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.

Factoring is multiplication run backwards: rewriting a polynomial as a product. Always pull out the greatest common factor first, then look for a pattern (difference of squares, perfect-square trinomial) or find two numbers with the right product and sum. Factoring matters because a product is zero exactly when a factor is zero — that is how quadratics get solved.

Factoring is worth the effort because products have a property sums lack: a product is zero only when a factor is zero. x2+2x−15=0x^2 + 2x - 15 = 0 is hard to read, but (x+5)(x−3)=0(x + 5)(x - 3) = 0 announces its solutions. Factored form also shows common pieces that cancel in fractions, the same way 1218\tfrac{12}{18} simplifies once 12=6⋅212 = 6 \cdot 2 and 18=6⋅318 = 6 \cdot 3.

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.

Side note
Where it goes wrong

Forgetting the GCF and then declaring 2x2+8x+82x^2 + 8x + 8 unfactorable, or "factoring" x2+4x^2 + 4 as (x+2)2(x + 2)^2. A sum of squares does not factor over the reals; check by expanding.

Try it
Try it: factor and check

Factor 8x+128x + 12 and x2−5xx^2 - 5x, and check each by multiplying back.

Show the answer

8x+12=4(2x+3)8x + 12 = 4(2x + 3); check 4⋅2x+4⋅3=8x+124 \cdot 2x + 4 \cdot 3 = 8x + 12. x2−5x=x(x−5)x^2 - 5x = x(x - 5); check x⋅x−5xx \cdot x - 5x.

Greatest Common Factor (GCF)

Concept
Method: Factor Out the GCF

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

  • 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).

Factoring out the GCF is the distributive property read right to left. The area picture makes it concrete: 12x3+18x212x^3 + 18x^2 is two rectangles sharing a height of 6x26x^2, with widths 2x2x and 33, so together they form one rectangle 6x26x^2 by (2x+3)(2x + 3). Taking the greatest common factor guarantees nothing common is left inside.

Example
Example: Factor 12x3+18x212x^3+18x^2
12x³18x²2x36x²6x²(2x + 3)
Two rectangles with the common height 6x26x^2 join into one.

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. ✓

Side note
Where it goes wrong

Factoring out a common factor that is not the greatest: 12x3+18x2=2x(6x2+9x)12x^3 + 18x^2 = 2x(6x^2 + 9x). It is correct but not finished: 6x2+9x6x^2 + 9x still shares 3x3x. The GCF is 6x26x^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.

Try it
Try it: GCF

Factor 14m4n2−21m2n3+7m2n214m^4n^2 - 21m^2n^3 + 7m^2n^2.

Show the answer

Numbers: GCF 77. Variables: smallest powers m2m^2 and n2n^2. So pull 7m2n27m^2n^2: 7m2n2(2m2−3n+1)7m^2n^2(2m^2 - 3n + 1). The last term was the GCF itself, so it leaves 11.

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).

Grouping works when the four terms came from multiplying two binomials: (x+3)(x2+2)(x + 3)(x^2 + 2) expands to x3+3x2+2x+6x^3 + 3x^2 + 2x + 6, and grouping simply undoes the two distributions in reverse. If the two parentheses do not match after pulling GCFs, the terms did not come from such a product in that order.

Example
Example: Factor x3+3x2+2x+6x^3+3x^2+2x+6
x+3x²+2x³3x²2x6
The four terms fill a 2×22 \times 2 box whose edges are the factors.
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. ✓

Side note
Where it goes wrong

Losing a sign when the third term is negative: x3+2x2−5x−10=x2(x+2)−5(x−2)x^3 + 2x^2 - 5x - 10 = x^2(x + 2) - 5(x - 2). Factoring −5-5 out of −5x−10-5x - 10 gives −5(x+2)-5(x + 2), and then the parentheses match: (x+2)(x2−5)(x + 2)(x^2 - 5).

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.

Try it
Try it: group

Factor 2ab+6a−5b−152ab + 6a - 5b - 15.

Show the answer

Group: (2ab+6a)+(−5b−15)=2a(b+3)−5(b+3)=(b+3)(2a−5)(2ab + 6a) + (-5b - 15) = 2a(b + 3) - 5(b + 3) = (b + 3)(2a - 5). Check: 2ab−5b+6a−152ab - 5b + 6a - 15.

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).

Expanding (x+p)(x+q)(x + p)(x + q) gives x2+(p+q)x+pqx^2 + (p + q)x + pq: the two numbers add to the middle coefficient and multiply to the constant. Factoring runs this backward. The area model shows the same thing: the x2x^2 square, strips of 3x3x and 4x4x, and a 3×4=123 \times 4 = 12 corner fit together into an (x+3)(x + 3)-by-(x+4)(x + 4) rectangle.

Example
Example: Factor x2+7x+12x^2+7x+12
x²4x3x12x+4x+3
x2+3x+4x+12x^2 + 3x + 4x + 12 tiles an (x+3)(x + 3)-by-(x+4)(x + 4) rectangle.

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. ✓

Side note
Where it goes wrong

Picking numbers that multiply to cc but not add to bb: x2+7x+12=(x+2)(x+6)x^2 + 7x + 12 = (x + 2)(x + 6). Check the middle: 2+6=8≠72 + 6 = 8 \ne 7. The pair must satisfy both: 3⋅4=123 \cdot 4 = 12 and 3+4=73 + 4 = 7.

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.

Try it
Try it: two numbers

Factor x2−11x+24x^2 - 11x + 24 and x2+2x−35x^2 + 2x - 35.

Show the answer

Product 2424, sum −11-11: −3-3 and −8-8, so (x−3)(x−8)(x - 3)(x - 8). Product −35-35, sum 22: 77 and −5-5, so (x+7)(x−5)(x + 7)(x - 5).

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

Concept
The “ac” / Grouping Method

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

  • 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.

Why multiply aa and cc? In (px+q)(rx+s)(px + q)(rx + s) the four products are prx2prx^2, psxpsx, qrxqrx and qsqs. The two middle ones, psps and qrqr, multiply to pqrspqrs, which is the product of the outer coefficients pr⋅qs=acpr \cdot qs = ac. So the middle term always splits into two pieces whose product is acac; finding them rebuilds the 2×22 \times 2 box, and grouping reads off its edges.

Example
Example: Factor 2x2+7x+32x^2+7x+3
x+32x+12x²6xx3
Split 7x7x into 6x+x6x + x; the box edges are the factors (2x+1)(x+3)(2x + 1)(x + 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. ✓

Side note
Where it goes wrong

Using the x2+bx+cx^2 + bx + c shortcut when a≠1a \ne 1: 2x2+7x+3=(x+6)(x+1)2x^2 + 7x + 3 = (x + 6)(x + 1). Expanding gives x2+7x+6x^2 + 7x + 6, a different polynomial. With a≠1a \ne 1, split the middle term using acac.

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. ✓

Try it
Try it: ac method

Factor 3x2−10x+83x^2 - 10x + 8.

Show the answer

ac=24ac = 24; two numbers with product 2424 and sum −10-10: −4-4 and −6-6. Split: 3x2−6x−4x+8=3x(x−2)−4(x−2)=(x−2)(3x−4)3x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (x - 2)(3x - 4). Check: 3x2−4x−6x+83x^2 - 4x - 6x + 8.

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.

Multiply (a+b)(a−b)(a + b)(a - b) and the cross terms −ab-ab and +ab+ab cancel, leaving a2−b2a^2 - b^2. So any difference of two squares splits into a sum times a difference. A sum of squares cannot: (a+b)(a+b)(a + b)(a + b) and (a−b)(a−b)(a - b)(a - b) both keep a middle term, and no real pair cancels it, since x2+4x^2 + 4 is never 00 for real xx.

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. ✓

Side note
Where it goes wrong

Factoring a sum: x2+25=(x+5)(x−5)x^2 + 25 = (x + 5)(x - 5). Expanding gives x2−25x^2 - 25. A sum of squares does not factor over the reals.

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.

Try it
Try it: squares

Factor 49a2−149a^2 - 1, x4−81x^4 - 81 and 12y2−7512y^2 - 75.

Show the answer

(7a+1)(7a−1)(7a + 1)(7a - 1). x4−81=(x2+9)(x2−9)=(x2+9)(x+3)(x−3)x^4 - 81 = (x^2 + 9)(x^2 - 9) = (x^2 + 9)(x + 3)(x - 3). 12y2−75=3(4y2−25)=3(2y+5)(2y−5)12y^2 - 75 = 3(4y^2 - 25) = 3(2y + 5)(2y - 5).

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.

A perfect-square trinomial is the area of a square: x2+6x+9x^2 + 6x + 9 is an xx-square, two 3x3x strips and a 3×33 \times 3 corner, which assemble into a square of side x+3x + 3. The middle term must be exactly twice the product of the square roots, because there are exactly two strips.

Example
Example: Factor x2+6x+9x^2+6x+9
x²3x3x9x+3x+3
Square plus two equal strips plus the corner: a square of side x+3x + 3.

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. ✓

Side note
Where it goes wrong

Assuming any trinomial with square ends is a perfect square: x2+10x+9=(x+3)2x^2 + 10x + 9 = (x + 3)^2. The middle term would have to be 2⋅x⋅3=6x2 \cdot x \cdot 3 = 6x. Here 10x10x does not fit; it factors as (x+1)(x+9)(x + 1)(x + 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. ✓

Try it
Try it: perfect squares

Decide which are perfect squares and factor them: 9x2−12x+49x^2 - 12x + 4, x2+8x+64x^2 + 8x + 64.

Show the answer

9x2=(3x)29x^2 = (3x)^2, 4=224 = 2^2, and 2⋅3x⋅2=12x2 \cdot 3x \cdot 2 = 12x matches: (3x−2)2(3x - 2)^2. For x2+8x+64x^2 + 8x + 64, a perfect square would need middle term 2⋅x⋅8=16x2 \cdot x \cdot 8 = 16x; 8x8x fails, and since D=64−256<0D = 64 - 256 < 0 it does not factor over the reals at all.

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.

Pulling the GCF first makes every later pattern easier to see, because the remaining numbers are smaller. 3x2−273x^2 - 27 does not look like a difference of squares until the 33 is removed and x2−9x^2 - 9 appears. "Completely" means no factor can be broken down further, like a prime factorization.

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. ✓

Side note
Where it goes wrong

Stopping one step early: x4−16=(x2+4)(x2−4)x^4 - 16 = (x^2 + 4)(x^2 - 4). The second factor is still a difference of squares: (x2+4)(x+2)(x−2)(x^2 + 4)(x + 2)(x - 2).

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!

Try it
Try it: completely

Factor completely: 5x3−45x5x^3 - 45x and 2x3−8x2−24x2x^3 - 8x^2 - 24x.

Show the answer

5x(x2−9)=5x(x+3)(x−3)5x(x^2 - 9) = 5x(x + 3)(x - 3). 2x(x2−4x−12)=2x(x−6)(x+2)2x(x^2 - 4x - 12) = 2x(x - 6)(x + 2).

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.

If neither factor were zero, their product would be a product of two nonzero numbers, which is never zero. So (x+5)(x−3)=0(x + 5)(x - 3) = 0 forces x+5=0x + 5 = 0 or x−3=0x - 3 = 0. On the graph of y=x2+2x−15y = x^2 + 2x - 15, those are the two places the curve meets the xx-axis.

Example
Example: Solve x2+2x−15=0x^2+2x-15=0
xy−7−6−5−4−3−2−11234−16−12−8−44
The factors vanish at x=−5x = -5 and x=3x = 3, where the graph crosses the axis.
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. ✓

Side note
Where it goes wrong

Dividing both sides by xx: from x2=4xx^2 = 4x writing x=4x = 4. That throws away x=0x = 0. Move everything over and factor: x(x−4)=0x(x - 4) = 0, so x=0x = 0 or x=4x = 4.

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.

Try it
Try it: solve

Solve 2x2=7x+152x^2 = 7x + 15.

Show the answer

Rewrite: 2x2−7x−15=02x^2 - 7x - 15 = 0. ac=−30ac = -30; numbers −10-10 and 33: 2x2−10x+3x−15=2x(x−5)+3(x−5)=(x−5)(2x+3)2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (x - 5)(2x + 3). So x=5x = 5 or x=−32x = -\tfrac32.

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.

Why It Works

To factor x2+bx+cx^2 + bx + c we need (x+p)(x+q)=x2+(p+q)x+pq(x + p)(x + q) = x^2 + (p + q)x + pq, so p+q=bp + q = b and pq=cpq = c — the search for two numbers is exactly matching coefficients. The zero product property is the reason factoring solves equations: if pq=0pq = 0 and p≠0p \ne 0, dividing by pp gives q=0q = 0. That property is special to zero (a product equal to 66 says nothing about the factors), which is why every equation must be set equal to zero before factoring.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.