Grade by Grade
A. Quadratic Functions · Quadratic Expressions and Equations

Factoring: Common Factors and Trinomials x² + bx + c

A1.3About 65 minutes
Learning goals
  • I can take out the greatest common factor, including a negative or a binomial common factor.
  • I can factor trinomials x² + bx + c using product and sum.
  • I can factor fully and check my answer by expanding.

Words to know

factor (verb)
Write an expression as a product.
greatest common factor (GCF)
The largest expression that divides every term.
factored fully
Written as a product in which no factor can be factored further.
trinomial
A polynomial with three terms.
leading coefficient
The coefficient a of the x² term in ax² + bx + c.

Factoring: expanding in reverse

Factoring rewrites a sum as a product. It undoes expanding:

(x + 3)(x + 4) → x² + 7x + 12 is expanding; x² + 7x + 12 → (x + 3)(x + 4) is factoring.

Why bother? Because a product tells you things a sum hides. In the lesson Solving Quadratic Equations by Factoring you will see that if (x + 3)(x + 4) = 0, then x = −3 or x = −4, instantly. The factors also tell you where a parabola crosses the x-axis.

Remember from Grade 10

You factored out common factors, factored trinomials x² + bx + c and ax² + bx + c, and used the difference of squares and perfect square patterns. Over the next three lessons you will review each type, practise choosing a strategy, and always factor fully.

Step 1, always: the greatest common factor (GCF)

Look for the largest number and the highest power of x that divide every term. Divide each term by it and write it in front of a bracket.

  • 6x² − 15x: the GCF is 3x, so 6x² − 15x = 3x(2x − 5).
  • −4x² + 12x: when the leading term is negative, take out a negative GCF: −4x(x − 3). (Check: −4x · x = −4x² and −4x · (−3) = +12x.)
  • The common factor can be a whole bracket: 3x(x − 2) + 5(x − 2) = (x − 2)(3x + 5). Think of (x − 2) as one object, like a ★: 3x★ + 5★ = ★(3x + 5).

Trinomials x² + bx + c

When you expand (x + m)(x + n) you get x² + (m + n)x + mn. So to factor x² + bx + c, look for two integers m and n with

  • product m × n = c, and
  • sum m + n = b.

List the factor pairs of c in an organized way and test their sums.

x²xxxxxxxx+ 4x+ 31 tile x²7 tiles x (4 + 3)12 unit tiles (4 × 3)x² + 7x + 12= (x + 3)(x + 4)
Algebra tiles for x² + 7x + 12 arranged as a rectangle. The side lengths are x + 3 and x + 4. The 12 unit tiles form a 3-by-4 block (product 12), and the x-tiles split 3 and 4 (sum 7).
Worked example

Example 1: Product and sum

Factor (a) x² + 7x + 12 (b) x² − 2x − 35

  1. (a) Product 12, sum 7. Factor pairs of 12: 1 × 12 (sum 13), 2 × 6 (sum 8), 3 × 4 (sum 7) ✓.
  2. So x² + 7x + 12 = (x + 3)(x + 4).
  3. (b) Product −35 (negative, so one number is negative), sum −2. Pairs: 1 and −35 (−34), −1 and 35 (34), 5 and −7 (−2) ✓, −5 and 7 (2).
  4. So x² − 2x − 35 = (x + 5)(x − 7).
  5. Check (b) by expanding: x² − 7x + 5x − 35 = x² − 2x − 35 ✓.

(a) (x + 3)(x + 4) (b) (x + 5)(x − 7)

c > 0, b > 0both numbers +x² + 7x + 12(x + 3)(x + 4)c > 0, b < 0both numbers −x² − 7x + 12(x − 3)(x − 4)c < 0one +, one −x² − 2x − 35(x − 7)(x + 5)
The signs of b and c tell you the signs of the two numbers. If c is positive, both numbers have the sign of b. If c is negative, the signs are different, and the number with the larger size has the sign of b.
Worked example

Example 2: Factor fully

Factor fully: (a) 2x² − 8x − 42 (b) −x² + x + 12

  1. (a) Common factor first: 2x² − 8x − 42 = 2(x² − 4x − 21).
  2. Now product −21 and sum −4: −7 and 3. So x² − 4x − 21 = (x − 7)(x + 3).
  3. Fully factored: 2(x − 7)(x + 3). Writing (2x − 14)(x + 3) is correct algebra but is not fully factored, because 2x − 14 still has a common factor of 2.
  4. (b) Take out −1: −x² + x + 12 = −(x² − x − 12).
  5. Product −12, sum −1: −4 and 3. So the answer is −(x − 4)(x + 3).
  6. Check (b) with x = 0: original 12; answer −(−4)(3) = 12 ✓.

(a) 2(x − 7)(x + 3) (b) −(x − 4)(x + 3)

Worked example

Example 3: When it does not factor

Factor x² + 4x + 6, if possible.

  1. Product 6, sum 4. Pairs with product 6: 1 and 6 (sum 7), 2 and 3 (sum 5), −1 and −6 (sum −7), −2 and −3 (sum −5).
  2. No pair has sum 4, and there is no common factor. So x² + 4x + 6 cannot be factored over the integers.
  3. That is a perfectly good answer. In the lesson Real Roots, x-Intercepts and the Discriminant, the discriminant gives a quick test for whether a trinomial factors.

x² + 4x + 6 does not factor over the integers.

Factoring checklist

  1. Common factor? Take it out first (negative if the leading term is negative).
  2. Trinomial with leading coefficient 1? Find two numbers: product c, sum b.
  3. Check by expanding, or by substituting a number such as x = 1.
  4. Is every factor fully factored? If a bracket still has a common factor, keep going.
Just for fun

Factoring is the only time a maths teacher will cheer when you take something apart and then refuse to put it back together.

Hands-on activity

Product-and-sum card game

You need: a deck of playing cards with the face cards removed (aces count as 1), paper and a pencil. Play with a family member.

  1. Deal two cards face down to your partner. Red cards are negative, black cards positive.
  2. Your partner tells you only the product and the sum of the two cards.
  3. You write the trinomial x² + (sum)x + (product) and factor it. Your factors reveal the cards!
  4. Switch roles. Score 1 point for each correct factoring; first to 5 wins.

What to look for: if the cards are black 4 and red 6 (that is, 4 and −6), the clues are product −24 and sum −2, the trinomial is x² − 2x − 24 and it factors as (x + 4)(x − 6). The numbers in the factors are exactly the two cards.

Your turn

1

Factor 5x² − 20x.

2

Factor x² + 9x + 20.

3

Factor x² − 11x + 24.

4

Factor x² + 3x − 40.

5

Factor x² − x − 56.

6

Factor fully: 3x² + 6x − 72.

7

Factor fully: −2x² + 10x + 28.

More practice: warm-up

1

Factor 8x² + 12x.

2

Factor x² + 5x + 6.

3

Factor x² − 8x + 15.

4

Factor x² + 2x − 15.

5

Factor x² − 6x − 27.

More practice: on your own

1

Factor x² + 13x + 42.

2

Factor x² − 4x − 60.

3

Factor fully: 4x² − 4x − 48.

4

Factor 2x(x + 1) − 7(x + 1).

5

Mia factored 3x² − 15x + 18 as (3x − 6)(x − 3). Is it correct? Is it fully factored?

Challenge

1

Factor fully: 0.5x² + 3x + 4.

2

Find all integers k so that x² + kx + 12 can be factored over the integers.

3

Factor (x + 2)² + 4(x + 2) + 3 by letting u = x + 2.

Check your understanding

Quiz

1.

What is the first thing to check when factoring any expression?

2.

Factor x² + 10x + 21.

3.

Factor x² − 3x − 18.

4.

Which is 4x² + 8x − 60 factored fully?

5.

x² + bx + c with c > 0 and b < 0 factors as:

6.

Which trinomial cannot be factored over the integers?

For parents and teachers

Big idea (A1.3): factoring reverses expanding. This lesson covers common factors (including a common bracket) and trinomials x² + bx + c, always factored fully.

How to help: quiz your teen on product-and-sum pairs ("two numbers that multiply to −24 and add to 5?" Answer: 8 and −3). Encourage them to check every factoring by expanding it back.

Common difficulties:

  • Skipping the common factor, so the answer is not fully factored.
  • Sign errors: when c is negative, the two numbers have opposite signs.
  • Giving up too early. A systematic list of factor pairs finds the answer or shows there is none.
  • Taking out a negative common factor and forgetting to change every sign in the bracket.