MPM2D Factoring Trinomials: Notes and Practice

A rectangular patio has an area of $x^{2} + 10 x + 16$ square metres. What could its length and width be? In the first lesson you saw that $(x + m) (x + n) = x^{2} + (m + n) x + mn$ : the middle coefficient is the sum of the two numbers and the constant is their product. So you need two numbers that multiply to 16 and add to 10. Those are 2 and 8, so the patio could be $(x + 2)$ m by $(x + 8)$ m. Check: $(x + 2) (x + 8) = x^{2} + 8 x + 2 x + 16 = x^{2} + 10 x + 16 .$ Turning a trinomial back into a product of two binomials is called factoring a trinomial. It is a key tool for solving quadratic equations and for finding where a parabola crosses the $x$ -axis, which you will do in the next two units.

Trinomials of the form x² + bx + c

x^{2} + b x + c = (x + m) (x + n) where m + n = b and mn = c

Search for $m$ and $n$ by listing the factor pairs of $c$ and adding each pair. The product condition limits you to a short list, and the sum picks out the right pair. For $x^{2} + 10 x + 16$ , the positive factor pairs of 16 are $1 \times 16$ (sum 17), $2 \times 8$ (sum 10) and $4 \times 4$ (sum 8). Only $2$ and $8$ add to 10, so $x^{2} + 10 x + 16 = (x + 2) (x + 8)$ . The order of the brackets does not matter: $(x + 8) (x + 2)$ is the same answer, because multiplication can be done in either order.

Fig 1Left: the factor pairs of −6 and their sums; only 2 and −3 add to −1, so x² − x − 6 = (x + 2)(x − 3). Right: a 2 by 2 box for 2x² + 11x + 5 with the middle term split into 10x and x, read off as (2x + 1)(x + 5).

The signs of $b$ and $c$ tell you the signs of the two numbers before you start listing. A positive product means the numbers have the same sign, and then the sum shows which sign. A negative product means opposite signs, and the number farther from zero takes the sign of $b$ (if $b = 0$ , the two numbers are opposites, such as $3$ and $- 3$ for $x^{2} - 9$ ). For $x^{2} - x - 6$ : the product $- 6$ is negative, so one number is positive and one negative, and the sum $- 1$ is negative, so the negative number is the larger in size. The pair is $2$ and $- 3$ : $2 \times (- 3) = - 6$ and $2 + (- 3) = - 1$ . So $x^{2} - x - 6 = (x + 2) (x - 3)$ .

Sign of c	Sign of b	The two numbers	Example
positive	positive	both positive	$x^{2} + 7 x + 12 = (x + 3) (x + 4)$
positive	negative	both negative	$x^{2} - 9 x + 20 = (x - 4) (x - 5)$
negative	positive	opposite signs; the one farther from zero is positive	$x^{2} + 2 x - 24 = (x + 6) (x - 4)$
negative	negative	opposite signs; the one farther from zero is negative	$x^{2} - x - 6 = (x + 2) (x - 3)$

Choosing the signs for x squared plus bx plus c

Common factor first, then check

Before anything else, take out any common factor.

3 x^{2} - 15 x + 18

looks hard, but

3 x^{2} - 15 x + 18 = 3 (x^{2} - 5 x + 6) = 3 (x - 2) (x - 3) .

If the

x^{2}

term is negative, take out

- 1

- x^{2} + 2 x + 15 = - (x^{2} - 2 x - 15) = - (x - 5) (x + 3) .

Finish every factorization by expanding it back to the original.

The same search works with two variables. In $x^{2} + 5 x y + 6 y^{2}$ you need two numbers with product 6 and sum 5, which are 2 and 3, and each one carries a $y$ : $x^{2} + 5 x y + 6 y^{2} = (x + 2 y) (x + 3 y) .$ Not every trinomial factors using integers. For $x^{2} + 4 x + 6$ the integer pairs with product 6 are $1, 6$ (sum 7), $2, 3$ (sum 5), $- 1, - 6$ (sum $- 7$ ) and $- 2, - 3$ (sum $- 5$ ). None adds to 4, so $x^{2} + 4 x + 6$ cannot be factored with integers. Once you have checked every pair, you can say so with confidence.

VideoAn Ontario teacher finds two numbers with the right product and sum to factor trinomials that start with x squared.Watch on YouTube

Trinomials of the form ax² + bx + c: decomposition

When the leading coefficient $a$ is not 1 (and there is no common factor to take out first), the sum and product search changes a little. Look at what expanding does: $(2 x + 1) (x + 5) = 2 x^{2} + 10 x + x + 5 = 2 x^{2} + 11 x + 5 .$ The two middle products, $10 x$ and $x$ , add to the middle term, and their coefficients multiply to $10 \times 1 = 10$ , which is also $2 \times 5$ , the first coefficient times the last. This works for every product of two binomials: in $(p x + q) (r x + s)$ the middle products have coefficients $p s$ and $q r$ , whose product $p s \cdot q r = (p r) (q s)$ is exactly $a \times c$ . So to factor $a x^{2} + b x + c$ , find two numbers whose product is $a c$ and whose sum is $b$ , use them to split (decompose) the middle term into two terms, and then factor by grouping, as in the last lesson.

Take out any common factor. Identify $a$ , $b$ and $c$ .
Calculate $a c$ . Find two integers with product $a c$ and sum $b$ . If no pair exists, the trinomial does not factor using integers.
Rewrite the middle term $b x$ as the sum of two terms using those numbers.
Group the four terms in pairs and take out the common factor of each pair. Both pairs should leave the same bracket.
Take out the common bracket, then check by expanding.

2 x^{2} + 11 x + 5 = 2 x^{2} + 10 x + x + 5 = 2 x (x + 5) + 1 (x + 5) = (x + 5) (2 x + 1)

The box on the right of the figure is the same idea drawn as an area model: put $a x^{2}$ and $c$ in opposite corners, the two split middle terms in the other corners, and read the common factor of each row and column. Either way, the most common slip is with signs, so the closing check is not optional. For $6 x^{2} + 7 x - 3$ : $a c = - 18$ , and the pair with product $- 18$ and sum $7$ is $9$ and $- 2$ . Then $6 x^{2} + 9 x - 2 x - 3 = 3 x (2 x + 3) - 1 (2 x + 3) = (2 x + 3) (3 x - 1) .$

Taking out a 1 or a negative 1

When a pair has no common factor except 1, write the 1 anyway:

x + 5 = 1 (x + 5)

. When the third term of the split is negative, as in

- 2 x - 3

, take out

- 1

so the bracket matches:

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

Factoring Trinomials

Key things to remember

Trinomials of the form x² + bx + c

Trinomials of the form ax² + bx + c: decomposition

Worked examples

Where marks get lost

Flashcards 8

Practice