Solving Quadratic Equations by Factorisation

What a quadratic equation is, and how to solve one by factorising it: when a product of factors is $0$ , one of the factors must be $0$ .

Before you start

In chapter 3 you factorised $x^2 + bx + c$ by finding two numbers whose product is $c$ and whose sum is $b$ : $x^2 + 5x + 6 = (x + 2)(x + 3)\text{,}$ because $2 \times 3 = 6$ and $2 + 3 = 5$ .
You also took out a common factor: $x^2 - 5x = x(x - 5)$ .
In Secondary 1 you solved linear equations such as $x + 3 = 0$ (so $x = -3$ ) and $2x - 1 = 0$ (so $x = \frac{1}{2}$ ). Every quadratic equation in this lesson ends as two equations like these.

What Is a Quadratic Equation?

In a linear equation such as $2x + 3 = 0$ , the highest power of $x$ is $1$ . In a quadratic equation the highest power of $x$ is $2$ : there is an $x^2$ term, and there may also be an $x$ term and a number.

A quadratic equation in one variable

A quadratic equation in $x$ is an equation that, after expanding and collecting terms, can be written as

$ax^2 + bx + c = 0$ , where $a$ , $b$ and $c$ are numbers and $a \neq 0$ .

$a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ and $c$ is the constant term. The sign in front of a term belongs to its coefficient: in $5x^2 - 8x + 3 = 0$ , $b = -8$ .

$b$ or $c$ (or both) may be $0$ . $a$ may not: with no $x^2$ term the equation is linear.

Quadratic or not?

Quadratic: $x^2 + 5x + 6 = 0$ , $2x^2 - 3x + 1 = 0$ , $x^2 - 4 = 0$ (here $b = 0$ ), $3x^2 + 7x = 0$ (here $c = 0$ ).

Not quadratic: $2x + 3 = 0$ (the highest power of $x$ is $1$ ); $x^3 - 2x = 0$ (the highest power is $3$ ); $x^2 + y = 3$ (two variables, not one).

Always collect the terms before you decide. An equation can show $x^2$ and still be linear, or show no $x^2$ at all and still be quadratic.

Example 1: Collect the terms first

Which of these are quadratic equations in $x$ ?
(a) $x^2 + 2x = x^2 + 5$
(b) $x(x + 3) = 10$
(c) $4 - x^2 = 0$

Solution

(a) Subtract $x^2$ from both sides: $2x = 5$ . The $x^2$ terms cancel, so this is a linear equation, not a quadratic one.

(b) Expand the bracket: $x^2 + 3x = 10$ . Subtract $10$ from both sides: $x^2 + 3x - 10 = 0$ . This is quadratic, with $a = 1$ , $b = 3$ and $c = -10$ .

(b) and (c) are quadratic equations; (a) is linear, since it becomes $2x = 5$ .

Example 2: Naming $a$ , $b$ and $c$

State the values of $a$ , $b$ and $c$ in
(a) $5x^2 - 8x + 3 = 0$
(b) $x^2 - 4 = 0$
(c) $3x^2 + 7x = 0$ .

Solution

(a) Compare with $ax^2 + bx + c = 0$ . The coefficient of $x^2$ is $5$ ; the term in $x$ is $-8x$ , so its coefficient is $-8$ ; the constant term is $3$ .

(b) $x^2$ is $1x^2$ , so $a = 1$ . There is no $x$ term, so $b = 0$ . The constant term is $-4$ .

(a) $a = 5$ , $b = -8$ , $c = 3$
(b) $a = 1$ , $b = 0$ , $c = -4$
(c) $a = 3$ , $b = 7$ , $c = 0$

When a Product Is Zero

Multiply any two numbers that are not zero: $3 \times 4 = 12$ , $(-2) \times 5 = -10$ , $\frac{1}{2} \times 6 = 3$ . The answer is never $0$ . The only way to get a product of $0$ is for one of the numbers to be $0$ : $0 \times 7 = 0$ and $5 \times 0 = 0$ .

This is the key to the whole chapter. If an equation says that two factors multiply to give $0$ , then one of them must be $0$ , and each of those is a simple linear equation.

If a product is zero

If $A \times B = 0$ , then $A = 0$ or $B = 0$ (or both).

$A$ and $B$ may be numbers or expressions such as $x + 3$ and $x - 5$ .

This is true only when the product is $0$ . If $A \times B = 6$ , neither $A$ nor $B$ has to be any particular number ( $2 \times 3$ , $1 \times 6$ , $12 \times \frac{1}{2}$ …).

The values of $x$ that make an equation true are its solutions, also called its roots.

Example 3: An equation already in factors

Solve $(x + 3)(x - 5) = 0$ .

Solution

The product of $(x + 3)$ and $(x - 5)$ is $0$ , so one of the factors is $0$ .

Either $x + 3 = 0$ , so $x = -3$ (subtract $3$ from both sides),

or $x - 5 = 0$ , so $x = 5$ (add $5$ to both sides).

Check $x = -3$ : $(-3 + 3)(-3 - 5) = 0 \times (-8) = 0$ ✓

Check $x = 5$ : $(5 + 3)(5 - 5) = 8 \times 0 = 0$ ✓

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

Notice the signs: the factor $x + 3$ gives the root $-3$ , and the factor $x - 5$ gives the root $5$ . Solve each factor; do not just copy its number.

Example 4: A lone $x$ , and a coefficient in a factor

Solve
(a) $x(x + 5) = 0$
(b) $3x(x - 2) = 0$
(c) $(2x - 1)(x + 3) = 0$ .

Solution

(a) The two factors are $x$ and $x + 5$ . Either $x = 0$ , or $x + 5 = 0$ , so $x = -5$ .

(b) The factors are $3x$ and $x - 2$ . Either $3x = 0$ , so $x = 0$ (divide both sides by $3$ ), or $x - 2 = 0$ , so $x = 2$ .

Check (c), $x = \frac{1}{2}$ : $(2 \times \frac{1}{2} - 1)(\frac{1}{2} + 3) = 0 \times 3\frac{1}{2} = 0$ ✓. And $x = -3$ : $(-6 - 1)(-3 + 3) = (-7) \times 0 = 0$ ✓

(a) $x = 0$ or $x = -5$
(b) $x = 0$ or $x = 2$
(c) $x = \frac{1}{2}$ or $x = -3$

In (a) and (b), $x = 0$ is a root like any other. Do not leave it out.

Example 5: Two equal roots

Solve $(x - 2)^2 = 0$ .

Solution

$(x - 2)^2 = 0$ means $(x - 2)(x - 2) = 0$ .

Either $x - 2 = 0$ or $x - 2 = 0$ . Both factors give the same equation, so $x = 2$ .

Check: $(2 - 2)^2 = 0^2 = 0$ ✓

$x = 2$ . The equation has two equal roots, and the answer is written once.

Solving x² + bx + c = 0 by Factorising

Most quadratic equations do not come in factors. Factorise the expression first; then the rule for a zero product finishes the work.

Solving a quadratic equation by factorisation

1. Make sure one side of the equation is $0$ . (Lesson 5.2 shows what to do when it is not.)

2. Factorise the other side completely, by any method from chapter 3.

3. Set each factor equal to $0$ .

4. Solve each of these linear equations.

5. Check each root by substituting it into the equation as it was first given.

The signs of the two numbers, for x² + bx + c

You need two numbers whose product is $c$ and whose sum is $b$ .

If $c$ is positive and $b$ is positive, both numbers are positive: $x^2 + 6x + 8 = (x + 2)(x + 4)\text{.}$

If $c$ is positive and $b$ is negative, both numbers are negative: $x^2 - 7x + 12 = (x - 3)(x - 4)\text{.}$

If $c$ is negative, the numbers have opposite signs, and the one with the larger size (ignoring its sign) has the same sign as $b$ : $x^2 + x - 12 = (x + 4)(x - 3)\text{.}$

If $b = 0$ , the two numbers are equal in size, as in $x^2 - 4 = (x + 2)(x - 2)\text{;}$ lesson 5.3 solves these.

Example 6: $b$ and $c$ both positive

Solve $x^2 + 6x + 8 = 0$ .

Solution

One side is already $0$ . Find two numbers with product $8$ and sum $6$ . Both are positive.

Factor pairs of $8$ : $1 \times 8$ (sum $9$ ), $2 \times 4$ (sum $6$ ) ✓.

$(x + 2)(x + 4) = 0$

Either $x + 2 = 0$ , so $x = -2$ , or $x + 4 = 0$ , so $x = -4$ .

Check $x = -2$ : $(-2)^2 + 6(-2) + 8 = 4 - 12 + 8 = 0$ ✓

Check $x = -4$ : $(-4)^2 + 6(-4) + 8 = 16 - 24 + 8 = 0$ ✓

$x = -2$ or $x = -4$

Example 7: $b$ negative, $c$ positive

Solve $x^2 - 7x + 12 = 0$ .

Solution

Two numbers with product $12$ and sum $-7$ : the product is positive and the sum negative, so both are negative.

Negative pairs with product $12$ : $-1$ and $-12$ (sum $-13$ ), $-2$ and $-6$ (sum $-8$ ), $-3$ and $-4$ (sum $-7$ ) ✓.

$(x - 3)(x - 4) = 0$

Either $x - 3 = 0$ , so $x = 3$ , or $x - 4 = 0$ , so $x = 4$ .

Check $x = 3$ : $9 - 21 + 12 = 0$ ✓. Check $x = 4$ : $16 - 28 + 12 = 0$ ✓

$x = 3$ or $x = 4$

Example 8: $c$ negative

Solve $x^2 - 2x - 15 = 0$ .

Solution

Two numbers with product $-15$ and sum $-2$ . The product is negative, so they have opposite signs, and their sizes differ by $2$ .

Pairs of $15$ : $1$ and $15$ (they differ by $14$ ), $3$ and $5$ (they differ by $2$ ) ✓. The larger, $5$ , takes the sign of $b$ , which is negative: the numbers are $-5$ and $3$ .

$(x - 5)(x + 3) = 0$

Either $x - 5 = 0$ , so $x = 5$ , or $x + 3 = 0$ , so $x = -3$ .

Check $x = 5$ : $25 - 10 - 15 = 0$ ✓. Check $x = -3$ : $9 + 6 - 15 = 0$ ✓

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

Equations with a Common Factor

When there is no constant term, as in $x^2 - 5x = 0$ , every term has the factor $x$ . Take it out, and the equation is in factors at once.

Example 9: Take out the common factor

Solve
(a) $x^2 - 5x = 0$
(b) $3x^2 + 12x = 0$ .

Solution

(a) Take out the common factor $x$ : $x(x - 5) = 0$ . Either $x = 0$ , or $x - 5 = 0$ , so $x = 5$ .

Check $x = 0$ : $0 - 0 = 0$ ✓. Check $x = 5$ : $25 - 25 = 0$ ✓

(b) The highest common factor of $3x^2$ and $12x$ is $3x$ : $3x(x + 4) = 0$ . Either $3x = 0$ , so $x = 0$ , or $x + 4 = 0$ , so $x = -4$ .

Check $x = -4$ : $3(-4)^2 + 12(-4) = 48 - 48 = 0$ ✓

(a) $x = 0$ or $x = 5$
(b) $x = 0$ or $x = -4$

Do not divide by x

It is tempting to divide $x^2 - 5x = 0$ by $x$ and get $x - 5 = 0$ . That loses the root $x = 0$ , because dividing by $x$ is dividing by $0$ when $x = 0$ . Take $x$ out as a factor instead.

Common Mistakes

Copying the numbers from the factors without solving: from $(x + 3)(x - 5) = 0$ , writing $x = 3$ or $x = -5$ . Solve each factor: $x + 3 = 0$ gives $x = -3$ .
Forgetting to divide: from $(2x - 1)(x + 3) = 0$ , writing $x = 1$ . The factor $2x - 1 = 0$ gives $x = \frac{1}{2}$ .
Giving only one root. A quadratic equation usually has two roots; give both, unless they are equal.
Losing $x = 0$ by dividing by $x$ . Take out $x$ as a common factor instead.
Factorising when the other side is not $0$ . The rule works only for a product equal to $0$ (lesson 5.2).
Not checking. Substitute each root into the equation as it was given; a wrong sign shows up at once.

Practice Questions

Work each question on paper before you open its solution. Check every root by substituting it back. The questions get harder as you go.

Question 1

Which of these are quadratic equations in $x$ ? For each one that is, state $a$ , $b$ and $c$ .
(a) $2x^2 - 7 = 0$
(b) $x^2 + 3x = x^2 - 6$
(c) $x(2x - 1) = 0$
(d) $6 + x - x^2 = 0$
(e) $x^3 + x^2 = 0$

ShowHide solution

(a) Quadratic: $a = 2$ , $b = 0$ (no $x$ term), $c = -7$ .

(b) Subtract $x^2$ from both sides: $3x = -6$ . Linear, not quadratic.

(d) In order of power: $-x^2 + x + 6 = 0$ . Quadratic: $a = -1$ , $b = 1$ , $c = 6$ .

(e) The highest power of $x$ is $3$ , so it is not quadratic.

(a), (c) and (d) are quadratic; (b) is linear and (e) has an $x^3$ term.

Question 2

Solve $(x - 4)(x + 1) = 0$ .

ShowHide solution

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

Check $x = 4$ : $0 \times 5 = 0$ ✓. Check $x = -1$ : $(-5) \times 0 = 0$ ✓

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

Question 3

Solve
(a) $x(x - 9) = 0$
(b) $5x(x + 2) = 0$
(c) $(3x + 2)(x - 4) = 0$ .

ShowHide solution

(a) Either $x = 0$ , or $x - 9 = 0$ , so $x = 9$ .

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

Check (c), $x = -\frac{2}{3}$ : $3 \times (-\frac{2}{3}) + 2 = -2 + 2 = 0\text{,}$ so the first factor is $0$ ✓.

(a) $x = 0$ or $x = 9$
(b) $x = 0$ or $x = -2$
(c) $x = -\frac{2}{3}$ or $x = 4$

Question 4

Solve $(x + 6)^2 = 0$ .

ShowHide solution

$(x + 6)(x + 6) = 0$ , and both factors give $x + 6 = 0$ , so $x = -6$ .

Check: $(-6 + 6)^2 = 0$ ✓

$x = -6$ (two equal roots)

Question 5

Solve $x^2 + 5x + 4 = 0$ .

ShowHide solution

Product $4$ , sum $5$ : both positive. $1 \times 4$ has sum $5$ ✓ ( $2 \times 2$ has sum $4$ ).

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

Check $x = -1$ : $1 - 5 + 4 = 0$ ✓. Check $x = -4$ : $16 - 20 + 4 = 0$ ✓

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

Question 6

Solve $x^2 - 8x + 15 = 0$ .

ShowHide solution

Product $15$ , sum $-8$ : both negative. $-1$ and $-15$ (sum $-16$ ); $-3$ and $-5$ (sum $-8$ ) ✓.

$(x - 3)(x - 5) = 0$ , so $x = 3$ or $x = 5$ .

Check $x = 3$ : $9 - 24 + 15 = 0$ ✓. Check $x = 5$ : $25 - 40 + 15 = 0$ ✓

$x = 3$ or $x = 5$

Question 7

Solve $x^2 + x - 30 = 0$ .

ShowHide solution

Product $-30$ , sum $1$ : opposite signs, sizes differing by $1$ . Pairs of $30$ : $1, 30$ ; $2, 15$ ; $3, 10$ ; $5, 6$ ✓. The larger, $6$ , takes the sign of $b$ (positive): $6$ and $-5$ .

$(x + 6)(x - 5) = 0$ , so $x = -6$ or $x = 5$ .

Check $x = -6$ : $36 - 6 - 30 = 0$ ✓. Check $x = 5$ : $25 + 5 - 30 = 0$ ✓

$x = -6$ or $x = 5$

Question 8

Solve $x^2 - 4x - 21 = 0$ .

ShowHide solution

Product $-21$ , sum $-4$ : opposite signs, sizes differing by $4$ . Pairs of $21$ : $1, 21$ ; $3, 7$ ✓. The larger, $7$ , takes the sign of $b$ (negative): $-7$ and $3$ .

$(x - 7)(x + 3) = 0$ , so $x = 7$ or $x = -3$ .

Check $x = 7$ : $49 - 28 - 21 = 0$ ✓. Check $x = -3$ : $9 + 12 - 21 = 0$ ✓

$x = 7$ or $x = -3$

Question 9

Solve
(a) $x^2 + 7x = 0$
(b) $4x^2 - 10x = 0$ .

ShowHide solution

(a) $x(x + 7) = 0$ , so $x = 0$ or $x = -7$ .

(b) The highest common factor of $4x^2$ and $10x$ is $2x$ : $2x(2x - 5) = 0$ . Either $2x = 0$ , so $x = 0$ , or $2x - 5 = 0$ , so $x = \frac{5}{2}$ .

Check (b), $x = \frac{5}{2}$ : $4 \times \frac{25}{4} - 10 \times \frac{5}{2} = 25 - 25 = 0$ ✓

(a) $x = 0$ or $x = -7$
(b) $x = 0$ or $x = \frac{5}{2}$

Question 10

Solve $x^2 + 2x - 48 = 0$ .

ShowHide solution

Product $-48$ , sum $2$ : opposite signs, sizes differing by $2$ . Pairs of $48$ : $1, 48$ ; $2, 24$ ; $3, 16$ ; $4, 12$ ; $6, 8$ ✓. The larger, $8$ , is positive like $b$ : $8$ and $-6$ .

$(x + 8)(x - 6) = 0$ , so $x = -8$ or $x = 6$ .

Check $x = -8$ : $64 - 16 - 48 = 0$ ✓. Check $x = 6$ : $36 + 12 - 48 = 0$ ✓

$x = -8$ or $x = 6$

Question 11

Mei solved $x^2 - x - 20 = 0$ like this: “ $(x - 5)(x + 4) = 0$ , so $x = -5$ or $x = 4$ .” Find her mistake and give the correct roots.

ShowHide solution

Her factorisation is right: $(x - 5)(x + 4) = x^2 + 4x - 5x - 20 = x^2 - x - 20\text{.}$

Her mistake is in the last step: she copied the numbers from the brackets with the wrong signs. $x - 5 = 0$ gives $x = 5$ , and $x + 4 = 0$ gives $x = -4$ .

Her answer fails the check: $x = -5$ gives $25 + 5 - 20 = 10$ , not $0$ .

Check $x = 5$ : $25 - 5 - 20 = 0$ ✓. Check $x = -4$ : $16 + 4 - 20 = 0$ ✓

The roots are $x = 5$ or $x = -4$ .

Question 12

Write down a quadratic equation, in the form $x^2 + bx + c = 0$ , whose roots are $2$ and $-5$ .

ShowHide solution

Work backwards. The root $2$ comes from the factor $x - 2$ , and the root $-5$ from the factor $x + 5$ .

So $(x - 2)(x + 5) = 0$ . Expand: $x^2 + 5x - 2x - 10 = 0\text{,}$ which is $x^2 + 3x - 10 = 0$ .

Check $x = 2$ : $4 + 6 - 10 = 0$ ✓. Check $x = -5$ : $25 - 15 - 10 = 0$ ✓

$x^2 + 3x - 10 = 0$

Key Points

A quadratic equation in $x$ is one that, after expanding and collecting terms, can be written as $ax^2 + bx + c = 0$ with $a \neq 0$ .
$a$ is the coefficient of $x^2$ , $b$ the coefficient of $x$ and $c$ the constant term; the sign in front of a term belongs to its coefficient.
If $A \times B = 0$ , then $A = 0$ or $B = 0$ . This works only when the product is $0$ .
To solve by factorisation: make one side $0$ , factorise the other side completely, set each factor equal to $0$ and solve, then check each root.
For $x^2 + bx + c$ , find two numbers whose product is $c$ and whose sum is $b$ .
A quadratic equation has at most two roots. $(x - 2)^2 = 0$ has two equal roots, $x = 2$ .
When every term has the factor $x$ , take it out: $x = 0$ is then one of the roots. Never divide by $x$ .

Before you start

What Is a Quadratic Equation?

A quadratic equation in one variable

Quadratic or not?

Example 1: Collect the terms first

Example 2: Naming aaa, bbb and ccc

When a Product Is Zero

If a product is zero

Example 3: An equation already in factors

Example 4: A lone xxx, and a coefficient in a factor

Example 5: Two equal roots

Solving x² + bx + c = 0 by Factorising

Solving a quadratic equation by factorisation

The signs of the two numbers, for x² + bx + c

Example 6: bbb and ccc both positive

Example 7: bbb negative, ccc positive

Example 8: ccc negative

Equations with a Common Factor

Example 9: Take out the common factor

Do not divide by x

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

Key Points

Example 2: Naming $a$ , $b$ and $c$

Example 4: A lone $x$ , and a coefficient in a factor

Example 6: $b$ and $c$ both positive

Example 7: $b$ negative, $c$ positive

Example 8: $c$ negative