take the additional step:

1. Factor the left-hand side:
   $$
   2x^2 + 7x + 3 = (2x + 1)(x + 3).
   $$
2. Use the fact that a product equals zero only when at least one factor is zero:
   $$
   2x + 1 = 0 \quad \text{or} \quad x + 3 = 0.
   $$

The detailed process of solving quadratic equations is handled in the chapters on quadratic equations; here, we only emphasize that accurate factoring is a key step in that process.