Factoring — Print

Factoring is multiplication run backwards. Always pull out the common factor first — it makes everything after it easier.

What the practice questions cover

These questions use 7 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Someone hands you one big square tile, seven long strips and twelve small squares. Make one rectangle from all of them.

you are handed the tiles — rebuild the rectangle x² seven x tiles twelve units x² 3x 4x 12 x 3 x 4 only one split of the seven x tiles leaves a clean rectangle: 3 and 4

The big tile is x2, each strip is x, each little square is 1. You are holding x2 + 7x + 12.

Split the seven strips between the right edge and the bottom. Try 2 and 5: the corner then needs 10 small squares, and you have twelve. It does not close.

Try 3 and 4: the corner needs exactly 12. It closes.

So the sides are x + 3 and x + 4 — and the reason 3 and 4 worked is that 3 × 4 = 12 and 3 + 4 = 7.

Factoring is the area model run backwards. You are given what is inside and asked for the sides — and only one split of the strips leaves a clean rectangle.

Factoring a quadratic by finding the side lengths factoring runs the area model backwards: you know the inside, find the sides x3 x4 x^23x 4x12 x^2 + 7x + 12 need a . b = 12 and a + b = 7 3 and 4 (x + 3)(x + 4) list the factor pairs of 12: 1 and 12, 2 and 6, 3 and 4 — only the last pair adds to 7

The order to try things

1. Pull out the GCF. 2. Count the terms. 3. Use the matching pattern. 4. Check whether any factor can go further.

Two terms: look for a difference of squares. Three terms: look for a pair of numbers. Four terms: try grouping.

Then multiply your answer back out. Factoring is the one topic where checking is completely mechanical.

GCF first, always

2x2 + 10x + 12 looks awkward until you take out the 2: 2(x2 + 5x + 6) = 2(x + 2)(x + 3). Skipping the GCF makes easy problems hard.

Trinomials with leading coefficient 1

For x2 + bx + c, find two numbers that multiply to c and add to b.

x2 + 7x + 12: the pair is 3 and 4, so it factors as (x + 3)(x + 4).

If c is positive, both numbers share b's sign. If c is negative, the numbers have opposite signs.

List the factor pairs

For c = 12: 1&12, 2&6, 3&4. Check which pair adds to b. Writing all the pairs down beats staring at the problem.

When the leading coefficient is not 1

For ax2 + bx + c, find two numbers multiplying to a · c and adding to b. Split the middle term with them, then group.

6x2 + 7x - 3: a · c = -18, and 9 × (-2) = -18 with 9 + (-2) = 7.

6x2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3) = (2x + 3)(3x - 1).

Multiply a by c, not just use c

With a leading coefficient, the target product is a · c. Using c alone is the reason this method appears not to work.

The patterns

Difference of squares: 9x2 - 25 = (3x - 5)(3x + 5). Both terms must be perfect squares, with a minus between them.

A sum of squares, like x2 + 25, does not factor over the real numbers.

Perfect square trinomial: x2 - 10x + 25 = (x - 5)2. Check that the middle term really is twice the product.

Factoring by grouping

Four terms: group in pairs, factor each pair, and hope the same bracket appears twice. x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3).

Worked examples

GCF then trinomial

Factor 3x2 - 27x + 60 completely.

  1. GCF is 3: 3(x2 - 9x + 20) Always first.
  2. Need two numbers multiplying to 20, adding to -9. Both negative, since the product is positive and the sum negative.
  3. -4 and -5 work. (-4)(-5) = 20 and -4 - 5 = -9.
  4. 3(x - 4)(x - 5) Check: 3(x2 - 9x + 20) = 3x2 - 27x + 60 ✓

3(x - 4)(x - 5)

Leading coefficient

Factor 2x2 + 11x + 12.

  1. a · c = 2 × 12 = 24; need a pair adding to 11. The split-the-middle method.
  2. 3 and 8: 3 × 8 = 24 and 3 + 8 = 11 ✓ Found the pair.
  3. 2x2 + 3x + 8x + 12 Split the middle term.
  4. x(2x + 3) + 4(2x + 3) = (2x + 3)(x + 4) Group and factor out the shared bracket.

(2x + 3)(x + 4)

Difference of squares twice

Factor 16x4 - 81 completely.

  1. 16x4 = (4x2)2 and 81 = 92 Both perfect squares.
  2. (4x2 - 9)(4x2 + 9) First application of the pattern.
  3. 4x2 - 9 is also a difference of squares: (2x - 3)(2x + 3) Keep going while you can.
  4. (2x - 3)(2x + 3)(4x2 + 9) 4x2 + 9 is a sum of squares, so it stops here.

(2x - 3)(2x + 3)(4x2 + 9)

Words to know

Factor
Something being multiplied.
GCF
Greatest common factor: the largest thing dividing every term.
Difference of squares
a2 - b2 = (a - b)(a + b).
Perfect square trinomial
a2 ± 2ab + b2 = (a ± b)2.
Prime polynomial
One that will not factor any further over the integers.

Mistakes to watch for

Skipping the GCF. 2x2 + 10x + 12 looks awkward until you take out the 2, leaving 2(x + 2)(x + 3). Going straight to the trinomial makes an easy problem hard and often unfactorable-looking.

Trying to factor a sum of squares. x2 + 9 has no real factors — there is no pair of numbers multiplying to 9 and adding to 0. Only a difference of squares splits.

Using c instead of a · c when the leading coefficient is not 1. For 6x2 + 7x - 3 the target product is -18, not -3. Using c alone is why this method appears not to work.

Stopping too early. After the first split, check each factor again — x2 - 9 inside your answer still factors. "Completely" means until nothing can be taken further.

Factoring is easy for small numbers and brutally hard for large ones — and modern banking depends on exactly that.

RSA encryption multiplies two enormous primes together. Anyone can see the product; nobody can factor it back in reasonable time.

In 2020 it took roughly 2,700 core-years of computing to factor a single 250-digit number. Multiplying those primes took a fraction of a second.

Problem 1

Factor 6x + 15.

Worked answer

3(2x + 5)

GCF is 3

3(2x + 5)

Problem 2

Factor 4x2 - 10x.

Worked answer

2x(2x - 5)

GCF is 2x

2x(2x - 5)

Problem 3

Factor x2 + 7x + 10.

Worked answer

(x + 2)(x + 5)

2 and 5

(x + 2)(x + 5)

Problem 4

Factor x2 - 9x + 20.

Worked answer

(x - 4)(x - 5)

-4 and -5: product 20, sum -9

(x - 4)(x - 5)

Problem 5

Factor x2 - 36.

Worked answer

(x - 6)(x + 6)

x2 - 62

(x - 6)(x + 6)

Problem 6

Factor x2 + 2x - 15.

Worked answer

(x + 5)(x - 3)

5 and -3: product -15, sum 2

(x + 5)(x - 3)

Problem 7

Factor 25x2 - 49.

Worked answer

(5x - 7)(5x + 7)

(5x)2 - 72

(5x - 7)(5x + 7)

Problem 8

Factor 2x2 + 10x + 12 completely.

Worked answer

2(x + 2)(x + 3)

2(x2 + 5x + 6)

2(x + 2)(x + 3)

Problem 9

Factor 2x2 + 7x + 3.

Worked answer

(2x + 1)(x + 3)

6 and 1

2x2 + 6x + x + 3

2x(x+3) + 1(x+3) = (2x+1)(x+3)

Problem 10

Factor x2 - 14x + 49.

Worked answer

(x - 7)^2

49 = 72 and 2 × 7 = 14 ✓

(x - 7)2

Problem 11

Which expression will not factor over the integers?

  1. x2 + 9
  2. x2 - 9
  3. x2 + 6x + 9
  4. x2 - 6x + 9
Worked answer

x2 + 9

x2 + 9 is a sum of squares

It has no real factors

Problem 12

Factor x3 + 3x2 + 2x + 6 by grouping.

Worked answer

(x + 3)(x^2 + 2)

x2(x + 3) + 2(x + 3)

(x + 3)(x2 + 2)

Problem 13

Factor 6x2 - 7x - 3.

Worked answer

(3x + 1)(2x - 3)

-9 and 2

6x2 - 9x + 2x - 3

3x(2x - 3) + 1(2x - 3) = (3x+1)(2x-3)

Problem 14

Factor 3x3 - 27x completely.

Worked answer

3x(x - 3)(x + 3)

3x(x2 - 9)

3x(x - 3)(x + 3)

Problem 15

Factor x4 - 81 completely.

Worked answer

(x - 3)(x + 3)(x^2 + 9)

(x2 - 9)(x2 + 9)

(x-3)(x+3)(x2+9)

Problem 16

A rectangle has area x2 + 11x + 24. Give expressions for its two sides.

Worked answer

x + 3 and x + 8

3 and 8: product 24, sum 11

(x + 3)(x + 8)

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