Factorising x² + bx + c When c Is Positive

How to factorise a quadratic such as $x^2 + 7x + 12$ or $x^2 - 5x + 6$ into two brackets: two numbers whose product is $c$ and whose sum is $b$ , found with the frame and a list of factor pairs.

Before you start

In lesson 3.3 you expanded $(x + p)(x + q) = x^2 + (p + q)x + pq\text{.}$ For example, $(x + 2)(x + 3) = x^2 + 5x + 6\text{:}$ the number term $6$ is $2 \times 3$ , and the coefficient of $x$ , $5$ , is $2 + 3$ .

Factorising $x^2 + bx + c$ runs this backwards: we look for the two numbers.

The Frame Read Backwards

To factorise $x^2 + 5x + 6$ , start a multiplication frame. The $x^2$ can only come from $x \times x$ , so it goes in the top left cell with $x$ on both headers. The number $6$ goes in the bottom right cell: it is the product of the two numbers we are looking for.

Start: $x^2$ and $6$ placed, the two numbers still to be found.

×	$x$
$x$	$x^2$
		$+6$

The two empty cells will hold the two $x$ terms, which must add up to $5x$ . So we need two numbers whose product is $6$ and whose sum is $5$ . List the factor pairs of $6$ : $1 \times 6$ (sum $7$ ) and $2 \times 3$ (sum $5$ ). The pair is $2$ and $3$ .

Finished: every cell is its row times its column.

×	$x$	$+3$
$x$	$x^2$	$+3x$
$+2$	$+2x$	$+6$

$x^2 + 3x + 2x + 6 = x^2 + 5x + 6\text{,}$ so $x^2 + 5x + 6 = (x + 2)(x + 3)$

Factorising $x^2 + bx + c$

Find two numbers $p$ and $q$ with $p \times q = c$ and $p + q = b$ . Then $x^2 + bx + c = (x + p)(x + q)\text{.}$

List the factor pairs of $c$ in order, and stop at the pair with the right sum. Check by expanding.

Listing Factor Pairs

Example 1: Three pairs to try

Factorise $x^2 + 7x + 12$ .

Solution

We need two numbers with product $12$ and sum $7$ .

Factor pairs of $12$ :

Pair	Product	Sum
$1, 12$	$12$	$13$
$2, 6$	$12$	$8$
$3, 4$	$12$	$7$ ✓

The pair is $3$ and $4$ .

Check: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12\text{.}$ ✓

$x^2 + 7x + 12 = (x + 3)(x + 4)$

Example 2: Many pairs

Factorise $x^2 + 13x + 36$ .

Solution

Pairs of $36$ , with their sums: $1, 36$ ( $37$ ); $2, 18$ ( $20$ ); $3, 12$ ( $15$ ); $4, 9$ ( $13$ ) ✓.

Listing in order means no pair is missed; we can stop at $4, 9$ .

$x^2 + 13x + 36 = (x + 4)(x + 9)$

When $b$ Is Negative

In $x^2 - 5x + 6$ the product is still $+6$ , but the sum must be $-5$ . Reason with the signs:

The signs when $c$ is positive

The product $c$ is positive, so the two numbers have the same sign.

If $b$ is positive, both numbers are positive: $(x + p)(x + q)$ .

If $b$ is negative, both numbers are negative: $(x - p)(x - q)$ .

Example 3: Both numbers negative

Factorise $x^2 - 5x + 6$ .

Solution

The product $+6$ is positive, so the numbers have the same sign; the sum $-5$ is negative, so both are negative.

Negative pairs of $6$ : $-1, -6$ (sum $-7$ ); $-2, -3$ (sum $-5$ ) ✓.

×	$x$	$-3$
$x$	$x^2$	$-3x$
$-2$	$-2x$	$+6$

$x^2 - 3x - 2x + 6 = x^2 - 5x + 6$

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Example 4: A larger number

Factorise $x^2 - 20x + 96$ .

Solution

Both numbers are negative, with product $96$ and sum $-20$ .

$-1, -96$ ( $-97$ ); $-2, -48$ ( $-50$ ); $-3, -32$ ( $-35$ ); $-4, -24$ ( $-28$ ); $-6, -16$ ( $-22$ ); $-8, -12$ ( $-20$ ) ✓.

Check: $(x - 8)(x - 12) = x^2 - 12x - 8x + 96 = x^2 - 20x + 96\text{.}$ ✓

$x^2 - 20x + 96 = (x - 8)(x - 12)$

Other Letters, Other Orders and Two Variables

Example 5: Another letter, and the number first

Factorise (a) $y^2 - 9y + 20$ ; (b) $12 + 7x + x^2$ .

Solution

(a) Both negative, product $20$ , sum $-9$ : $-4$ and $-5$ . So $(y - 4)(y - 5)$ .

(b) Write it in descending powers first: $x^2 + 7x + 12$ . From Example 1, $(x + 3)(x + 4)$ .

(a) $(y - 4)(y - 5)$ ; (b) $(x + 3)(x + 4)$

Example 6: Two variables

Factorise (a) $x^2 + 5xy + 6y^2$ ; (b) $a^2 - 7ab + 10b^2$ .

Solution

(a) The same numbers as $x^2 + 5x + 6$ , but each number now goes with $y$ : $2y \times 3y = 6y^2$ and $2xy + 3xy = 5xy$ .

×	$x$	$+3y$
$x$	$x^2$	$+3xy$
$+2y$	$+2xy$	$+6y^2$

$x^2 + 3xy + 2xy + 6y^2 = x^2 + 5xy + 6y^2$

(b) Product $10$ , sum $-7$ : $-2$ and $-5$ . So $(a - 2b)(a - 5b)$ .

(a) $(x + 2y)(x + 3y)$ ; (b) $(a - 2b)(a - 5b)$

A Perfect Square, and Quadratics That Do Not Factorise

Example 7: The two numbers are equal

Factorise $x^2 - 10x + 25$ .

Solution

Both negative, product $25$ , sum $-10$ : $-5$ and $-5$ .

$(x - 5)(x - 5) = (x - 5)^2\text{,}$ a perfect square, as in lesson 3.7.

$x^2 - 10x + 25 = (x - 5)^2$

Example 8: No pair works

Can $x^2 + 4x + 5$ be factorised?

Solution

We need product $5$ and sum $4$ . The pairs of $5$ are $1, 5$ (sum $6$ ) and $-1, -5$ (sum $-6$ ).

The list is complete and no pair has sum $4$ .

$x^2 + 4x + 5$ cannot be factorised into brackets with integers.

Unknown Coefficients and Areas

Example 9: All the values of $k$

Find all the integers $k$ for which $x^2 + kx + 12$ can be factorised into two brackets with integers.

Solution

$k$ is the sum of two integers whose product is $12$ . The pairs are $1, 12$ ; $2, 6$ ; $3, 4$ and their negatives.

Sums: $13$ , $8$ , $7$ , and $-13$ , $-8$ , $-7$ .

$k = 7, 8, 13, -7, -8$ or $-13$

Example 10: The other side of a rectangle

A rectangle has area $(x^2 + 7x + 12)$ cm² and length $(x + 4)$ cm. Find its breadth.

Solution

$x^2 + 7x + 12 = (x + 3)(x + 4)\text{,}$ so the breadth is $(x + 3)$ cm.

Check with $x = 2$ : area $4 + 14 + 12 = 30$ cm², and $6 \times 5 = 30$ . ✓

The breadth is $(x + 3)$ cm.

Common Mistakes

Using a pair with the right product but the wrong sum: $x^2 + 7x + 10 \ne (x + 1)(x + 10)\text{,}$ since $1 + 10 = 11$ . It is $(x + 2)(x + 5)$ .
Using positive numbers when $b$ is negative: $x^2 - 6x + 8 = (x - 2)(x - 4)\text{,}$ not $(x + 2)(x + 4)$ .
Writing $(x - 2)(x + 2)$ for $x^2 - 4x + 4$ . That is $x^2 - 4$ ; the answer is $(x - 2)^2$ .
Missing a pair by not listing them in order, and then saying a quadratic cannot be factorised.
Forgetting to put the terms in descending powers first: $12 + 7x + x^2$ is $x^2 + 7x + 12$ .

Practice

Try each question before you open its solution. Check each answer by expanding it.

Question 1

Factorise (a) $x^2 + 6x + 8$ ; (b) $x^2 + 11x + 24$ ; (c) $x^2 + 9x + 20$ .

ShowHide solution

(a) Product $8$ , sum $6$ : $2, 4$ .

(b) Product $24$ , sum $11$ : $1, 24$ ( $25$ ); $2, 12$ ( $14$ ); $3, 8$ ( $11$ ) ✓.

(a) $(x + 2)(x + 4)$ ; (b) $(x + 3)(x + 8)$ ; (c) $(x + 4)(x + 5)$

Question 2

Factorise (a) $x^2 - 7x + 10$ ; (b) $x^2 - 8x + 12$ ; (c) $x^2 - 9x + 8$ .

ShowHide solution

(a) Both negative, product $10$ , sum $-7$ : $-2, -5$ .

(b) Product $12$ , sum $-8$ : $-2, -6$ .

(a) $(x - 2)(x - 5)$ ; (b) $(x - 2)(x - 6)$ ; (c) $(x - 1)(x - 8)$

Question 3

Factorise (a) $x^2 + 13x + 40$ ; (b) $x^2 - 15x + 56$ ; (c) $x^2 - 14x + 48$ .

ShowHide solution

(a) $1, 40$ ( $41$ ); $2, 20$ ( $22$ ); $4, 10$ ( $14$ ); $5, 8$ ( $13$ ) ✓.

(b) Both negative: $(-7) \times (-8) = 56$ and $-7 - 8 = -15$ .

(a) $(x + 5)(x + 8)$ ; (b) $(x - 7)(x - 8)$ ; (c) $(x - 6)(x - 8)$

Question 4

Factorise $x^2 + 8x + 15$ using a multiplication frame.

ShowHide solution

Product $15$ , sum $8$ : $3$ and $5$ .

×	$x$	$+5$
$x$	$x^2$	$+5x$
$+3$	$+3x$	$+15$

$x^2 + 5x + 3x + 15 = x^2 + 8x + 15$

$(x + 3)(x + 5)$

Question 5

Factorise (a) $p^2 + 11p + 24$ ; (b) $21 - 10x + x^2$ ; (c) $18 + 9y + y^2$ .

ShowHide solution

(a) $3 \times 8 = 24$ and $3 + 8 = 11$ .

(b) $x^2 - 10x + 21$ : $(-3) \times (-7) = 21$ and $-3 - 7 = -10$ .

(a) $(p + 3)(p + 8)$ ; (b) $(x - 3)(x - 7)$ ; (c) $(y + 3)(y + 6)$

Question 6

Factorise (a) $x^2 + 7xy + 12y^2$ ; (b) $m^2 - 9mn + 14n^2$ .

ShowHide solution

(a) Product $12$ , sum $7$ : $3$ and $4$ , each with $y$ .

(b) Product $14$ , sum $-9$ : $-2$ and $-7$ , each with $n$ .

(a) $(x + 3y)(x + 4y)$ ; (b) $(m - 2n)(m - 7n)$

Question 7

Factorise (a) $x^2 + 12x + 36$ ; (b) $y^2 - 16y + 64$ .

ShowHide solution

(a) $6 \times 6 = 36$ and $6 + 6 = 12$ : $(x + 6)(x + 6)$ .

(b) $(-8) \times (-8) = 64$ and $-8 - 8 = -16$ : $(y - 8)(y - 8)$ .

(a) $(x + 6)^2$ ; (b) $(y - 8)^2$

Question 8

Which of these can be factorised? Factorise those that can. (a) $x^2 + 5x + 3$ ; (b) $x^2 - 13x + 30$ ; (c) $x^2 + 2x + 2$

ShowHide solution

(a) Pairs of $3$ : $1, 3$ (sum $4$ ) and $-1, -3$ (sum $-4$ ). None has sum $5$ : it cannot be factorised.

(b) $(-3) \times (-10) = 30$ and $-3 - 10 = -13$ : $(x - 3)(x - 10)$ .

Only (b): $(x - 3)(x - 10)$ .

Question 9

Find all the integers $k$ for which $x^2 + kx + 18$ can be factorised into two brackets with integers.

ShowHide solution

Pairs of $18$ : $1, 18$ (sum $19$ ); $2, 9$ (sum $11$ ); $3, 6$ (sum $9$ ); and their negatives (sums $-19$ , $-11$ , $-9$ ).

$k = 9, 11, 19, -9, -11$ or $-19$

Question 10

Find all the positive integers $c$ for which $x^2 + 8x + c$ can be factorised into two brackets with integers.

ShowHide solution

We need $p + q = 8$ with $pq = c$ positive, so $p$ and $q$ are both positive integers.

$1 + 7$ : $c = 7$ . $2 + 6$ : $c = 12$ . $3 + 5$ : $c = 15$ . $4 + 4$ : $c = 16$ .

$c = 7, 12, 15$ or $16$

Question 11

A rectangular garden has area $(x^2 + 9x + 14)$ m² and width $(x + 2)$ m. (a) Find its length. (b) Find its perimeter. (c) Check both when $x = 3$ .

ShowHide solution

(a) $x^2 + 9x + 14 = (x + 2)(x + 7)\text{,}$ so the length is $(x + 7)$ m.

(b) $2(x + 2) + 2(x + 7) = 4x + 18$ m.

(a) $(x + 7)$ m; (b) $(4x + 18)$ m

Question 12

Each line has one mistake. Find it and correct it. (a) $x^2 + 10x + 16 = (x + 1)(x + 16)\text{;}$ (b) $x^2 - 12x + 27 = (x + 3)(x + 9)\text{;}$ (c) $x^2 - 6x + 9 = (x - 3)(x + 3)$

ShowHide solution

(a) $1 \times 16 = 16$ , but $1 + 16 = 17$ , not $10$ . The pair is $2, 8$ : $(x + 2)(x + 8)$ .

(b) $(x + 3)(x + 9) = x^2 + 12x + 27\text{.}$ For $-12x$ both numbers are negative: $(x - 3)(x - 9)$ .

(a) $(x + 2)(x + 8)$ ; (b) $(x - 3)(x - 9)$ ; (c) $(x - 3)^2$

Key Points

$x^2 + bx + c = (x + p)(x + q)\text{,}$ where $pq = c$ and $p + q = b$ .
In the frame, $x^2$ goes top left and $c$ bottom right; the two $x$ terms on the other diagonal add to $bx$ .
When $c$ is positive the two numbers have the same sign: both positive if $b > 0$ , both negative if $b < 0$ .
List the factor pairs of $c$ in order so that none is missed.
Equal numbers give a perfect square, such as $(x - 5)^2$ ; if no pair has sum $b$ , the quadratic does not factorise with integers.
Check by expanding.

Factorising x² + bx + c When c Is Positive

Before you start

The Frame Read Backwards

Factorising x2+bx+cx^2 + bx + cx2+bx+c

Listing Factor Pairs

Example 1: Three pairs to try

Example 2: Many pairs

When bbb Is Negative

The signs when ccc is positive

Example 3: Both numbers negative

Example 4: A larger number

Other Letters, Other Orders and Two Variables

Example 5: Another letter, and the number first

Example 6: Two variables

A Perfect Square, and Quadratics That Do Not Factorise

Example 7: The two numbers are equal

Example 8: No pair works

Unknown Coefficients and Areas

Example 9: All the values of kkk

Example 10: The other side of a rectangle

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

Factorising $x^2 + bx + c$

When $b$ Is Negative

The signs when $c$ is positive

Example 9: All the values of $k$