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Quadratic Relations of the Form y = ax² + bx + c · Quadratic Expressions and Equations

Factoring Trinomials of the Form ax² + bx + c

3.2About 70 minutes
Learning goals
  • I can factor ax² + bx + c by decomposition (splitting the middle term).
  • I can use algebra tiles or guess and check to factor when a ≠ 1.
  • I can take out a common factor before factoring a trinomial.
  • I can explain why the decomposition method works.

Words to know

decomposition
A factoring method for ax² + bx + c: split bx into two terms whose coefficients multiply to ac, then factor by grouping.
leading coefficient
The coefficient of the highest-degree term. In 6x² − 11x + 4 it is 6.
complex trinomial
A trinomial ax² + bx + c in which a is not 1 (and not a common factor).
guess and check
Trying possible pairs of binomial factors and expanding to see which one gives the trinomial.

When the x² term has a coefficient

In the last lesson every trinomial started with 1x². But many real quadratics look like 2a² + 11a + 5 or 6x² − 11x + 4. A sporting-goods store might model profit with 3x² terms, and a physics formula for a thrown ball starts with −4.9t².

The product-sum puzzle still works, with one twist. It is called decomposition, because you decompose (split) the middle term into two parts.

Factoring ax² + bx + c by decomposition

  1. Take out any common factor first.
  2. Multiply a × c.
  3. Find two integers whose product is ac and whose sum is b.
  4. Split the middle term bx into two terms using those numbers.
  5. Factor by grouping (from the common-factoring lesson).
  6. Check by expanding.

When a = 1, step 3 is exactly the product-sum puzzle you already know.

Worked example

Example 1: Factor 2a² + 11a + 5

Factor 2a² + 11a + 5.

  1. No common factor. a × c = 2 × 5 = 10.
  2. Two numbers with product 10 and sum 11: 1 and 10.
  3. Split the middle term: 2a² + 1a + 10a + 5.
  4. Group: a(2a + 1) + 5(2a + 1).
  5. Factor out (2a + 1): (2a + 1)(a + 5).
  6. Check: 2a² + 10a + a + 5 = 2a² + 11a + 5 ✓.

(2a + 1)(a + 5)

Decomposition: 6x² − 11x + 41. Find ac6 × 4 = 242. Product 24, sum −11−3 and −83. Split the middle term6x² − 3x − 8x + 44. Factor each pair3x(2x − 1) − 4(2x − 1)5. Factor out the bracket(2x − 1)(3x − 4)
The decomposition method step by step for 6x² − 11x + 4. In step 4 the second pair needs a negative common factor, −4, so the brackets match.
Worked example

Example 2: Both numbers negative

Factor 6x² − 11x + 4.

  1. ac = 24. Product +24 and sum −11, so both numbers are negative.
  2. Pairs: −1 and −24 (sum −25); −2 and −12 (−14); −3 and −8 (−11 ✓).
  3. Split: 6x² − 3x − 8x + 4.
  4. Group: 3x(2x − 1) − 4(2x − 1).
  5. Factor: (2x − 1)(3x − 4).
  6. Check: 6x² − 8x − 3x + 4 = 6x² − 11x + 4 ✓.

(2x − 1)(3x − 4)

Worked example

Example 3: A negative constant

Factor 3x² + 4x − 15.

  1. ac = 3 × (−15) = −45. Need product −45 and sum +4.
  2. Try 9 and −5: 9 × (−5) = −45 and 9 + (−5) = 4 ✓.
  3. Split: 3x² + 9x − 5x − 15.
  4. Group: 3x(x + 3) − 5(x + 3).
  5. Factor: (x + 3)(3x − 5).

(x + 3)(3x − 5)

Worked example

Example 4: Common factor first

Factor 12x² + 2x − 4 fully.

  1. Common factor 2: 2(6x² + x − 2).
  2. For 6x² + x − 2: ac = −12; need product −12, sum 1: 4 and −3.
  3. Split: 6x² + 4x − 3x − 2 = 2x(3x + 2) − 1(3x + 2).
  4. Factor: (3x + 2)(2x − 1).
  5. Do not forget the 2 you took out.

2(3x + 2)(2x − 1)

x²x²2x + 3x + 12x² + 5x + 32 x²-tiles, 5 x-tiles, 3 unit tilesSides: 2x + 3 and x + 12x² + 5x + 3 = (2x + 3)(x + 1)
Algebra tiles for 2x² + 5x + 3. The two x²-tiles sit side by side, so one side is 2x + 3 and the other is x + 1. Count the x-tiles: 3 standing up + 2 lying down = 5x.
Worked example

Example 5: Two variables

Factor 2x² + 7xy + 3y².

  1. ac = 2 × 3 = 6; need product 6, sum 7: 1 and 6.
  2. Split: 2x² + 1xy + 6xy + 3y².
  3. Group: x(2x + y) + 3y(2x + y).
  4. Factor: (2x + y)(x + 3y).

(2x + y)(x + 3y)

Another way: guess and check

Some people prefer to guess and check. For 3x² + 4x − 15, the first terms must be 3x and x, and the last terms must multiply to −15. Try the arrangements and expand the middle term each time: (3x − 5)(x + 3) gives 9x − 5x = 4x ✓. This works well when a and c have few factors. Decomposition always works and never needs luck.

Hands-on activity

Tile it out

You need: the paper tiles from the common-factoring lesson (or make new ones: squares for x², long strips for x, small squares for 1), and a flat table.

  1. Collect 2 x²-tiles, 7 x-tiles and 3 unit tiles.
  2. Try to arrange them into a single rectangle with no gaps. Hint: place the two x²-tiles side by side.
  3. Write the length and width. That is the factorization of 2x² + 7x + 3.
  4. Repeat with 3 x²-tiles, 7 x-tiles and 2 unit tiles.
  5. Check both answers using decomposition.

Model answers: 2x² + 7x + 3 = (2x + 1)(x + 3) and 3x² + 7x + 2 = (3x + 1)(x + 2).

Just for fun

Decomposition is also what happens to fallen maple leaves every autumn. The leaves break down into soil; trinomials break down into binomials. Only one of them smells like fall.

Your turn

1

Factor 2x² + 7x + 3.

2

Factor 3x² − 10x + 8.

3

Factor 5x² + 13x − 6.

4

Factor 4x² − 4x − 15.

5

Factor 6x² + 5x + 1.

6

Factor 8x² − 2x − 6 fully.

7

Factor 3x² + 8xy + 4y².

More practice: warm-up

1

For 2x² + 9x + 4, what is ac, and which two numbers have product ac and sum 9?

2

Finish factoring 2x² + 9x + 4.

3

Factor 3x² + 7x + 2.

4

Factor 2x² − 5x + 3.

5

Factor 5x² + 4x − 1.

More practice: on your own

1

Factor 10x² − 11x − 6.

2

Factor 9x² + 12x + 4. What do you notice?

3

Factor −6x² + x + 2 fully.

4

Factor 12x² − 34x + 14 fully.

5

A rectangle has area 6x² + 7x − 3 and one side 3x − 1. Find the other side.

Challenge

1

Find all integers k so that 2x² + kx + 3 can be factored over the integers.

2

Factor 4x² + 4xy − 15y².

3

Explain why decomposition works. (Expand (px + q)(rx + s) and compare with ax² + bx + c.)

Check your understanding

Quiz

1.

Factor 2a² + 11a + 5.

2.

To factor 6x² + x − 15 by decomposition, which two numbers split the middle term?

3.

Factor 3x² − 14x + 8.

4.

Which is a factor of 4x² − 8x − 5?

5.

Factor 6x² + 15x + 6 fully.

6.

(2x + 3)(3x − 1) is the factored form of:

For parents and teachers

Big idea: This lesson finishes the trinomial part of expectation 3.2 with "complex" trinomials ax² + bx + c where a ≠ 1, such as the curriculum's example 2a² + 11a + 5. Decomposition turns the problem into the factoring-by-grouping skill from two lessons ago.

How to help: Ask your teen to say the steps out loud: common factor, ac, two numbers, split, group, check. If the grouping step gives brackets that do not match, the usual fix is to take out a negative from the second pair.

Common difficulties:

  • Using c instead of ac when looking for the two numbers.
  • Forgetting the 1 in a grouping like 2x(3x + 2) − 1(3x + 2).
  • Dropping the common factor that was taken out at the start.