Factorising ax² + bx + c

How to factorise a quadratic such as $2x^2 + 7x - 15$ , where the coefficient of $x^2$ is not 1: by guess and check in the frame, by two numbers with product $ac$ and sum $b$ , and by splitting the middle term.

Before you start

In lessons 3.9 and 3.10 you factorised $x^2 + bx + c$ with two numbers whose product is $c$ and whose sum is $b$ . In lesson 3.8 you factorised four terms by grouping: $2x(x + 5) - 3(x + 5) = (2x - 3)(x + 5)\text{.}$

Now the $x^2$ term has a coefficient $a$ other than $1$ . Both ideas come back.

Guess and Check with the Frame

To factorise $2x^2 + 7x - 15$ : since $2x^2 = 2x \times x$ , the brackets look like $(2x \;\square)(x \;\square)$ . The two numbers multiply to $-15$ . We try a pair in the frame and check the $x$ terms.

Example 1: Two trials

Factorise $2x^2 + 7x - 15$ by guess and check.

Solution

Trial: $(2x + 5)(x - 3)$ .

×	$x$	$-3$
$2x$	$2x^2$	$-6x$
$+5$	$+5x$	$-15$

$x$ terms: $-6x + 5x = -x$ , not $7x$ ✗

Swap the numbers: $(2x - 3)(x + 5)$ .

×	$x$	$+5$
$2x$	$2x^2$	$+10x$
$-3$	$-3x$	$-15$

$x$ terms: $10x - 3x = 7x$ ✓

The two trials use the same numbers but give different middle terms, because the number in the second bracket is multiplied by $2x$ , not by $x$ : $2x \times 5 = 10x$ , but $2x \times (-3) = -6x$ .

In the worst case there are $8$ trials here: four pairs with product $-15$ ( $1, -15$ ; $-1, 15$ ; $3, -5$ ; $-3, 5$ ), each placed two ways. The next method avoids the guessing.

$2x^2 + 7x - 15 = (2x - 3)(x + 5)$

Two Numbers with Product $ac$ and Sum $b$

Look at the finished frame for $(2x - 3)(x + 5)$ again. Multiply the two cells on each diagonal: $2x^2 \times (-15) = -30x^2$ and $10x \times (-3x) = -30x^2\text{.}$ They are equal.

This happens in every frame. For $(px + q)(rx + s)$ the cells are $prx^2$ , $psx$ , $qrx$ and $qs$ , and each diagonal multiplies to $pqrs\,x^2$ . So the two $x$ terms have coefficients whose product is $pr \times qs = a \times c$ and whose sum is $b$ .

Factorising $ax^2 + bx + c$ with the frame

1. Take out the highest common factor (HCF) of all three terms first, if there is one (see Example 9), and work on the quadratic left in the bracket.

2. Work out $ac$ . Find two numbers whose product is $ac$ and whose sum is $b$ .

3. Put $ax^2$ in the top left cell and $c$ in the bottom right cell. Put the two $x$ terms in the other two cells.

4. The highest common factor (HCF) of the top row is the first header on the left. Divide each top-row cell by it to get the headers along the top; divide the bottom-left cell by the first top header to get the last header.

5. Check that every cell is its row times its column. The headers are the two brackets.

Example 2: The frame method

Factorise $2x^2 + 7x - 15$ using $ac$ .

Solution

Step 1: $2$ , $7$ and $-15$ have no common factor other than $1$ , so there is nothing to take out.

Step 2: $ac = 2 \times (-15) = -30\text{.}$ Pairs with product $-30$ : $1, -30$ ( $-29$ ); $-1, 30$ ( $29$ ); $2, -15$ ( $-13$ ); $-2, 15$ ( $13$ ); $3, -10$ ( $-7$ ); $-3, 10$ ( $7$ ) ✓. The numbers are $-3$ and $10$ .

Step 3: $2x^2$ top left, $-15$ bottom right, $10x$ and $-3x$ on the other diagonal. The headers are not known yet:

×
	$2x^2$	$+10x$
	$-3x$	$-15$

Step 4: The top row $2x^2$ , $10x$ has HCF $2x$ : that is the first left header. Then $2x^2 \div 2x = x$ and $10x \div 2x = 5$ are the top headers. Then $-3x \div x = -3$ is the second left header.

×	$x$	$+5$
$2x$	$2x^2$	$+10x$
$-3$	$-3x$	$-15$

Check the last cell: $(-3) \times 5 = -15$ ✓

$2x^2 + 7x - 15 = (2x - 3)(x + 5)$

Example 3: A positive number term

Factorise $3x^2 + 10x + 8$ .

Solution

$ac = 3 \times 8 = 24$ ; product $24$ , sum $10$ : $4$ and $6$ .

Top row $3x^2$ , $6x$ : HCF $3x$ . Top headers $x$ and $2$ . Then $4x \div x = 4$ ; check $4 \times 2 = 8$ . ✓

×	$x$	$+2$
$3x$	$3x^2$	$+6x$
$+4$	$+4x$	$+8$

$3x^2 + 6x + 4x + 8 = 3x^2 + 10x + 8$

$3x^2 + 10x + 8 = (3x + 4)(x + 2)$

Splitting the Middle Term

The same two numbers can be used without a frame. Write the middle term as the two $x$ terms, then factorise the four terms by grouping, as in lesson 3.8.

Splitting the middle term

First take out the HCF of all three terms, if there is one. Find two numbers with product $ac$ and sum $b$ . Use them to split $bx$ into two terms. Then group in pairs and take out the common bracket.

Example 4: Split, then group

Factorise (a) $2x^2 + 7x - 15$ ; (b) $6x^2 - 13x + 6$ by splitting the middle term.

Solution

(a) The numbers are $10$ and $-3$ (Example 2): $7x = 10x - 3x$ .

$2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5) = (2x - 3)(x + 5)$

(b) $ac = 36$ ; product $36$ , sum $-13$ : $-4$ and $-9$ . So $-13x = -4x - 9x$ .

$6x^2 - 4x - 9x + 6 = 2x(3x - 2) - 3(3x - 2) = (2x - 3)(3x - 2)$

Check (b): $(2x - 3)(3x - 2) = 6x^2 - 4x - 9x + 6 = 6x^2 - 13x + 6\text{.}$ ✓

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

The order of the split does not matter

$2x^2 - 3x + 10x - 15 = x(2x - 3) + 5(2x - 3) = (x + 5)(2x - 3)\text{:}$ the same answer.

When $a$ and $c$ Have Several Factor Pairs

Example 5: Larger values of $ac$

Factorise (a) $6x^2 + x - 2$ ; (b) $12x^2 - 17x + 6$ ; (c) $4x^2 - 4x - 15$ .

Solution

(a) $ac = -12$ ; product $-12$ , sum $1$ : $4$ and $-3$ .

$6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2)$

(b) $ac = 72$ ; product $72$ , sum $-17$ : both negative, $-8$ and $-9$ .

$12x^2 - 8x - 9x + 6 = 4x(3x - 2) - 3(3x - 2) = (4x - 3)(3x - 2)$

$4x^2 - 10x + 6x - 15 = 2x(2x - 5) + 3(2x - 5) = (2x + 3)(2x - 5)$

(a) $(2x - 1)(3x + 2)$ ; (b) $(4x - 3)(3x - 2)$ ; (c) $(2x + 3)(2x - 5)$

A Negative $x^2$ Term, and the Number First

When the $x^2$ term is negative, take out $-1$ first, so that the quadratic inside the bracket has a positive $x^2$ term. Write the terms in descending powers before you start.

Example 6: Take out $-1$

Factorise (a) $-2x^2 - 7x + 15$ ; (b) $3 + 5x - 2x^2$ .

Solution

(a) $-2x^2 - 7x + 15 = -(2x^2 + 7x - 15) = -(2x - 3)(x + 5)\text{.}$

The minus sign can go into the first bracket: $-(2x - 3) = 3 - 2x$ . So the answer can also be written $(3 - 2x)(x + 5)$ .

(b) In descending powers: $-2x^2 + 5x + 3 = -(2x^2 - 5x - 3)\text{.}$

$2x^2 - 5x - 3$ : $ac = -6$ , sum $-5$ : $-6$ and $1$ . $2x^2 - 6x + x - 3 = 2x(x - 3) + (x - 3) = (2x + 1)(x - 3)\text{.}$

So $3 + 5x - 2x^2 = -(2x + 1)(x - 3) = (2x + 1)(3 - x)\text{.}$

Check (b): $(2x + 1)(3 - x) = 6x - 2x^2 + 3 - x = -2x^2 + 5x + 3\text{.}$ ✓

(a) $-(2x - 3)(x + 5)$ , or $(3 - 2x)(x + 5)$ ; (b) $(2x + 1)(3 - x)$

Two Variables, and a Perfect Square

Example 7: Two variables

Factorise (a) $2x^2 + 5xy - 3y^2$ ; (b) $6a^2 - ab - 2b^2$ .

Solution

(a) Work with the numbers $2$ , $5$ , $-3$ : $ac = -6$ , sum $5$ : $6$ and $-1$ . Each $x$ term now carries $y$ .

$2x^2 + 6xy - xy - 3y^2 = 2x(x + 3y) - y(x + 3y) = (2x - y)(x + 3y)$

(b) $ac = -12$ , sum $-1$ : $-4$ and $3$ .

$6a^2 - 4ab + 3ab - 2b^2 = 2a(3a - 2b) + b(3a - 2b) = (2a + b)(3a - 2b)$

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

Example 8: An identity met this way

Factorise $4x^2 + 12x + 9$ .

Solution

$ac = 36$ , sum $12$ : $6$ and $6$ . $4x^2 + 6x + 6x + 9 = 2x(2x + 3) + 3(2x + 3) = (2x + 3)(2x + 3)\text{.}$

This is the perfect square $(2x + 3)^2$ , which lesson 3.7 finds at once: $4x^2 = (2x)^2$ , $9 = 3^2$ and $2 \times 2x \times 3 = 12x$ .

$4x^2 + 12x + 9 = (2x + 3)^2$

A Common Factor First

Before you look for two brackets, look for a common factor of all three terms and take it out. Otherwise the numbers are larger than they need to be, and a bracket can be left with a common factor inside it.

Example 9: Take out the common factor

Factorise (a) $4c^2 + 10c + 4$ ; (b) $5d^2 - 145d + 600$ .

Solution

(a) Each term is even: $4c^2 + 10c + 4 = 2(2c^2 + 5c + 2)\text{.}$ For $2c^2 + 5c + 2$ : $ac = 4$ , sum $5$ : $1$ and $4$ .

$2c^2 + c + 4c + 2 = c(2c + 1) + 2(2c + 1) = (c + 2)(2c + 1)\text{.}$ So $4c^2 + 10c + 4 = 2(2c + 1)(c + 2)\text{.}$

Note: $(2c + 1)(2c + 4)$ also expands to $4c^2 + 10c + 4$ , but it is not finished: $2c + 4$ still has the common factor $2$ .

(b) Each term is a multiple of $5$ : $5(d^2 - 29d + 120)$ . Product $120$ , sum $-29$ : both negative; $(-5) \times (-24) = 120$ and $-5 - 24 = -29$ .

So $5d^2 - 145d + 600 = 5(d - 5)(d - 24)\text{.}$ Check: $5(d^2 - 29d + 120) = 5d^2 - 145d + 600\text{.}$ ✓

(a) $2(2c + 1)(c + 2)$ ; (b) $5(d - 5)(d - 24)$

Quadratics That Do Not Factorise, Unknowns and Areas

Example 10: No pair works

Can $2x^2 + 3x + 4$ be factorised?

Solution

$ac = 8$ . Pairs with product $8$ : $1, 8$ (sum $9$ ); $2, 4$ (sum $6$ ); $-1, -8$ (sum $-9$ ); $-2, -4$ (sum $-6$ ). None has sum $3$ .

$2x^2 + 3x + 4$ cannot be factorised into brackets with integers.

Example 11: The missing coefficient

Given that $3x^2 + kx - 10 = (3x + 5)(x - 2)$ for all values of $x$ , find $k$ .

Solution

$(3x + 5)(x - 2) = 3x^2 - 6x + 5x - 10 = 3x^2 - x - 10\text{.}$

Compare the $x$ terms: $k = -1$ .

$k = -1$

Example 12: The other side of a rectangle

A rectangle has area $(2x^2 + 7x + 3)$ cm² and one side $(2x + 1)$ cm. Find the other side.

Solution

$ac = 6$ , sum $7$ : $1$ and $6$ . $2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)\text{.}$

Check with $x = 3$ : area $18 + 21 + 3 = 42$ cm², and $7 \times 6 = 42$ . ✓

The other side is $(x + 3)$ cm.

Common Mistakes

Using product $c$ instead of $ac$ : for $2x^2 + 7x - 15$ the two numbers multiply to $2 \times (-15) = -30$ .
Putting the numbers in the wrong brackets: $(2x + 5)(x - 3) = 2x^2 - x - 15\text{,}$ not $2x^2 + 7x - 15$ . Check by expanding.
Leaving a common factor in a bracket: $(2c + 1)(2c + 4)$ is not finished; the answer is $2(2c + 1)(c + 2)$ .
Dropping the minus sign taken out: $-2x^2 - 7x + 15 = -(2x - 3)(x + 5)\text{,}$ not $(2x - 3)(x + 5)$ .
In the $ac$ frame, writing the headers before the cells are filled. Fill the cells first; the headers come from the common factors. (In guess and check the headers are the guess, and the cells test it.)

Practice

Try each question before you open its solution. Use whichever method you like, and check each answer by expanding it.

Question 1

Factorise $2x^2 + 5x + 3$ by guess and check in a frame.

ShowHide solution

$2x^2 = 2x \times x$ ; the numbers multiply to $3$ : try $1$ and $3$ , both positive.

Trial $(2x + 1)(x + 3)$ : $x$ terms $6x + x = 7x$ ✗. Trial $(2x + 3)(x + 1)$ :

×	$x$	$+1$
$2x$	$2x^2$	$+2x$
$+3$	$+3x$	$+3$

$x$ terms: $2x + 3x = 5x$ ✓

$(2x + 3)(x + 1)$

Question 2

Factorise (a) $3x^2 + 7x - 6$ ; (b) $5x^2 - 13x - 6$ .

ShowHide solution

(a) $ac = -18$ , sum $7$ : $9$ and $-2$ . $3x^2 + 9x - 2x - 6 = 3x(x + 3) - 2(x + 3) = (3x - 2)(x + 3)\text{.}$

(b) $ac = -30$ , sum $-13$ : $-15$ and $2$ . $5x^2 - 15x + 2x - 6 = 5x(x - 3) + 2(x - 3) = (5x + 2)(x - 3)\text{.}$

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

Question 3

Factorise $2x^2 + 11x + 12$ using the frame method with $ac$ .

ShowHide solution

$ac = 24$ , sum $11$ : $3$ and $8$ . Top row $2x^2$ , $8x$ : HCF $2x$ ; top headers $x$ and $4$ ; then $3x \div x = 3$ ; check $3 \times 4 = 12$ . ✓

×	$x$	$+4$
$2x$	$2x^2$	$+8x$
$+3$	$+3x$	$+12$

$2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12$

$(2x + 3)(x + 4)$

Question 4

Factorise by splitting the middle term: (a) $6x^2 + 7x + 2$ ; (b) $10x^2 - 11x - 6$ .

ShowHide solution

(a) $ac = 12$ , sum $7$ : $3$ and $4$ . $6x^2 + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1)\text{.}$

(b) $ac = -60$ , sum $-11$ : $-15$ and $4$ . $10x^2 - 15x + 4x - 6 = 5x(2x - 3) + 2(2x - 3) = (5x + 2)(2x - 3)\text{.}$

(a) $(2x + 1)(3x + 2)$ ; (b) $(5x + 2)(2x - 3)$

Question 5

Factorise (a) $6x^2 - 5x - 6$ ; (b) $8x^2 - 14x + 3$ .

ShowHide solution

(a) $ac = -36$ , sum $-5$ : $-9$ and $4$ . $6x^2 - 9x + 4x - 6 = 3x(2x - 3) + 2(2x - 3) = (3x + 2)(2x - 3)\text{.}$

(b) $ac = 24$ , sum $-14$ : $-12$ and $-2$ . $8x^2 - 12x - 2x + 3 = 4x(2x - 3) - 1(2x - 3) = (4x - 1)(2x - 3)\text{.}$

(a) $(2x - 3)(3x + 2)$ ; (b) $(2x - 3)(4x - 1)$

Question 6

Factorise (a) $-3x^2 + 5x + 2$ ; (b) $2 + x - 6x^2$ .

ShowHide solution

(a) $-(3x^2 - 5x - 2)$ . $ac = -6$ , sum $-5$ : $-6$ and $1$ . $3x^2 - 6x + x - 2 = 3x(x - 2) + (x - 2) = (3x + 1)(x - 2)\text{.}$

So $-(3x + 1)(x - 2) = (3x + 1)(2 - x)\text{.}$

(b) $-6x^2 + x + 2 = -(6x^2 - x - 2)\text{.}$ $ac = -12$ , sum $-1$ : $-4$ and $3$ . $6x^2 - 4x + 3x - 2 = 2x(3x - 2) + (3x - 2) = (2x + 1)(3x - 2)\text{.}$

So $-(2x + 1)(3x - 2) = (2x + 1)(2 - 3x)\text{.}$

(a) $(3x + 1)(2 - x)$ ; (b) $(2x + 1)(2 - 3x)$

Question 7

Factorise (a) $3a^2 + 5ab - 2b^2$ ; (b) $4x^2 - 9xy + 2y^2$ .

ShowHide solution

(a) $ac = -6$ , sum $5$ : $6$ and $-1$ . $3a^2 + 6ab - ab - 2b^2 = 3a(a + 2b) - b(a + 2b) = (3a - b)(a + 2b)\text{.}$

(b) $ac = 8$ , sum $-9$ : $-8$ and $-1$ . $4x^2 - 8xy - xy + 2y^2 = 4x(x - 2y) - y(x - 2y) = (4x - y)(x - 2y)\text{.}$

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

Question 8

Factorise (a) $9x^2 - 12x + 4$ ; (b) $4x^2 - 25$ , using $ac$ (think of (b) as $4x^2 + 0x - 25$ ). Then say which identity each one is.

ShowHide solution

(a) $ac = 36$ , sum $-12$ : $-6$ and $-6$ . $9x^2 - 6x - 6x + 4 = 3x(3x - 2) - 2(3x - 2) = (3x - 2)^2\text{:}$ a perfect square.

(b) $ac = -100$ , sum $0$ : $10$ and $-10$ . $4x^2 + 10x - 10x - 25 = 2x(2x + 5) - 5(2x + 5) = (2x - 5)(2x + 5)\text{:}$ a difference of two squares.

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

Question 9

Take out the common factor first, then factorise: (a) $6x^2 + 9x + 3$ ; (b) $4x^2 - 2x - 12$ .

ShowHide solution

(a) $3(2x^2 + 3x + 1)$ . $ac = 2$ , sum $3$ : $1$ and $2$ . $2x^2 + x + 2x + 1 = x(2x + 1) + (2x + 1)\text{.}$ So $3(2x + 1)(x + 1)$ .

(b) $2(2x^2 - x - 6)$ . $ac = -12$ , sum $-1$ : $-4$ and $3$ . $2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2)\text{.}$ So $2(2x + 3)(x - 2)$ .

(a) $3(2x + 1)(x + 1)$ ; (b) $2(2x + 3)(x - 2)$

Question 10

Which of these can be factorised? Factorise those that can. (a) $3x^2 + 2x + 5$ ; (b) $3x^2 + 8x + 5$

ShowHide solution

(a) $ac = 15$ : pairs $1, 15$ (sum $16$ ), $3, 5$ (sum $8$ ) and their negatives. None has sum $2$ : it cannot be factorised.

(b) $ac = 15$ , sum $8$ : $3$ and $5$ . $3x^2 + 3x + 5x + 5 = 3x(x + 1) + 5(x + 1) = (3x + 5)(x + 1)\text{.}$

Only (b): $(3x + 5)(x + 1)$ .

Question 11

Given that $2x^2 + kx - 12 = (2x + 3)(x - 4)$ for all values of $x$ , find $k$ .

ShowHide solution

$(2x + 3)(x - 4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12\text{,}$ so $k = -5$ .

$k = -5$

Question 12

A rectangular poster has area $(6x^2 + 11x + 3)$ cm² and width $(2x + 3)$ cm. Find its length, and check your answer when $x = 1$ .

ShowHide solution

$ac = 18$ , sum $11$ : $2$ and $9$ . $6x^2 + 2x + 9x + 3 = 2x(3x + 1) + 3(3x + 1) = (2x + 3)(3x + 1)\text{.}$

The length is $(3x + 1)$ cm. When $x = 1$ : area $6 + 11 + 3 = 20$ cm², and $(2 + 3)(3 + 1) = 5 \times 4 = 20\text{.}$ ✓

The length is $(3x + 1)$ cm.

Key Points

Take out any common factor of all three terms first.
For $ax^2 + bx + c$ , find two numbers whose product is $ac$ and whose sum is $b$ .
In the frame, put $ax^2$ and $c$ on one diagonal and the two $x$ terms on the other; the headers come from the common factors.
Or split $bx$ into the two $x$ terms and factorise by grouping.
If the $x^2$ term is negative, take out $-1$ first.
Check by expanding.

Factorising ax² + bx + c

Before you start

Guess and Check with the Frame

Example 1: Two trials

Two Numbers with Product acacac and Sum bbb

Factorising ax2+bx+cax^2 + bx + cax2+bx+c with the frame

Example 2: The frame method

Example 3: A positive number term

Splitting the Middle Term

Splitting the middle term

Example 4: Split, then group

The order of the split does not matter

When aaa and ccc Have Several Factor Pairs

Example 5: Larger values of acacac

A Negative x2x^2x2 Term, and the Number First

Example 6: Take out −1-1−1

Two Variables, and a Perfect Square

Example 7: Two variables

Example 8: An identity met this way

A Common Factor First

Example 9: Take out the common factor

Quadratics That Do Not Factorise, Unknowns and Areas

Example 10: No pair works

Example 11: The missing coefficient

Example 12: 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

Two Numbers with Product $ac$ and Sum $b$

Factorising $ax^2 + bx + c$ with the frame

When $a$ and $c$ Have Several Factor Pairs

Example 5: Larger values of $ac$

A Negative $x^2$ Term, and the Number First

Example 6: Take out $-1$