Common Factoring

Factoring is expanding in reverse: instead of multiplying brackets out, you write an expression as a product. The first thing to look for, every single time, is a common factor that every term shares. It’s the simplest kind of factoring, and it’s also the first step of every other kind you’ll learn in this unit.

Key ideas

Factoring undoes expanding

When you expand, you multiply a factor into a bracket. When you factor, you pull it back out:

$$\underbrace{3x(x + 4)}_{\text{factored}} \quad \xrightarrow{\ \text{expand}\ } \quad \underbrace{3x^2 + 12x}_{\text{expanded}} \quad \xrightarrow{\ \text{factor}\ } \quad 3x(x + 4)$$

If you need a refresher on expanding, see polynomial operations.

The greatest common factor

The greatest common factor (GCF) of some terms is the largest expression that divides into every one of them.

• Numbers: find the largest number that divides all the coefficients. The GCF of $12$, $18$ and $30$ is $6$.
• Variables: a variable is in the GCF only if it’s in every term, and it gets the lowest exponent that appears. The GCF of $x^5$, $x^3$ and $x^2$ is $x^2$.

So the GCF of $12x^5$, $18x^3$ and $30x^2$ is $6x^2$.

Taking out the GCF

1. Find the GCF of all the terms.
2. Divide each term by the GCF. The results go inside the bracket.
3. Write the answer as GCF $\times$ (bracket).
4. Check by expanding.

$$12x^5 + 18x^3 - 30x^2 = 6x^2(2x^3 + 3x - 5)$$

because $\dfrac{12x^5}{6x^2} = 2x^3$, $\dfrac{18x^3}{6x^2} = 3x$, and $\dfrac{-30x^2}{6x^2} = -5$.

If a term is the GCF, it leaves a $1$ behind, not nothing: $4x^2 + 4x = 4x(x + 1)$.

When the first term is negative, it’s common to take out a negative GCF so the bracket starts with a positive term. Every sign inside flips: $-6x^2 + 9x = -3x(2x - 3)$.

A common binomial factor

The common factor doesn’t have to be a single term. In

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

both terms contain the bracket $(x - 2)$. Treat the whole bracket like one thing and take it out: the leftovers $5x$ and $3$ go together in a second bracket.

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

Watch for brackets that are opposites, like $(x - 2)$ and $(2 - x)$. Since $2 - x = -(x - 2)$, you can rewrite one to match the other.

Factoring by grouping

A four-term expression with no common factor in all four terms can often be factored by grouping:

1. Group the terms in pairs.
2. Take out the GCF of each pair.
3. If the two brackets match, take out that common bracket.

$$\begin{aligned} 3m + 3n + km + kn &= (3m + 3n) + (km + kn) \\ &= 3(m + n) + k(m + n) \\ &= (m + n)(3 + k) \end{aligned}$$

If the brackets don’t match, try a different pairing (or take out a negative from the second pair).

On the SAT

Desmos can’t factor, so on the SAT common factoring is a by-hand skill, and it’s quick once you spot the GCF. The fastest check is to expand your answer in your head and compare. If you want Desmos to confirm, graph the original and your factored form (with $x$ as the variable): if the graphs overlap exactly, they’re equivalent. See using Desmos on the SAT.

Worked examples

Example 1: A single common factor

Factor.

• (a) $6x + 15$
• (b) $8x^2 - 12x$

Solution.

(a) The GCF of $6$ and $15$ is $3$, and $x$ isn’t in both terms.

$$6x + 15 = 3(2x + 5)$$

(b) The GCF of $8$ and $12$ is $4$, and both terms have at least one $x$. So the GCF is $4x$.

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

Check: $4x(2x - 3) = 8x^2 - 12x$. ✓

Example 2: Three terms, two variables

Factor $10a^3b^2 - 15a^2b + 5ab$.

Solution. Find the GCF piece by piece:

• numbers: the GCF of $10$, $15$ and $5$ is $5$
• $a$: the exponents are $3$, $2$ and $1$, so take $a^1 = a$
• $b$: the exponents are $2$, $1$ and $1$, so take $b$

The GCF is $5ab$. Divide each term by it:

$$\frac{10a^3b^2}{5ab} = 2a^2b \qquad \frac{-15a^2b}{5ab} = -3a \qquad \frac{5ab}{5ab} = 1$$ $$10a^3b^2 - 15a^2b + 5ab = 5ab(2a^2b - 3a + 1)$$

Check: $5ab(2a^2b) - 5ab(3a) + 5ab(1) = 10a^3b^2 - 15a^2b + 5ab$. ✓

Example 3: Common binomial factors

Factor.

• (a) $4x(x + 3) - 7(x + 3)$
• (b) $2y(y - 5) + 3(5 - y)$

Solution.

(a) Both terms contain $(x + 3)$. Take it out, and the leftovers $4x$ and $-7$ form the other factor:

$$4x(x + 3) - 7(x + 3) = (x + 3)(4x - 7)$$

(b) The brackets $(y - 5)$ and $(5 - y)$ are opposites. Rewrite $3(5 - y)$ as $-3(y - 5)$:

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

Check (b) with $y = 1$: the original is $2(1)(-4) + 3(4) = -8 + 12 = 4$, and $(1 - 5)(2 - 3) = (-4)(-1) = 4$. ✓

Example 4: Factoring by grouping

Factor.

• (a) $x^3 - 2x^2 + 5x - 10$
• (b) $6mn - 3m - 4n + 2$

Solution.

(a) Group the first two terms and the last two terms:

$$\begin{aligned} x^3 - 2x^2 + 5x - 10 &= (x^3 - 2x^2) + (5x - 10) \\ &= x^2(x - 2) + 5(x - 2) \\ &= (x - 2)(x^2 + 5) \end{aligned}$$

(b) From the first pair, take out $3m$. From the second pair, take out $-2$ (not $2$), so the brackets match:

$$\begin{aligned} 6mn - 3m - 4n + 2 &= 3m(2n - 1) - 2(2n - 1) && -2(2n - 1) = -4n + 2 \\ &= (2n - 1)(3m - 2) \end{aligned}$$

Check (b): $(2n - 1)(3m - 2) = 6mn - 4n - 3m + 2$. ✓

Common mistakes

Not taking out the greatest common factor. $8x^2 - 12x = 2x(4x - 6)$ is true, but it isn’t fully factored, because $4x - 6$ still has a common factor of $2$. Always check the bracket: if its terms still share a factor, you didn’t take out the GCF.

Leaving out the 1. $5x^2 + 5x$ is $5x(x + 1)$, not $5x(x)$. When a term equals the GCF, dividing leaves $1$. Expanding $5x(x)$ gives only $5x^2$, which shows something’s missing.

Sign errors with a negative factor. When you take out $-3x$ from $-6x^2 + 9x$, every term inside changes sign: $-3x(2x - 3)$. Check by expanding: $-3x(2x) = -6x^2$ and $-3x(-3) = 9x$. ✓

Grouping with the wrong sign. In $6mn - 3m - 4n + 2$, taking out $+2$ from the second pair gives $2(-2n + 1)$, which doesn’t match $(2n - 1)$. Take out $-2$ instead to get $-2(2n - 1)$.

Treating opposite brackets as the same. $(x - 3)$ and $(3 - x)$ are not equal: $3 - x = -(x - 3)$. Rewrite one of them, with the sign change, before taking out the common bracket.

Skipping the check. Factoring is easy to check: expand your answer and compare. It takes ten seconds and catches almost every error.

Practice

1. (Warm-up) Find the greatest common factor.

• (a) $18$ and $24$
• (b) $12x^3$ and $20x^2$
• (c) $9a^2b$ and $15ab^3$

Solution

(a) $6$

(b) The GCF of $12$ and $20$ is $4$, and the lower power of $x$ is $x^2$: the GCF is $4x^2$.

(c) The GCF of $9$ and $15$ is $3$; the lowest powers are $a$ and $b$: the GCF is $3ab$.

2. (Warm-up) Factor.

• (a) $7x - 21$
• (b) $3x^2 + 9x$
• (c) $5y^2 - 5y$

Solution

(a) $7(x - 3)$

(b) $3x(x + 3)$

(c) $5y(y - 1)$. The second term is the GCF itself, so it leaves $1$.

3. (Warm-up) Factor $12m^2 + 8m + 4$.

Solution

The GCF is $4$ ($m$ is not in the last term):

$$12m^2 + 8m + 4 = 4(3m^2 + 2m + 1)$$

4. (Core) Factor $-6x^2 + 9x$ by taking out a negative common factor.

Solution

Take out $-3x$, which flips the sign of every term:

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

Check: $-3x(2x) + (-3x)(-3) = -6x^2 + 9x$. ✓ (Taking out $3x$ instead gives $3x(-2x + 3)$, which is also correct.)

5. (Core) Factor $14x^3y^2 - 21x^2y^3 + 7x^2y^2$.

Solution

The GCF is $7x^2y^2$:

$$14x^3y^2 - 21x^2y^3 + 7x^2y^2 = 7x^2y^2(2x - 3y + 1)$$

Check: $7x^2y^2(2x) = 14x^3y^2$, $7x^2y^2(-3y) = -21x^2y^3$, and $7x^2y^2(1) = 7x^2y^2$. ✓

6. (Core) Factor.

• (a) $5x(x - 4) + 3(x - 4)$
• (b) $3a(2a + 1) - (2a + 1)$
• (c) $2x(x - 3) + 5(3 - x)$

Solution

(a) $(x - 4)(5x + 3)$

(b) The second term is $-1(2a + 1)$, so the leftovers are $3a$ and $-1$:

$$3a(2a + 1) - (2a + 1) = (2a + 1)(3a - 1)$$

(c) Rewrite $5(3 - x)$ as $-5(x - 3)$:

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

7. (Core) Factor by grouping.

• (a) $xy + 3x + 2y + 6$
• (b) $10ab - 15a - 4b + 6$

Solution

(a)

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

(b) Take out $-2$ from the second pair so the brackets match:

$$10ab - 15a - 4b + 6 = 5a(2b - 3) - 2(2b - 3) = (2b - 3)(5a - 2)$$

Check (b): $(2b - 3)(5a - 2) = 10ab - 4b - 15a + 6$. ✓

8. (Challenge) A rectangular garden has an area of $(4x^2 + 10x)$ m². One side is $2x$ m long.

• (a) Find an expression for the other side.
• (b) Find both dimensions and the area when $x = 3$.

Solution

(a) Factor out $2x$: $4x^2 + 10x = 2x(2x + 5)$. The other side is $(2x + 5)$ m.

(b) When $x = 3$, the sides are $2(3) = 6$ m and $2(3) + 5 = 11$ m. The area is $6 \times 11 = 66$ m².

Check with the original: $4(3)^2 + 10(3) = 36 + 30 = 66$. ✓

9. (Challenge) Factor each by grouping. In (b), you’ll need to rearrange the terms first.

• (a) $x^3 + 4x^2 - 3x - 12$
• (b) $3xy - 8 + 6x - 4y$

Solution

(a)

$$\begin{aligned} x^3 + 4x^2 - 3x - 12 &= x^2(x + 4) - 3(x + 4) \\ &= (x + 4)(x^2 - 3) \end{aligned}$$

$x^2 - 3$ can’t be factored any further using integers.

(b) As written, the first pair $3xy - 8$ has no common factor. Rearrange so each pair shares something:

$$\begin{aligned} 3xy - 8 + 6x - 4y &= 3xy + 6x - 4y - 8 \\ &= 3x(y + 2) - 4(y + 2) \\ &= (y + 2)(3x - 4) \end{aligned}$$

Check: $(y + 2)(3x - 4) = 3xy - 4y + 6x - 8$. ✓