Factoring

Study Sheet

Factoring

Turning sums into products --- the reverse of multiplying

What Factoring Is (and Why It Helps)

Concept
The Big Idea

Factoring means writing an expression as a product (things multiplied together). It is the reverse of distributing/multiplying.

3(x+2)⏟factored form → multiply  3x+6⏟expanded form → factor  3(x+2)⏟factored form\underbrace{3(x+2)}_{\text{factored form}} \ \xrightarrow{\ \text{multiply}\ }\ \underbrace{3x+6}_{\text{expanded form}} \ \xrightarrow{\ \text{factor}\ }\ \underbrace{3(x+2)}_{\text{factored form}}

When we distribute we go forward; when we factor we go backward.

Why it is useful: a factored form shows the hidden building blocks of an expression. It lets us simplify fractions, and --- most importantly --- it lets us solve equations that equal zero (see the last section). Factoring is the key that unlocks quadratics.

Tip

Always check your factoring by multiplying back out. If the product equals what you started with, you are correct. This one habit catches almost every mistake.

Greatest Common Factor (GCF)

Concept
Method: Factor Out the GCF

The GCF of several terms is the largest factor they all share.

  • [leftmargin=*]
  • Numbers: take the largest number that divides every coefficient.
  • Variables: take each shared variable to its smallest exponent that appears.
  • Write GCF×(what is left in each term)\text{GCF}\times(\text{what is left in each term}).

This uses the distributive property in reverse: ab+ac=a(b+c)ab+ac=a(b+c).

Example
Example: Factor 12x3+18x212x^3+18x^2

Coefficients 1212 and 1818 share GCF 66. Both terms have xx; the smaller power is x2x^2. So the GCF is 6x26x^2.

12x3+18x2=6x2(2x)+6x2(3)=6x2(2x+3).\begin{aligned} 12x^3+18x^2 &= 6x^2(2x)+6x^2(3)\\ &= 6x^2(2x+3). \end{aligned}

Check: 6x2(2x+3)=12x3+18x26x^2(2x+3)=12x^3+18x^2. ✓

Example
Example: Factor 15a2b−20ab2+5ab15a^2b-20ab^2+5ab

Numbers 15,20,515,20,5 share GCF 55. Every term has at least one aa and one bb, so pull 5ab5ab.

15a2b−20ab2+5ab=5ab(3a−4b+1).15a^2b-20ab^2+5ab = 5ab(3a-4b+1).

Check: 5ab(3a−4b+1)=15a2b−20ab2+5ab5ab(3a-4b+1)=15a^2b-20ab^2+5ab. ✓ (Do not forget the +1+1!)

Tip

Tip: When one whole term is the GCF, a 11 is left behind in its place. Missing that 11 is the most common GCF error.

Factoring by Grouping (Four Terms)

Concept
Method

With four terms, group them in pairs, factor the GCF from each pair, and then factor out the matching binomial.

ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b).ax+ay+bx+by = a(x+y)+b(x+y)=(x+y)(a+b).
Example
Example: Factor x3+3x2+2x+6x^3+3x^2+2x+6
x3+3x2+2x+6=(x3+3x2)+(2x+6)=x2(x+3)+2(x+3)=(x+3)(x2+2).\begin{aligned} x^3+3x^2+2x+6 &= (x^3+3x^2)+(2x+6)\\ &= x^2(x+3)+2(x+3)\\ &= (x+3)(x^2+2). \end{aligned}

Check: (x+3)(x2+2)=x3+2x+3x2+6(x+3)(x^2+2)=x^3+2x+3x^2+6. ✓

Tip

Tip: The two parentheses must match after step two. If they do not, try grouping the terms in a different order, or factor out a negative so the signs line up.

Trinomials x2+bx+cx^2+bx+c (Leading Coefficient 11)

Concept
Method: Find Two Numbers

To factor x2+bx+cx^2+bx+c, find two numbers that

multiply to candadd to b.\textbf{multiply to } c \qquad\text{and}\qquad \textbf{add to } b.

If those numbers are pp and qq, then x2+bx+c=(x+p)(x+q)x^2+bx+c=(x+p)(x+q).

Example
Example: Factor x2+7x+12x^2+7x+12

Need two numbers with product 1212 and sum 77: those are 33 and 44.

x2+7x+12=(x+3)(x+4).x^2+7x+12=(x+3)(x+4).

Check: (x+3)(x+4)=x2+4x+3x+12=x2+7x+12(x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12. ✓

Example
Example: Factor x2−x−12x^2-x-12

Product −12-12, sum −1-1: those are −4-4 and 33.

x2−x−12=(x−4)(x+3).x^2-x-12=(x-4)(x+3).

Check: (x−4)(x+3)=x2+3x−4x−12=x2−x−12(x-4)(x+3)=x^2+3x-4x-12=x^2-x-12. ✓

Tip

Sign patterns: If c>0c>0, the two numbers share the sign of bb. If c<0c<0, they have opposite signs, and the bigger one carries the sign of bb.

Trinomials ax2+bx+cax^2+bx+c (Leading Coefficient ≠1\neq 1)

Concept
The “ac” / Grouping Method

For ax2+bx+cax^2+bx+c:

  • [leftmargin=*]
  • Multiply a⋅ca\cdot c.
  • Find two numbers that multiply to acac and add to bb.
  • Split the middle term bxbx using those two numbers.
  • Group and factor.
Example
Example: Factor 2x2+7x+32x^2+7x+3

Here ac=2⋅3=6ac=2\cdot 3=6. Two numbers multiplying to 66, adding to 77: 66 and 11.

2x2+7x+3=2x2+6x+1x+3=2x(x+3)+1(x+3)=(x+3)(2x+1).\begin{aligned} 2x^2+7x+3 &= 2x^2+6x+1x+3\\ &= 2x(x+3)+1(x+3)\\ &= (x+3)(2x+1). \end{aligned}

Check: (x+3)(2x+1)=2x2+x+6x+3=2x2+7x+3(x+3)(2x+1)=2x^2+x+6x+3=2x^2+7x+3. ✓

Example
Example: Factor 6x2−x−26x^2-x-2

ac=6⋅(−2)=−12ac=6\cdot(-2)=-12. Numbers multiplying to −12-12, adding to −1-1: −4-4 and 33.

6x2−x−2=6x2−4x+3x−2=2x(3x−2)+1(3x−2)=(3x−2)(2x+1).\begin{aligned} 6x^2-x-2 &= 6x^2-4x+3x-2\\ &= 2x(3x-2)+1(3x-2)\\ &= (3x-2)(2x+1). \end{aligned}

Check: (3x−2)(2x+1)=6x2+3x−4x−2=6x2−x−2(3x-2)(2x+1)=6x^2+3x-4x-2=6x^2-x-2. ✓

Difference of Squares

Concept
The Pattern
a2−b2=(a+b)(a−b).a^2-b^2=(a+b)(a-b).

A difference of two perfect squares always factors this way.

Example
Example: Factor x2−25x^2-25

Here x2=(x)2x^2=(x)^2 and 25=(5)225=(5)^2, so a=xa=x, b=5b=5:

x2−25=(x+5)(x−5).x^2-25=(x+5)(x-5).

Check: (x+5)(x−5)=x2−5x+5x−25=x2−25(x+5)(x-5)=x^2-5x+5x-25=x^2-25. ✓

Example
Example: Factor 9y2−169y^2-16

9y2=(3y)29y^2=(3y)^2 and 16=(4)216=(4)^2:

9y2−16=(3y+4)(3y−4).9y^2-16=(3y+4)(3y-4).

Check: (3y+4)(3y−4)=9y2−16(3y+4)(3y-4)=9y^2-16. ✓

Tip

Important: A sum of squares a2+b2a^2+b^2 does not factor over the real numbers. Only the difference does.

Perfect-Square Trinomials

Concept
The Patterns
a2+2ab+b2=(a+b)2,a2−2ab+b2=(a−b)2.\begin{aligned} a^2+2ab+b^2 &= (a+b)^2,\\ a^2-2ab+b^2 &= (a-b)^2. \end{aligned}

Spot these when the first and last terms are perfect squares and the middle term is twice the product of their roots.

Example
Example: Factor x2+6x+9x^2+6x+9

x2=(x)2x^2=(x)^2, 9=(3)29=(3)^2, and the middle term 6x=2⋅x⋅36x=2\cdot x\cdot 3. It fits!

x2+6x+9=(x+3)2.x^2+6x+9=(x+3)^2.

Check: (x+3)2=x2+6x+9(x+3)^2=x^2+6x+9. ✓

Example
Example: Factor 4x2−20x+254x^2-20x+25

4x2=(2x)24x^2=(2x)^2, 25=(5)225=(5)^2, middle −20x=−2⋅2x⋅5-20x=-2\cdot 2x\cdot 5.

4x2−20x+25=(2x−5)2.4x^2-20x+25=(2x-5)^2.

Check: (2x−5)2=4x2−20x+25(2x-5)^2=4x^2-20x+25. ✓

Factoring Completely

Concept
Method

Always pull out the GCF first, then keep factoring each piece until nothing else factors. An expression is factored completely when every factor is prime.

Example
Example: Factor completely 3x2−273x^2-27

First the GCF 33, then a difference of squares:

3x2−27=3(x2−9)=3(x+3)(x−3).\begin{aligned} 3x^2-27 &= 3(x^2-9)\\ &= 3(x+3)(x-3). \end{aligned}

Check: 3(x+3)(x−3)=3(x2−9)=3x2−273(x+3)(x-3)=3(x^2-9)=3x^2-27. ✓

Example
Example: Factor completely 2x3+10x2+12x2x^3+10x^2+12x

GCF is 2x2x, then factor the trinomial:

2x3+10x2+12x=2x(x2+5x+6)=2x(x+2)(x+3).\begin{aligned} 2x^3+10x^2+12x &= 2x(x^2+5x+6)\\ &= 2x(x+2)(x+3). \end{aligned}

Check: 2x(x+2)(x+3)=2x(x2+5x+6)=2x3+10x2+12x2x(x+2)(x+3)=2x(x^2+5x+6)=2x^3+10x^2+12x. ✓

Tip

Tip: After you finish, glance at every factor once more. Students often stop one step early --- a leftover x2−9x^2-9 still factors!

Zero Product Property (Solving by Factoring)

Concept
The Property

If a product equals zero, then at least one factor must be zero:

if A⋅B=0,then A=0 or B=0.\text{if } A\cdot B=0,\quad\text{then } A=0 \text{ or } B=0.

So to solve an equation: get one side =0=0, factor, then set each factor equal to zero.

Example
Example: Solve x2+2x−15=0x^2+2x-15=0
x2+2x−15=0(x+5)(x−3)=0x+5=0  or  x−3=0x=−5  or  x=3.\begin{aligned} x^2+2x-15 &= 0\\ (x+5)(x-3) &= 0\\ x+5=0 \ \ &\text{or}\ \ x-3=0\\ x=-5 \ \ &\text{or}\ \ x=3. \end{aligned}

Check: (−5)2+2(−5)−15=25−10−15=0(-5)^2+2(-5)-15=25-10-15=0. ✓

Tip

Tip: The equation must equal zero before you factor. (x+5)(x−3)=7(x+5)(x-3)=7 does not mean x+5=7x+5=7; move everything to one side first.

Going Deeper: Advanced Factoring

Concept
Sum and Difference of Cubes

Two more patterns join the difference of squares. Notice the sign rule: the binomial matches the sign in the middle, and the trinomial has the opposite middle sign and no doubling.

a3+b3=(a+b)(a2−ab+b2),a3−b3=(a−b)(a2+ab+b2).\begin{aligned} a^3+b^3 &= (a+b)(a^2-ab+b^2),\\ a^3-b^3 &= (a-b)(a^2+ab+b^2). \end{aligned}

A memory aid is SOAP: Same sign, Opposite sign, Always Positive (for the last term).

Example
Example: Factor 8x3+278x^3+27

Recognize the cubes: 8x3=(2x)38x^3=(2x)^3 and 27=(3)327=(3)^3, so a=2xa=2x, b=3b=3. This is a sum of cubes.

8x3+27=(2x)3+(3)3=(2x+3)((2x)2−(2x)(3)+32)=(2x+3)(4x2−6x+9).\begin{aligned} 8x^3+27 &= (2x)^3+(3)^3\\ &= (2x+3)\big((2x)^2-(2x)(3)+3^2\big)\\ &= (2x+3)(4x^2-6x+9). \end{aligned}

Check: (2x+3)(4x2−6x+9)=8x3−12x2+18x+12x2−18x+27=8x3+27(2x+3)(4x^2-6x+9)=8x^3-12x^2+18x+12x^2-18x+27=8x^3+27. ✓

Example
Example: Factor x3−64x^3-64

Here x3=(x)3x^3=(x)^3 and 64=(4)364=(4)^3, a difference of cubes with a=xa=x, b=4b=4.

x3−64=(x−4)(x2+4x+16).x^3-64=(x-4)(x^2+4x+16).

Check: (x−4)(x2+4x+16)=x3+4x2+16x−4x2−16x−64=x3−64(x-4)(x^2+4x+16)=x^3+4x^2+16x-4x^2-16x-64=x^3-64. ✓

Concept
Quartics in Quadratic Form

An expression like x4+Bx2+Cx^4+Bx^2+C is secretly a quadratic in disguise. Substitute u=x2u=x^2 (so u2=x4u^2=x^4), factor the ordinary quadratic in uu, then swap x2x^2 back in and keep factoring.

Example
Example: Factor completely x4−13x2+36x^4-13x^2+36

Let u=x2u=x^2. Then we need two numbers multiplying to 3636 and adding to −13-13: those are −4-4 and −9-9.

x4−13x2+36=u2−13u+36=(u−4)(u−9)=(x2−4)(x2−9)=(x−2)(x+2)(x−3)(x+3).\begin{aligned} x^4-13x^2+36 &= u^2-13u+36\\ &= (u-4)(u-9)\\ &= (x^2-4)(x^2-9)\\ &= (x-2)(x+2)(x-3)(x+3). \end{aligned}

Each x2−4x^2-4 and x2−9x^2-9 is itself a difference of squares --- do not stop early! Check: (x2−4)(x2−9)=x4−9x2−4x2+36=x4−13x2+36(x^2-4)(x^2-9)=x^4-9x^2-4x^2+36=x^4-13x^2+36. ✓

Concept
The Add-and-Subtract-a-Square Trick (Sophie Germain)

A sum of squares usually will not factor --- but sometimes we can create a difference of squares by adding and subtracting a middle term. For x4+4x^4+4, add and subtract 4x24x^2 to complete a perfect square:

x4+4=x4+4x2+4⏟(x2+2)2−4x2=(x2+2)2−(2x)2.x^4+4 = \underbrace{x^4+4x^2+4}_{(x^2+2)^2}-4x^2=(x^2+2)^2-(2x)^2.

Now it is a difference of squares! This is the Sophie Germain identity.

Example
Example: Factor x4+4x^4+4

Using the difference of squares A2−B2=(A+B)(A−B)A^2-B^2=(A+B)(A-B) with A=x2+2A=x^2+2 and B=2xB=2x:

x4+4=(x2+2)2−(2x)2=(x2+2+2x)(x2+2−2x)=(x2+2x+2)(x2−2x+2).\begin{aligned} x^4+4 &= (x^2+2)^2-(2x)^2\\ &= (x^2+2+2x)(x^2+2-2x)\\ &= (x^2+2x+2)(x^2-2x+2). \end{aligned}

Check: (x2+2x+2)(x2−2x+2)=x4−2x3+2x2+2x3−4x2+4x+2x2−4x+4=x4+4(x^2+2x+2)(x^2-2x+2)=x^4-2x^3+2x^2+2x^3-4x^2+4x+2x^2-4x+4=x^4+4. ✓

Concept
Factoring as a Calculator Shortcut

The difference of squares works on numbers, not just variables. It can turn a scary subtraction into an easy multiplication:

a2−b2=(a+b)(a−b).a^2-b^2=(a+b)(a-b).
Example
Example: Evaluate 20242−202322024^2-2023^2 without a calculator

Let a=2024a=2024 and b=2023b=2023:

20242−20232=(2024+2023)(2024−2023)=(4047)(1)=4047.2024^2-2023^2=(2024+2023)(2024-2023)=(4047)(1)=4047.

No large squaring needed --- the difference of consecutive squares is just their sum. ✓

Concept
Reading Off the Sum and Product of the Solutions

If a quadratic factors as x2+bx+c=(x−r)(x−s)x^2+bx+c=(x-r)(x-s), then multiplying out shows

x2+bx+c=x2−(r+s)x+rs.x^2+bx+c=x^2-(r+s)x+rs.

Matching coefficients gives Vieta's relations:

r+s=−bandrs=c.r+s=-b \qquad\text{and}\qquad rs=c.

So you can find the sum and product of the roots straight from the equation, without solving it.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}
Tip

Watch the signs. In (x−r)(x−s)(x-r)(x-s) the roots are rr and ss, but the coefficient is −(r+s)-(r+s). For x2−7x+12x^2-7x+12, the roots sum to +7+7 (not −7-7) and multiply to 1212; indeed x2−7x+12=(x−3)(x−4)x^2-7x+12=(x-3)(x-4) with 3+4=73+4=7 and 3⋅4=123\cdot 4=12.

Concept
Preview: The Factor Theorem

How do you factor something that is not a nice pattern, like x3−7x+6x^3-7x+6? The Factor Theorem (studied fully in Algebra 2) says:

(x−r) is a factor of a polynomial p(x)   ⟺   p(r)=0.(x-r)\text{ is a factor of a polynomial }p(x)\ \iff\ p(r)=0.

In words: if plugging in x=rx=r makes the whole polynomial equal zero, then (x−r)(x-r) divides it evenly. So you can test candidate roots to discover a factor.

Reminder — Remainder and Factor theorems:  P(x)=(x−c)Q(x)+P(c)⟹(x−c)∣P(x)  ⟺  P(c)=0P(x)=(x-c)Q(x)+P(c) \quad\Longrightarrow\quad (x-c)\mid P(x) \iff P(c)=0
Example
Example: Use the Factor Theorem to start factoring x3−7x+6x^3-7x+6

Test small integers. Try x=1x=1:

p(1)=13−7(1)+6=1−7+6=0.p(1)=1^3-7(1)+6=1-7+6=0.

Since p(1)=0p(1)=0, the theorem guarantees (x−1)(x-1) is a factor. Dividing out (x−1)(x-1) leaves x2+x−6x^2+x-6, which factors normally:

x3−7x+6=(x−1)(x2+x−6)=(x−1)(x+3)(x−2).x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x+3)(x-2).

Check: (x−1)(x+3)(x−2)=(x−1)(x2+x−6)=x3+x2−6x−x2−x+6=x3−7x+6(x-1)(x+3)(x-2)=(x-1)(x^2+x-6)=x^3+x^2-6x-x^2-x+6=x^3-7x+6. ✓

Tip

Where to look for roots: for a polynomial with integer coefficients, any integer root must divide the constant term. For x3−7x+6x^3-7x+6 the constant is 66, so only ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6 are worth testing. This is a first taste of the Rational Root Theorem.

Formulas, Proofs & Tips

Tip
The factoring patterns
a2−b2=(a−b)(a+b),a2±2ab+b2=(a±b)2,a3±b3=(a±b)(a2∓ab+b2)a^2-b^2=(a-b)(a+b),\quad a^2\pm 2ab+b^2=(a\pm b)^2,\quad a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)

What it means. Three shapes worth recognising instantly.

Example. x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3) and x2+6x+9=(x+3)2x^2+6x+9=(x+3)^2.

Why it works. Each is verified by expanding. For the difference of squares, (a−b)(a+b)=a2+ab−ab−b2=a2−b2(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2 — the middle terms cancel. For cubes, expanding (a−b)(a2+ab+b2)(a-b)(a^2+ab+b^2) cancels everything except a3−b3a^3-b^3.

Tip. There is no factorisation of a2+b2a^2+b^2 over the real numbers. Always pull out the greatest common factor before hunting for a pattern.