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

Factoring Trinomials of the Form x² + bx + c

3.2About 65 minutes
Learning goals
  • I can explain why factoring x² + bx + c means finding two numbers with product c and sum b.
  • I can factor trinomials of the form x² + bx + c, including ones with negative terms.
  • I can take out a common factor first and then factor the trinomial.
  • I can recognize when a trinomial cannot be factored over the integers.

Words to know

trinomial
A polynomial with three terms, such as x² − x − 6.
simple trinomial
A trinomial of the form x² + bx + c, where the coefficient of x² is 1.
factor pair
Two integers that multiply to give a number. The factor pairs of 6 include 2 and 3, and −1 and −6.
factor over the integers
Factor using only integer coefficients.
algebra tiles
Tiles that model x², x and 1. Arranging them in a rectangle shows a factorization.

A number puzzle

I am thinking of two numbers. Their product is 12 and their sum is 7. What are they?

If you said 3 and 4, you have already factored your first trinomial:

x² + 7x + 12 = (x + 3)(x + 4).

In this lesson you will learn why that puzzle works, and how to solve it quickly even when negative numbers are involved.

Why the product-sum puzzle works

Expand a general product of two binomials:

(x + m)(x + n) = x² + nx + mx + mn = x² + (m + n)x + mn.

Compare this with x² + bx + c:

  • the constant c is the product mn;
  • the coefficient b of x is the sum m + n.

So to factor x² + bx + c, find two integers whose product is c and whose sum is b. Then write (x + m)(x + n).

x²x + 3x + 2x² + 5x + 61 x²-tile, 5 x-tiles, 6 unit tilesThe 6 unit tiles make a 2 by 3 block:2 × 3 = 6 and 2 + 3 = 5x² + 5x + 6 = (x + 2)(x + 3)
Algebra tiles for x² + 5x + 6 arranged as a rectangle. The side lengths are the factors. The unit tiles form a 2 by 3 block: that is why you need two numbers that multiply to 6 and add to 5.
c (product)b (sum)Signs of the two numbersExample
positivepositiveboth positivex² + 7x + 12 = (x + 3)(x + 4)
positivenegativeboth negativex² − 10x + 16 = (x − 2)(x − 8)
negativepositive or negativeone positive, one negative; the larger one (ignoring sign) has the same sign as bx² − x − 6 = (x − 3)(x + 2)
Use the signs of b and c to narrow down your search.
Worked example

Example 1: Both numbers positive

Factor x² + 7x + 12.

  1. Need product +12 and sum +7. Both numbers are positive.
  2. Factor pairs of 12: 1 and 12 (sum 13), 2 and 6 (sum 8), 3 and 4 (sum 7) ✓.
  3. Write (x + 3)(x + 4).
  4. Check: x² + 4x + 3x + 12 = x² + 7x + 12 ✓.

(x + 3)(x + 4)

Worked example

Example 2: Both numbers negative

Factor x² − 10x + 16.

  1. Product +16, sum −10: both numbers are negative.
  2. Pairs: −1 and −16 (sum −17); −2 and −8 (sum −10) ✓; −4 and −4 (sum −8).
  3. Check: (x − 2)(x − 8) = x² − 8x − 2x + 16 = x² − 10x + 16 ✓.

(x − 2)(x − 8)

Worked example

Example 3: One positive, one negative

Factor x² − x − 6.

  1. Product −6 means the signs are different. Sum −1 means the number with the larger size (ignoring its sign) is negative.
  2. Pairs with product −6: 1 and −6 (sum −5); 2 and −3 (sum −1) ✓; 3 and −2 (sum 1); 6 and −1 (sum 5).
  3. Write (x + 2)(x − 3).
  4. Check: x² − 3x + 2x − 6 = x² − x − 6 ✓.

(x + 2)(x − 3)

The X diagram for x² − x − 6−6−12−3Top: the product cBottom: the sum bSides: the two numbers2 × (−3) = −62 + (−3) = −1
Many students organize the search with an X diagram: the product goes on top, the sum on the bottom, and the two numbers you find go on the sides.
Worked example

Example 4: Common factor first

Factor 2x² − 6x − 56 fully.

  1. Always look for a common factor first. GCF = 2: 2(x² − 3x − 28).
  2. Factor x² − 3x − 28: product −28, sum −3. Try 4 and −7: 4 × (−7) = −28 and 4 + (−7) = −3 ✓.
  3. So 2(x + 4)(x − 7).
  4. Check with x = 0: 2(4)(−7) = −56 ✓ (the constant term).

2(x + 4)(x − 7)

Worked example

Example 5: Two variables

Factor x² + 2xy − 15y².

  1. Think of it as x² + (2y)x − 15y². The numbers must multiply to −15 and add to 2, then each gets a y.
  2. 5 and −3: product −15, sum 2 ✓.
  3. So (x + 5y)(x − 3y).
  4. Check: x² − 3xy + 5xy − 15y² = x² + 2xy − 15y² ✓.

(x + 5y)(x − 3y)

Worked example

Example 6: When it will not factor

Can x² + 3x + 5 be factored using integers?

  1. Product +5, sum +3. The only integer pairs with product 5 are 1 and 5 (sum 6) and −1 and −5 (sum −6).
  2. No pair has sum 3.

No. Not every trinomial factors over the integers. Later in this unit you will meet the quadratic formula, which works even when factoring does not.

Watch out

  • Common factor first, then factor the trinomial that is left.
  • Check the middle term when you expand: the sum is where most mistakes hide.
  • (x + 2)(x − 3) and (x − 3)(x + 2) are the same answer: the order of factors does not matter.
Hands-on activity

Product-sum card game

You need: a deck of playing cards with the face cards (jacks, queens, kings) removed, a partner, paper and a pencil.

Rules: black cards are positive and red cards are negative. An ace is 1.

  1. Player A secretly draws two cards and calculates their sum and product.
  2. Player A writes the trinomial x² + (sum)x + (product). For example, a black 4 and a red 6 give sum −2 and product −24, so x² − 2x − 24.
  3. Player B factors it. The answer must match the two cards: (x + 4)(x − 6).
  4. Swap roles. Score a point for each correct factoring. Play 10 rounds.

Model answer: if the cards are a red 3 and a red 5, the trinomial is x² − 8x + 15 = (x − 3)(x − 5).

Just for fun

Why do mathematicians love the X diagram?

Because in factoring, X really does mark the spot where the treasure (the two numbers) is hiding.

Your turn

1

Factor x² + 9x + 20.

2

Factor x² − 9x + 14.

3

Factor x² + 4x − 21.

4

Factor x² − 2x − 24.

5

Factor x² − 13x + 36.

6

Factor 3x² + 3x − 60 fully.

7

Factor a² − 5ab − 14b².

8

Can x² + 6x + 10 be factored over the integers? Explain.

More practice: warm-up

1

Find two numbers with product 15 and sum 8.

2

Find two numbers with product −12 and sum 1.

3

Factor x² + 6x + 8.

4

Factor x² − 8x + 15.

5

Factor x² + x − 2.

More practice: on your own

1

Factor x² + 10x + 16.

2

Factor −x² + 4x + 12 fully.

3

Factor x³ − 5x² − 36x fully.

4

Factor x² − 12xy + 35y².

5

A rectangle has area x² + 11x + 28 cm² and length (x + 7) cm. Find the width. Then find both dimensions and the area when x = 5.

Challenge

1

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

2

Factor (x + 1)² + 5(x + 1) + 6. (Hint: let u = x + 1.)

3

Factor x⁴ + 5x² + 4. (Hint: think of x⁴ as (x²)².)

Check your understanding

Quiz

1.

Factor x² + 8x + 15.

2.

Factor x² − 3x − 18.

3.

To factor x² − 11x + 24, which two numbers do you need?

4.

Which trinomial cannot be factored over the integers?

5.

Factor 2x² − 2x − 40 fully.

6.

If x² + bx + c = (x + m)(x + n), which statement is true?

For parents and teachers

Big idea: This lesson covers the "simple trinomial" part of expectation 3.2: factoring x² + bx + c by finding two numbers with product c and sum b, after taking out any common factor.

How to help: Play the card game together; it builds fast recall of factor pairs. When your teen is stuck, ask them to list all factor pairs of c and their sums in a little table.

Common difficulties:

  • Sign errors, especially when c is negative. The sign table in the lesson helps.
  • Forgetting to take out a common factor first (e.g. 3x² + 3x − 60).
  • Not checking the middle term. A quick expansion catches almost every mistake.