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.
When you expand, you multiply a factor into a bracket. When you factor, you pull it back out:
3 x ( x + 4 ) ⏟ factored → expand 3 x 2 + 12 x ⏟ expanded → factor 3 x ( x + 4 ) \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) factored 3 x ( x + 4 ) expand expanded 3 x 2 + 12 x factor 3 x ( x + 4 ) If you need a refresher on expanding, see polynomial operations .
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 12 12 18 18 18 30 30 30 6 6 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^5 x 5 x 3 x^3 x 3 x 2 x^2 x 2 x 2 x^2 x 2
So the GCF of 12 x 5 12x^5 12 x 5 18 x 3 18x^3 18 x 3 30 x 2 30x^2 30 x 2 6 x 2 6x^2 6 x 2
Find the GCF of all the terms.
Divide each term by the GCF. The results go inside the bracket.
Write the answer as GCF × \times ×
Check by expanding.
12 x 5 + 18 x 3 − 30 x 2 = 6 x 2 ( 2 x 3 + 3 x − 5 ) 12x^5 + 18x^3 - 30x^2 = 6x^2(2x^3 + 3x - 5) 12 x 5 + 18 x 3 − 30 x 2 = 6 x 2 ( 2 x 3 + 3 x − 5 ) because 12 x 5 6 x 2 = 2 x 3 \dfrac{12x^5}{6x^2} = 2x^3 6 x 2 12 x 5 = 2 x 3 18 x 3 6 x 2 = 3 x \dfrac{18x^3}{6x^2} = 3x 6 x 2 18 x 3 = 3 x − 30 x 2 6 x 2 = − 5 \dfrac{-30x^2}{6x^2} = -5 6 x 2 − 30 x 2 = − 5
If a term is the GCF, it leaves a 1 1 1 4 x 2 + 4 x = 4 x ( x + 1 ) 4x^2 + 4x = 4x(x + 1) 4 x 2 + 4 x = 4 x ( 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: − 6 x 2 + 9 x = − 3 x ( 2 x − 3 ) -6x^2 + 9x = -3x(2x - 3) − 6 x 2 + 9 x = − 3 x ( 2 x − 3 )
The common factor doesn’t have to be a single term. In
5 x ( x − 2 ) + 3 ( x − 2 ) 5x(x - 2) + 3(x - 2) 5 x ( x − 2 ) + 3 ( x − 2 ) both terms contain the bracket ( x − 2 ) (x - 2) ( x − 2 ) 5 x 5x 5 x 3 3 3
5 x ( x − 2 ) + 3 ( x − 2 ) = ( x − 2 ) ( 5 x + 3 ) 5x(x - 2) + 3(x - 2) = (x - 2)(5x + 3) 5 x ( x − 2 ) + 3 ( x − 2 ) = ( x − 2 ) ( 5 x + 3 ) Watch for brackets that are opposites , like ( x − 2 ) (x - 2) ( x − 2 ) ( 2 − x ) (2 - x) ( 2 − x ) 2 − x = − ( x − 2 ) 2 - x = -(x - 2) 2 − x = − ( x − 2 )
A four-term expression with no common factor in all four terms can often be factored by grouping :
Group the terms in pairs.
Take out the GCF of each pair.
If the two brackets match, take out that common bracket.
3 m + 3 n + k m + k n = ( 3 m + 3 n ) + ( k m + k n ) = 3 ( m + n ) + k ( m + n ) = ( m + n ) ( 3 + k ) \begin{aligned}
3m + 3n + km + kn &= (3m + 3n) + (km + kn) \\
&= 3(m + n) + k(m + n) \\
&= (m + n)(3 + k)
\end{aligned} 3 m + 3 n + k m + k n = ( 3 m + 3 n ) + ( k m + k n ) = 3 ( m + n ) + k ( m + n ) = ( m + n ) ( 3 + k ) If the brackets don’t match, try a different pairing (or take out a negative from the second pair).
Factor.
(a) 6 x + 15 6x + 15 6 x + 15
(b) 8 x 2 − 12 x 8x^2 - 12x 8 x 2 − 12 x
Solution.
(a) The GCF of 6 6 6 15 15 15 3 3 3 x x x
6 x + 15 = 3 ( 2 x + 5 ) 6x + 15 = 3(2x + 5) 6 x + 15 = 3 ( 2 x + 5 ) (b) The GCF of 8 8 8 12 12 12 4 4 4 x x x 4 x 4x 4 x
8 x 2 − 12 x = 4 x ( 2 x − 3 ) 8x^2 - 12x = 4x(2x - 3) 8 x 2 − 12 x = 4 x ( 2 x − 3 ) Check: 4 x ( 2 x − 3 ) = 8 x 2 − 12 x 4x(2x - 3) = 8x^2 - 12x 4 x ( 2 x − 3 ) = 8 x 2 − 12 x
Factor 10 a 3 b 2 − 15 a 2 b + 5 a b 10a^3b^2 - 15a^2b + 5ab 10 a 3 b 2 − 15 a 2 b + 5 ab
Solution. Find the GCF piece by piece:
numbers: the GCF of 10 10 10 15 15 15 5 5 5 5 5 5
a a a 3 3 3 2 2 2 1 1 1 a 1 = a a^1 = a a 1 = a b b b 2 2 2 1 1 1 1 1 1 b b b
The GCF is 5 a b 5ab 5 ab
10 a 3 b 2 5 a b = 2 a 2 b − 15 a 2 b 5 a b = − 3 a 5 a b 5 a b = 1 \frac{10a^3b^2}{5ab} = 2a^2b \qquad \frac{-15a^2b}{5ab} = -3a \qquad \frac{5ab}{5ab} = 1 5 ab 10 a 3 b 2 = 2 a 2 b 5 ab − 15 a 2 b = − 3 a 5 ab 5 ab = 1 10 a 3 b 2 − 15 a 2 b + 5 a b = 5 a b ( 2 a 2 b − 3 a + 1 ) 10a^3b^2 - 15a^2b + 5ab = 5ab(2a^2b - 3a + 1) 10 a 3 b 2 − 15 a 2 b + 5 ab = 5 ab ( 2 a 2 b − 3 a + 1 ) Check: 5 a b ( 2 a 2 b ) − 5 a b ( 3 a ) + 5 a b ( 1 ) = 10 a 3 b 2 − 15 a 2 b + 5 a b 5ab(2a^2b) - 5ab(3a) + 5ab(1) = 10a^3b^2 - 15a^2b + 5ab 5 ab ( 2 a 2 b ) − 5 ab ( 3 a ) + 5 ab ( 1 ) = 10 a 3 b 2 − 15 a 2 b + 5 ab
Factor.
(a) 4 x ( x + 3 ) − 7 ( x + 3 ) 4x(x + 3) - 7(x + 3) 4 x ( x + 3 ) − 7 ( x + 3 )
(b) 2 y ( y − 5 ) + 3 ( 5 − y ) 2y(y - 5) + 3(5 - y) 2 y ( y − 5 ) + 3 ( 5 − y )
Solution.
(a) Both terms contain ( x + 3 ) (x + 3) ( x + 3 ) 4 x 4x 4 x − 7 -7 − 7
4 x ( x + 3 ) − 7 ( x + 3 ) = ( x + 3 ) ( 4 x − 7 ) 4x(x + 3) - 7(x + 3) = (x + 3)(4x - 7) 4 x ( x + 3 ) − 7 ( x + 3 ) = ( x + 3 ) ( 4 x − 7 ) (b) The brackets ( y − 5 ) (y - 5) ( y − 5 ) ( 5 − y ) (5 - y) ( 5 − y ) 3 ( 5 − y ) 3(5 - y) 3 ( 5 − y ) − 3 ( y − 5 ) -3(y - 5) − 3 ( y − 5 )
2 y ( y − 5 ) + 3 ( 5 − y ) = 2 y ( y − 5 ) − 3 ( y − 5 ) = ( y − 5 ) ( 2 y − 3 ) \begin{aligned}
2y(y - 5) + 3(5 - y) &= 2y(y - 5) - 3(y - 5) \\
&= (y - 5)(2y - 3)
\end{aligned} 2 y ( y − 5 ) + 3 ( 5 − y ) = 2 y ( y − 5 ) − 3 ( y − 5 ) = ( y − 5 ) ( 2 y − 3 ) Check (b) with y = 1 y = 1 y = 1 2 ( 1 ) ( − 4 ) + 3 ( 4 ) = − 8 + 12 = 4 2(1)(-4) + 3(4) = -8 + 12 = 4 2 ( 1 ) ( − 4 ) + 3 ( 4 ) = − 8 + 12 = 4 ( 1 − 5 ) ( 2 − 3 ) = ( − 4 ) ( − 1 ) = 4 (1 - 5)(2 - 3) = (-4)(-1) = 4 ( 1 − 5 ) ( 2 − 3 ) = ( − 4 ) ( − 1 ) = 4
Factor.
(a) x 3 − 2 x 2 + 5 x − 10 x^3 - 2x^2 + 5x - 10 x 3 − 2 x 2 + 5 x − 10
(b) 6 m n − 3 m − 4 n + 2 6mn - 3m - 4n + 2 6 mn − 3 m − 4 n + 2
Solution.
(a) Group the first two terms and the last two terms:
x 3 − 2 x 2 + 5 x − 10 = ( x 3 − 2 x 2 ) + ( 5 x − 10 ) = x 2 ( x − 2 ) + 5 ( x − 2 ) = ( x − 2 ) ( x 2 + 5 ) \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} x 3 − 2 x 2 + 5 x − 10 = ( x 3 − 2 x 2 ) + ( 5 x − 10 ) = x 2 ( x − 2 ) + 5 ( x − 2 ) = ( x − 2 ) ( x 2 + 5 ) (b) From the first pair, take out 3 m 3m 3 m − 2 -2 − 2 2 2 2
6 m n − 3 m − 4 n + 2 = 3 m ( 2 n − 1 ) − 2 ( 2 n − 1 ) − 2 ( 2 n − 1 ) = − 4 n + 2 = ( 2 n − 1 ) ( 3 m − 2 ) \begin{aligned}
6mn - 3m - 4n + 2 &= 3m(2n - 1) - 2(2n - 1) && -2(2n - 1) = -4n + 2 \\
&= (2n - 1)(3m - 2)
\end{aligned} 6 mn − 3 m − 4 n + 2 = 3 m ( 2 n − 1 ) − 2 ( 2 n − 1 ) = ( 2 n − 1 ) ( 3 m − 2 ) − 2 ( 2 n − 1 ) = − 4 n + 2 Check (b): ( 2 n − 1 ) ( 3 m − 2 ) = 6 m n − 4 n − 3 m + 2 (2n - 1)(3m - 2) = 6mn - 4n - 3m + 2 ( 2 n − 1 ) ( 3 m − 2 ) = 6 mn − 4 n − 3 m + 2
Not taking out the greatest common factor. 8 x 2 − 12 x = 2 x ( 4 x − 6 ) 8x^2 - 12x = 2x(4x - 6) 8 x 2 − 12 x = 2 x ( 4 x − 6 ) 4 x − 6 4x - 6 4 x − 6 2 2 2
Leaving out the 1. 5 x 2 + 5 x 5x^2 + 5x 5 x 2 + 5 x 5 x ( x + 1 ) 5x(x + 1) 5 x ( x + 1 ) 5 x ( x ) 5x(x) 5 x ( x ) 1 1 1 5 x ( x ) 5x(x) 5 x ( x ) 5 x 2 5x^2 5 x 2
Sign errors with a negative factor. When you take out − 3 x -3x − 3 x − 6 x 2 + 9 x -6x^2 + 9x − 6 x 2 + 9 x every term inside changes sign: − 3 x ( 2 x − 3 ) -3x(2x - 3) − 3 x ( 2 x − 3 ) − 3 x ( 2 x ) = − 6 x 2 -3x(2x) = -6x^2 − 3 x ( 2 x ) = − 6 x 2 − 3 x ( − 3 ) = 9 x -3x(-3) = 9x − 3 x ( − 3 ) = 9 x
Grouping with the wrong sign. In 6 m n − 3 m − 4 n + 2 6mn - 3m - 4n + 2 6 mn − 3 m − 4 n + 2 + 2 +2 + 2 2 ( − 2 n + 1 ) 2(-2n + 1) 2 ( − 2 n + 1 ) ( 2 n − 1 ) (2n - 1) ( 2 n − 1 ) − 2 -2 − 2 − 2 ( 2 n − 1 ) -2(2n - 1) − 2 ( 2 n − 1 )
Treating opposite brackets as the same. ( x − 3 ) (x - 3) ( x − 3 ) ( 3 − x ) (3 - x) ( 3 − x ) 3 − x = − ( x − 3 ) 3 - x = -(x - 3) 3 − x = − ( x − 3 )
Skipping the check. Factoring is easy to check: expand your answer and compare. It takes ten seconds and catches almost every error.
1. (Warm-up) Find the greatest common factor.
(a) 18 18 18 24 24 24
(b) 12 x 3 12x^3 12 x 3 20 x 2 20x^2 20 x 2
(c) 9 a 2 b 9a^2b 9 a 2 b 15 a b 3 15ab^3 15 a b 3
Solution (a) 6 6 6
(b) The GCF of 12 12 12 20 20 20 4 4 4 x x x x 2 x^2 x 2 4 x 2 4x^2 4 x 2
(c) The GCF of 9 9 9 15 15 15 3 3 3 a a a b b b 3 a b 3ab 3 ab
2. (Warm-up) Factor.
(a) 7 x − 21 7x - 21 7 x − 21
(b) 3 x 2 + 9 x 3x^2 + 9x 3 x 2 + 9 x
(c) 5 y 2 − 5 y 5y^2 - 5y 5 y 2 − 5 y
Solution (a) 7 ( x − 3 ) 7(x - 3) 7 ( x − 3 )
(b) 3 x ( x + 3 ) 3x(x + 3) 3 x ( x + 3 )
(c) 5 y ( y − 1 ) 5y(y - 1) 5 y ( y − 1 ) 1 1 1
3. (Warm-up) Factor 12 m 2 + 8 m + 4 12m^2 + 8m + 4 12 m 2 + 8 m + 4
Solution The GCF is 4 4 4 m m m
12 m 2 + 8 m + 4 = 4 ( 3 m 2 + 2 m + 1 ) 12m^2 + 8m + 4 = 4(3m^2 + 2m + 1) 12 m 2 + 8 m + 4 = 4 ( 3 m 2 + 2 m + 1 ) 4. (Core) Factor − 6 x 2 + 9 x -6x^2 + 9x − 6 x 2 + 9 x
Solution Take out − 3 x -3x − 3 x
− 6 x 2 + 9 x = − 3 x ( 2 x − 3 ) -6x^2 + 9x = -3x(2x - 3) − 6 x 2 + 9 x = − 3 x ( 2 x − 3 ) Check: − 3 x ( 2 x ) + ( − 3 x ) ( − 3 ) = − 6 x 2 + 9 x -3x(2x) + (-3x)(-3) = -6x^2 + 9x − 3 x ( 2 x ) + ( − 3 x ) ( − 3 ) = − 6 x 2 + 9 x 3 x 3x 3 x 3 x ( − 2 x + 3 ) 3x(-2x + 3) 3 x ( − 2 x + 3 )
5. (Core) Factor 14 x 3 y 2 − 21 x 2 y 3 + 7 x 2 y 2 14x^3y^2 - 21x^2y^3 + 7x^2y^2 14 x 3 y 2 − 21 x 2 y 3 + 7 x 2 y 2
Solution The GCF is 7 x 2 y 2 7x^2y^2 7 x 2 y 2
14 x 3 y 2 − 21 x 2 y 3 + 7 x 2 y 2 = 7 x 2 y 2 ( 2 x − 3 y + 1 ) 14x^3y^2 - 21x^2y^3 + 7x^2y^2 = 7x^2y^2(2x - 3y + 1) 14 x 3 y 2 − 21 x 2 y 3 + 7 x 2 y 2 = 7 x 2 y 2 ( 2 x − 3 y + 1 ) Check: 7 x 2 y 2 ( 2 x ) = 14 x 3 y 2 7x^2y^2(2x) = 14x^3y^2 7 x 2 y 2 ( 2 x ) = 14 x 3 y 2 7 x 2 y 2 ( − 3 y ) = − 21 x 2 y 3 7x^2y^2(-3y) = -21x^2y^3 7 x 2 y 2 ( − 3 y ) = − 21 x 2 y 3 7 x 2 y 2 ( 1 ) = 7 x 2 y 2 7x^2y^2(1) = 7x^2y^2 7 x 2 y 2 ( 1 ) = 7 x 2 y 2
6. (Core) Factor.
(a) 5 x ( x − 4 ) + 3 ( x − 4 ) 5x(x - 4) + 3(x - 4) 5 x ( x − 4 ) + 3 ( x − 4 )
(b) 3 a ( 2 a + 1 ) − ( 2 a + 1 ) 3a(2a + 1) - (2a + 1) 3 a ( 2 a + 1 ) − ( 2 a + 1 )
(c) 2 x ( x − 3 ) + 5 ( 3 − x ) 2x(x - 3) + 5(3 - x) 2 x ( x − 3 ) + 5 ( 3 − x )
Solution (a) ( x − 4 ) ( 5 x + 3 ) (x - 4)(5x + 3) ( x − 4 ) ( 5 x + 3 )
(b) The second term is − 1 ( 2 a + 1 ) -1(2a + 1) − 1 ( 2 a + 1 ) 3 a 3a 3 a − 1 -1 − 1
3 a ( 2 a + 1 ) − ( 2 a + 1 ) = ( 2 a + 1 ) ( 3 a − 1 ) 3a(2a + 1) - (2a + 1) = (2a + 1)(3a - 1) 3 a ( 2 a + 1 ) − ( 2 a + 1 ) = ( 2 a + 1 ) ( 3 a − 1 ) (c) Rewrite 5 ( 3 − x ) 5(3 - x) 5 ( 3 − x ) − 5 ( x − 3 ) -5(x - 3) − 5 ( x − 3 )
2 x ( x − 3 ) + 5 ( 3 − x ) = 2 x ( x − 3 ) − 5 ( x − 3 ) = ( x − 3 ) ( 2 x − 5 ) 2x(x - 3) + 5(3 - x) = 2x(x - 3) - 5(x - 3) = (x - 3)(2x - 5) 2 x ( x − 3 ) + 5 ( 3 − x ) = 2 x ( x − 3 ) − 5 ( x − 3 ) = ( x − 3 ) ( 2 x − 5 ) 7. (Core) Factor by grouping.
(a) x y + 3 x + 2 y + 6 xy + 3x + 2y + 6 x y + 3 x + 2 y + 6
(b) 10 a b − 15 a − 4 b + 6 10ab - 15a - 4b + 6 10 ab − 15 a − 4 b + 6
Solution (a)
x y + 3 x + 2 y + 6 = x ( y + 3 ) + 2 ( y + 3 ) = ( y + 3 ) ( x + 2 ) xy + 3x + 2y + 6 = x(y + 3) + 2(y + 3) = (y + 3)(x + 2) x y + 3 x + 2 y + 6 = x ( y + 3 ) + 2 ( y + 3 ) = ( y + 3 ) ( x + 2 ) (b) Take out − 2 -2 − 2
10 a b − 15 a − 4 b + 6 = 5 a ( 2 b − 3 ) − 2 ( 2 b − 3 ) = ( 2 b − 3 ) ( 5 a − 2 ) 10ab - 15a - 4b + 6 = 5a(2b - 3) - 2(2b - 3) = (2b - 3)(5a - 2) 10 ab − 15 a − 4 b + 6 = 5 a ( 2 b − 3 ) − 2 ( 2 b − 3 ) = ( 2 b − 3 ) ( 5 a − 2 ) Check (b): ( 2 b − 3 ) ( 5 a − 2 ) = 10 a b − 4 b − 15 a + 6 (2b - 3)(5a - 2) = 10ab - 4b - 15a + 6 ( 2 b − 3 ) ( 5 a − 2 ) = 10 ab − 4 b − 15 a + 6
8. (Challenge) A rectangular garden has an area of ( 4 x 2 + 10 x ) (4x^2 + 10x) ( 4 x 2 + 10 x ) 2 x 2x 2 x
(a) Find an expression for the other side.
(b) Find both dimensions and the area when x = 3 x = 3 x = 3
Solution (a) Factor out 2 x 2x 2 x 4 x 2 + 10 x = 2 x ( 2 x + 5 ) 4x^2 + 10x = 2x(2x + 5) 4 x 2 + 10 x = 2 x ( 2 x + 5 ) ( 2 x + 5 ) (2x + 5) ( 2 x + 5 )
(b) When x = 3 x = 3 x = 3 2 ( 3 ) = 6 2(3) = 6 2 ( 3 ) = 6 2 ( 3 ) + 5 = 11 2(3) + 5 = 11 2 ( 3 ) + 5 = 11 6 × 11 = 66 6 \times 11 = 66 6 × 11 = 66
Check with the original: 4 ( 3 ) 2 + 10 ( 3 ) = 36 + 30 = 66 4(3)^2 + 10(3) = 36 + 30 = 66 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 + 4 x 2 − 3 x − 12 x^3 + 4x^2 - 3x - 12 x 3 + 4 x 2 − 3 x − 12
(b) 3 x y − 8 + 6 x − 4 y 3xy - 8 + 6x - 4y 3 x y − 8 + 6 x − 4 y
Solution (a)
x 3 + 4 x 2 − 3 x − 12 = x 2 ( x + 4 ) − 3 ( x + 4 ) = ( x + 4 ) ( x 2 − 3 ) \begin{aligned}
x^3 + 4x^2 - 3x - 12 &= x^2(x + 4) - 3(x + 4) \\
&= (x + 4)(x^2 - 3)
\end{aligned} x 3 + 4 x 2 − 3 x − 12 = x 2 ( x + 4 ) − 3 ( x + 4 ) = ( x + 4 ) ( x 2 − 3 ) x 2 − 3 x^2 - 3 x 2 − 3
(b) As written, the first pair 3 x y − 8 3xy - 8 3 x y − 8
3 x y − 8 + 6 x − 4 y = 3 x y + 6 x − 4 y − 8 = 3 x ( y + 2 ) − 4 ( y + 2 ) = ( y + 2 ) ( 3 x − 4 ) \begin{aligned}
3xy - 8 + 6x - 4y &= 3xy + 6x - 4y - 8 \\
&= 3x(y + 2) - 4(y + 2) \\
&= (y + 2)(3x - 4)
\end{aligned} 3 x y − 8 + 6 x − 4 y = 3 x y + 6 x − 4 y − 8 = 3 x ( y + 2 ) − 4 ( y + 2 ) = ( y + 2 ) ( 3 x − 4 ) Check: ( y + 2 ) ( 3 x − 4 ) = 3 x y − 4 y + 6 x − 8 (y + 2)(3x - 4) = 3xy - 4y + 6x - 8 ( y + 2 ) ( 3 x − 4 ) = 3 x y − 4 y + 6 x − 8
