Rearranging Quadratic Equations Before Factorising

How to make one side of a quadratic equation $0$ before you factorise: collecting terms, expanding brackets, a coefficient of $x^2$ other than $1$ , a common number factor, and why you never divide by $x$ .

Before you start

In lesson 5.1 you solved equations such as $x^2 - 2x - 15 = 0$ : factorise, $(x - 5)(x + 3) = 0$ ; set each factor equal to $0$ ; solve, $x = 5$ or $x = -3$ . That works only because one side is $0$ .
Doing the same to both sides keeps an equation true: adding or subtracting the same term, or multiplying or dividing by the same number that is not $0$ .
In chapter 3 you factorised $ax^2 + bx + c$ with $a \neq 1$ by finding two numbers whose product is $ac$ and whose sum is $b$ , and splitting the middle term.

Making One Side Zero

In $x^2 + 3x = 4$ the right side is $4$ , not $0$ , so factorising the left side tells you nothing: two numbers that multiply to $4$ could be anything. First collect every term on one side.

Rearranging before you factorise

1. Expand any brackets.

2. Collect all the terms on one side by adding or subtracting the same term on both sides, so that the other side is $0$ . Choose the side that keeps the $x^2$ term positive.

3. Write the terms in order: $ax^2 + bx + c = 0$ .

4. If all of $a$ , $b$ and $c$ have a common factor, divide both sides by it. Divide by a number, never by $x$ .

Then factorise and solve as in lesson 5.1, and check each root in the original equation.

Example 1: Terms on both sides

Solve $x^2 + 3x = 4$ .

Solution

Subtract $4$ from both sides: $x^2 + 3x - 4 = 0$ .

Two numbers with product $-4$ and sum $3$ : $4$ and $-1$ .

$(x + 4)(x - 1) = 0$ , so $x = -4$ or $x = 1$ .

Check in the original equation. $x = -4$ : $16 - 12 = 4$ ✓. $x = 1$ : $1 + 3 = 4$ ✓

$x = -4$ or $x = 1$

Example 2: $x^2$ on both sides

Solve $2x^2 + 4x = x^2 - 3x - 10\text{.}$

Solution

Subtract $x^2$ from both sides: $x^2 + 4x = -3x - 10$ .

Add $3x$ to both sides: $x^2 + 7x = -10$ .

Add $10$ to both sides: $x^2 + 7x + 10 = 0$ .

Product $10$ , sum $7$ : $2$ and $5$ . So $(x + 2)(x + 5) = 0$ , and $x = -2$ or $x = -5$ .

Check $x = -2$ . Left side: $2(4) + 4(-2) = 0$ . Right side: $4 + 6 - 10 = 0$ ✓

Check $x = -5$ . Left side: $2(25) + 4(-5) = 30\text{.}$ Right side: $25 + 15 - 10 = 30$ ✓

$x = -2$ or $x = -5$

Example 3: Keep the $x^2$ term positive

Solve $5x = x^2 + 6$ .

Solution

The $x^2$ term is on the right, so collect everything on the right. Subtract $5x$ from both sides: $0 = x^2 - 5x + 6$ , that is, $x^2 - 5x + 6 = 0$ .

Product $6$ , sum $-5$ : $-2$ and $-3$ . So $(x - 2)(x - 3) = 0$ , and $x = 2$ or $x = 3$ .

Check $x = 2$ : $5 \times 2 = 10$ and $4 + 6 = 10$ ✓. Check $x = 3$ : $15$ and $9 + 6 = 15$ ✓

$x = 2$ or $x = 3$

Collecting on the left instead gives $-x^2 + 5x - 6 = 0$ . That is not wrong, but a negative $x^2$ term is harder to factorise. Multiply both sides by $-1$ to make it positive, as in the next example.

Example 4: A negative $x^2$ term

Solve $-x^2 + 2x + 3 = 0$ .

Solution

Multiply both sides by $-1$ . Every sign changes, and $0 \times (-1) = 0$ : $x^2 - 2x - 3 = 0$ .

Product $-3$ , sum $-2$ : $-3$ and $1$ . So $(x - 3)(x + 1) = 0$ , and $x = 3$ or $x = -1$ .

Check in the original equation. $x = 3$ : $-9 + 6 + 3 = 0$ ✓. $x = -1$ : $-1 - 2 + 3 = 0$ ✓

$x = 3$ or $x = -1$

Never Divide by x

In $x^2 = 4x$ it is tempting to divide both sides by $x$ and write $x = 4$ . But $x = 0$ also makes the equation true ( $0 = 0$ ), and dividing by $x$ has lost it. Dividing by $x$ is not allowed when $x$ might be $0$ . Rearrange and take out the common factor instead.

Example 5: An $x$ on each side

Solve
(a) $x^2 = 4x$
(b) $3x^2 = 12x$ .

Solution

(a) Subtract $4x$ from both sides: $x^2 - 4x = 0$ . Take out $x$ : $x(x - 4) = 0$ . So $x = 0$ or $x = 4$ .

Check: $x = 0$ gives $0 = 0$ ✓; $x = 4$ gives $16 = 16$ ✓.

(b) Subtract $12x$ : $3x^2 - 12x = 0$ . Take out $3x$ : $3x(x - 4) = 0$ . So $x = 0$ or $x = 4$ .

Check $x = 4$ : $3 \times 16 = 48$ and $12 \times 4 = 48$ ✓

In (b) you may divide both sides by $3$ , a number that is not $0$ , to get $x^2 = 4x$ . You may not divide by $x$ .

(a) $x = 0$ or $x = 4$
(b) $x = 0$ or $x = 4$

Expanding Brackets First

An equation such as $x(x + 3) = 10$ looks as if it is already in factors. It is not ready: the right side is $10$ , not $0$ . Expand, then collect the terms on one side.

Example 6: A bracket equal to a number

Solve $x(x + 3) = 10$ .

Solution

Expand: $x^2 + 3x = 10$ . Subtract $10$ : $x^2 + 3x - 10 = 0$ .

Product $-10$ , sum $3$ : $5$ and $-2$ . So $(x + 5)(x - 2) = 0$ , and $x = -5$ or $x = 2$ .

Check $x = -5$ : $(-5)(-2) = 10$ ✓. Check $x = 2$ : $2 \times 5 = 10$ ✓

$x = -5$ or $x = 2$

Example 7: Two brackets equal to a number

Solve $(x - 1)(x - 2) = 6$ .

Solution

A wrong start: “ $x - 1 = 6$ or $x - 2 = 6$ , so $x = 7$ or $x = 8$ .” Test it: $x = 7$ gives $6 \times 5 = 30$ , not $6$ . Two numbers that multiply to $6$ need not include a $6$ ( $2 \times 3$ , $(-2) \times (-3)$ …). The rule for a product works only for $0$ .

The right way. Expand: $x^2 - 2x - x + 2 = 6$ , so $x^2 - 3x + 2 = 6$ . Subtract $6$ : $x^2 - 3x - 4 = 0$ .

Product $-4$ , sum $-3$ : $-4$ and $1$ . So $(x - 4)(x + 1) = 0$ , and $x = 4$ or $x = -1$ .

Check $x = 4$ : $3 \times 2 = 6$ ✓. Check $x = -1$ : $(-2) \times (-3) = 6$ ✓

$x = 4$ or $x = -1$

When the Coefficient of x² Is Not 1

Once one side is $0$ , factorise the other side by any method from chapter 3. When $a \neq 1$ , find two numbers whose product is $ac$ and whose sum is $b$ , split the middle term, and group. Each factor then gives a linear equation, and the roots may be fractions.

Example 8: Already equal to 0

Solve $2x^2 + 5x - 3 = 0$ .

Solution

$ac = 2 \times (-3) = -6\text{.}$ Two numbers with product $-6$ and sum $5$ : $6$ and $-1$ .

Split the middle term: $2x^2 + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3)\text{.}$

$(2x - 1)(x + 3) = 0$ . Either $2x - 1 = 0$ , so $x = \frac{1}{2}$ , or $x + 3 = 0$ , so $x = -3$ .

Check $x = \frac{1}{2}$ : $2 \times \frac{1}{4} + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 0$ ✓. Check $x = -3$ : $18 - 15 - 3 = 0$ ✓

$x = \frac{1}{2}$ or $x = -3$

Example 9: Rearrange, then $a \neq 1$

Solve $2x^2 = 5x + 3$ .

Solution

Subtract $5x$ and $3$ from both sides: $2x^2 - 5x - 3 = 0$ .

$ac = -6$ ; product $-6$ , sum $-5$ : $-6$ and $1$ .

$2x^2 - 6x + x - 3 = 2x(x - 3) + (x - 3) = (2x + 1)(x - 3)\text{.}$

$(2x + 1)(x - 3) = 0$ . Either $2x + 1 = 0$ , so $x = -\frac{1}{2}$ , or $x - 3 = 0$ , so $x = 3$ .

Check $x = -\frac{1}{2}$ : left side $2 \times \frac{1}{4} = \frac{1}{2}$ ; right side $-\frac{5}{2} + 3 = \frac{1}{2}$ ✓. Check $x = 3$ : $18$ and $15 + 3 = 18$ ✓

$x = -\frac{1}{2}$ or $x = 3$

A Common Number Factor

When every coefficient has the same number factor, divide both sides by it first. The numbers become smaller and the factorising easier, and the roots do not change.

Example 10: Divide by the common factor

Solve $2x^2 - 8x - 10 = 0$ .

Solution

Every coefficient is even. Divide both sides by $2$ (and $0 \div 2 = 0$ ): $x^2 - 4x - 5 = 0$ .

Product $-5$ , sum $-4$ : $-5$ and $1$ . So $(x - 5)(x + 1) = 0$ , and $x = 5$ or $x = -1$ .

Check in the original equation. $x = 5$ : $50 - 40 - 10 = 0$ ✓. $x = -1$ : $2 + 8 - 10 = 0$ ✓

You could also write $2(x - 5)(x + 1) = 0$ . The factor $2$ is never $0$ , so the roots come from the brackets, and they are the same.

$x = 5$ or $x = -1$

Challenge: x in a Denominator

In chapter 4 you multiplied both sides of a formula by a denominator to clear a fraction, and you did the same with equations in Secondary 1. Sometimes the equation that results is quadratic.

Example 11: Clear the fraction first

Solve $x = \frac{6}{x - 1}$ , where $x \neq 1$ .

Solution

Multiply both sides by $(x - 1)$ , which is not $0$ because $x \neq 1$ : $x(x - 1) = 6$ .

Expand and subtract $6$ : $x^2 - x - 6 = 0$ .

Product $-6$ , sum $-1$ : $-3$ and $2$ . So $(x - 3)(x + 2) = 0$ , and $x = 3$ or $x = -2$ .

Neither root is $1$ , so both are allowed. Check $x = 3$ : $\frac{6}{2} = 3$ ✓. Check $x = -2$ : $\frac{6}{-3} = -2$ ✓

$x = 3$ or $x = -2$

Common Mistakes

Using the factors before the other side is $0$ : $(x - 1)(x - 2) = 6$ does not give $x - 1 = 6$ . Expand and collect first.
Dividing both sides by $x$ , which loses the root $x = 0$ . Subtract, then take out $x$ as a common factor.
“Moving” a term across without changing its sign: from $x^2 - 12 = x$ , writing $x^2 + x - 12 = 0$ . Subtract $x$ from both sides: $x^2 - x - 12 = 0$ .
Leaving a negative $x^2$ term and then factorising wrongly. Collect on the other side, or multiply both sides by $-1$ .
Checking in the rearranged equation. A mistake made while rearranging is caught only by checking in the original equation.

Practice Questions

Rearrange each equation so that one side is 0 before you factorise, and check each root in the original equation. The questions get harder as you go.

Question 1

Solve $x^2 - 3x = 10$ .

ShowHide solution

Subtract $10$ : $x^2 - 3x - 10 = 0$ . Product $-10$ , sum $-3$ : $-5$ and $2$ .

$(x - 5)(x + 2) = 0$ , so $x = 5$ or $x = -2$ .

Check $x = 5$ : $25 - 15 = 10$ ✓. Check $x = -2$ : $4 + 6 = 10$ ✓

$x = 5$ or $x = -2$

Question 2

Solve $x^2 + 8 = 6x$ .

ShowHide solution

Subtract $6x$ from both sides: $x^2 - 6x + 8 = 0$ . Product $8$ , sum $-6$ : $-2$ and $-4$ .

$(x - 2)(x - 4) = 0$ , so $x = 2$ or $x = 4$ .

Check $x = 2$ : $4 + 8 = 12$ and $6 \times 2 = 12$ ✓. Check $x = 4$ : $16 + 8 = 24$ and $24$ ✓

$x = 2$ or $x = 4$

Question 3

Solve
(a) $x^2 = 7x$
(b) $5x^2 = 20x$ .

ShowHide solution

(a) $x^2 - 7x = 0$ , so $x(x - 7) = 0$ , and $x = 0$ or $x = 7$ . Check: $49 = 49$ ✓.

(b) $5x^2 - 20x = 0$ , so $5x(x - 4) = 0$ , and $x = 0$ or $x = 4$ . Check: $5 \times 16 = 80$ and $20 \times 4 = 80$ ✓.

(a) $x = 0$ or $x = 7$
(b) $x = 0$ or $x = 4$

Question 4

Solve $x(x + 7) = 60$ .

ShowHide solution

Expand and subtract $60$ : $x^2 + 7x - 60 = 0$ . Product $-60$ , sum $7$ : $12$ and $-5$ .

$(x + 12)(x - 5) = 0$ , so $x = -12$ or $x = 5$ .

Check $x = -12$ : $(-12)(-5) = 60$ ✓. Check $x = 5$ : $5 \times 12 = 60$ ✓

$x = -12$ or $x = 5$

Question 5

Solve $(x + 2)(x - 1) = 4$ .

ShowHide solution

Expand: $x^2 - x + 2x - 2 = 4$ , so $x^2 + x - 2 = 4$ . Subtract $4$ : $x^2 + x - 6 = 0$ .

Product $-6$ , sum $1$ : $3$ and $-2$ . $(x + 3)(x - 2) = 0$ , so $x = -3$ or $x = 2$ .

Check $x = -3$ : $(-1)(-4) = 4$ ✓. Check $x = 2$ : $4 \times 1 = 4$ ✓

$x = -3$ or $x = 2$

Question 6

Solve $18 + 3x - x^2 = 0$ .

ShowHide solution

Add $x^2$ to both sides, so that the $x^2$ term is positive: $18 + 3x = x^2$ , that is, $x^2 - 3x - 18 = 0$ .

Product $-18$ , sum $-3$ : $-6$ and $3$ . $(x - 6)(x + 3) = 0$ , so $x = 6$ or $x = -3$ .

Check $x = 6$ : $18 + 18 - 36 = 0$ ✓. Check $x = -3$ : $18 - 9 - 9 = 0$ ✓

$x = 6$ or $x = -3$

Question 7

Solve $2x^2 + 4x - 6 = 0$ .

ShowHide solution

Divide both sides by $2$ : $x^2 + 2x - 3 = 0$ . Product $-3$ , sum $2$ : $3$ and $-1$ .

$(x + 3)(x - 1) = 0$ , so $x = -3$ or $x = 1$ .

Check in the original. $x = -3$ : $18 - 12 - 6 = 0$ ✓. $x = 1$ : $2 + 4 - 6 = 0$ ✓

$x = -3$ or $x = 1$

Question 8

Solve $3x^2 - 10x + 8 = 0$ .

ShowHide solution

$ac = 24$ . Product $24$ , sum $-10$ : $-4$ and $-6$ .

$3x^2 - 6x - 4x + 8 = 3x(x - 2) - 4(x - 2) = (3x - 4)(x - 2)\text{.}$

$(3x - 4)(x - 2) = 0$ . Either $3x - 4 = 0$ , so $x = \frac{4}{3}$ , or $x - 2 = 0$ , so $x = 2$ .

Check $x = \frac{4}{3}$ : $3 \times \frac{16}{9} - \frac{40}{3} + 8 = \frac{16}{3} - \frac{40}{3} + \frac{24}{3} = 0$ ✓. Check $x = 2$ : $12 - 20 + 8 = 0$ ✓

$x = \frac{4}{3}$ or $x = 2$

Question 9

Solve $3x^2 + 2 = 7x$ .

ShowHide solution

Subtract $7x$ : $3x^2 - 7x + 2 = 0$ . $ac = 6$ ; product $6$ , sum $-7$ : $-6$ and $-1$ .

$3x^2 - 6x - x + 2 = 3x(x - 2) - (x - 2) = (3x - 1)(x - 2)\text{.}$

$(3x - 1)(x - 2) = 0$ , so $x = \frac{1}{3}$ or $x = 2$ .

Check $x = \frac{1}{3}$ : $3 \times \frac{1}{9} + 2 = \frac{7}{3}$ and $7 \times \frac{1}{3} = \frac{7}{3}$ ✓. Check $x = 2$ : $12 + 2 = 14$ and $14$ ✓

$x = \frac{1}{3}$ or $x = 2$

Question 10

Solve $(2x - 1)(x + 2) = 2x + 13\text{.}$

ShowHide solution

Expand the left side: $2x^2 + 4x - x - 2 = 2x^2 + 3x - 2\text{.}$ So $2x^2 + 3x - 2 = 2x + 13\text{.}$

Subtract $2x$ and $13$ from both sides: $2x^2 + x - 15 = 0$ .

$ac = -30$ ; product $-30$ , sum $1$ : $6$ and $-5$ . $2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)\text{.}$

$(2x - 5)(x + 3) = 0$ , so $x = \frac{5}{2}$ or $x = -3$ .

Check $x = \frac{5}{2}$ : left side $4 \times 4\frac{1}{2} = 18$ ; right side $5 + 13 = 18$ ✓. Check $x = -3$ : left side $(-7)(-1) = 7$ ; right side $-6 + 13 = 7$ ✓

$x = \frac{5}{2}$ or $x = -3$

Question 11

Siti solved $x^2 - 15 = 2x$ like this: “ $x^2 + 2x - 15 = 0$ , so $(x + 5)(x - 3) = 0$ , and $x = -5$ or $x = 3$ .” Find her mistake and solve the equation correctly.

ShowHide solution

She moved $2x$ to the left without changing its sign. Subtracting $2x$ from both sides gives $x^2 - 2x - 15 = 0$ .

Her answer fails the check: $x = 3$ gives $9 - 15 = -6$ on the left but $6$ on the right.

Product $-15$ , sum $-2$ : $-5$ and $3$ . $(x - 5)(x + 3) = 0$ , so $x = 5$ or $x = -3$ .

Check $x = 5$ : $25 - 15 = 10$ and $2 \times 5 = 10$ ✓. Check $x = -3$ : $9 - 15 = -6$ and $-6$ ✓

$x = 5$ or $x = -3$

Question 12 (challenge)

Solve $x + 2 = \frac{15}{x}$ , where $x \neq 0$ .

ShowHide solution

Multiply both sides by $x$ , which is not $0$ : $x^2 + 2x = 15$ .

Subtract $15$ : $x^2 + 2x - 15 = 0$ . Product $-15$ , sum $2$ : $5$ and $-3$ .

$(x + 5)(x - 3) = 0$ , so $x = -5$ or $x = 3$ . Neither is $0$ , so both are allowed.

Check $x = -5$ : $-5 + 2 = -3$ and $\frac{15}{-5} = -3$ ✓. Check $x = 3$ : $3 + 2 = 5$ and $\frac{15}{3} = 5$ ✓

$x = -5$ or $x = 3$

Key Points

Factorise only when one side of the equation is $0$ .
Expand brackets, then collect all the terms on one side by doing the same to both sides; keep the $x^2$ term positive.
A product equal to a number that is not $0$ , such as $(x - 1)(x - 2) = 6$ , must be expanded first.
Never divide by $x$ : subtract and take out $x$ as a common factor, so that $x = 0$ is not lost.
You may divide both sides by a common number factor, or multiply by $-1$ ; the roots do not change.
When $a \neq 1$ , use two numbers whose product is $ac$ and whose sum is $b$ ; the roots may be fractions.
Check each root in the original equation.

Before you start

Making One Side Zero

Rearranging before you factorise

Example 1: Terms on both sides

Example 2: x2x^2x2 on both sides

Example 3: Keep the x2x^2x2 term positive

Example 4: A negative x2x^2x2 term

Never Divide by x

Example 5: An xxx on each side

Expanding Brackets First

Example 6: A bracket equal to a number

Example 7: Two brackets equal to a number

When the Coefficient of x² Is Not 1

Example 8: Already equal to 0

Example 9: Rearrange, then a≠1a \neq 1a≠1

A Common Number Factor

Example 10: Divide by the common factor

Challenge: x in a Denominator

Example 11: Clear the fraction first

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12 (challenge)

Key Points

Example 2: $x^2$ on both sides

Example 3: Keep the $x^2$ term positive

Example 4: A negative $x^2$ term

Example 5: An $x$ on each side

Example 9: Rearrange, then $a \neq 1$