Solving quadratics by factorising — Mathematics

Before this

This leads to

What a zero tells you

I am thinking of two numbers. When I multiply them, I get 12. What are they? You cannot say. They could be 3 and 4, or 2 and 6, or 24 and one half, or −1 and −12. There are endless pairs.

Now I am thinking of two different numbers, and when I multiply them, I get 0. What do you know? Quite a lot. At least one of them must be 0. There is no other way to make a product of zero. That small fact, which sounds almost too obvious to say, is the whole key to this lecture.

The last unit gave you three ways to rewrite a quadratic: factorising into two brackets, the difference of two squares, and completing the square. It ended with a hint: each bracket is zero for one value of $x$ . This unit turns those rewriting tools into equation-solving tools, starting with factorising.

The zero product property

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

Why is it true? Suppose $A$ is not zero. Then you are allowed to divide both sides of $A \times B = 0$ by $A$ , and you get $B = 0$ . So if one of them is not zero, the other one has to be. Nothing like this works for any other number. If $A \times B = 12$ , knowing $A$ is not zero only tells you $B = 12 \div A$ , which could be anything.

Here is how it solves an equation. Take $x^{2} + 5 x + 6 = 0$ . Factorise the left side, as in the lecture on factorising quadratics: you need two numbers with product 6 and sum 5, which are 2 and 3.

(x + 2) (x + 3) = 0

Now read it as $A \times B = 0$ , with $A = x + 2$ and $B = x + 3$ . One of the brackets must be zero. So either $x + 2 = 0$ , which gives $x = - 2$ , or $x + 3 = 0$ , which gives $x = - 3$ . Two small linear equations replace one quadratic. That is it. That is the whole idea.

Check both answers in the original equation. For $x = - 2$ : $4 - 10 + 6 = 0$ . For $x = - 3$ : $9 - 15 + 6 = 0$ . Both work. The solutions of an equation are also called its roots.

Solving a quadratic by factorising

1. Rearrange so that one side is 0.

2. Factorise the other side.

3. Set each factor equal to 0.

4. Solve each small equation.

5. Check each answer in the original equation.

Why the zero matters so much

Step 1 is the step people skip, so look at what goes wrong without it.

Stop and think

A student solves $(x + 2) (x + 3) = 6$ like this: " $x + 2 = 6$ or $x + 3 = 6$ , so $x = 4$ or $x = 3$ ." Check the answers. What went wrong?

Show the answer

Try $x = 4$ : $(6) (7) = 42$ , not 6. Try $x = 3$ : $(5) (6) = 30$ , not 6. Both are wrong. A product of 6 does not mean one factor is 6; it could be 2 × 3, or 1 × 6, or 12 × 0.5. Only a product of 0 forces a factor to be 0. The correct method rearranges first. Expand: $x^{2} + 5 x + 6 = 6$ , so $x^{2} + 5 x = 0$ . Factorise: $x (x + 5) = 0$ . So $x = 0$ or $x = - 5$ . Check $x = 0$ : $2 \times 3 = 6$ . Check $x = - 5$ : $(- 3) (- 2) = 6$ . Both work.

Roots on a graph

The graph of $y = (x + 2) (x + 3)$ shows the same story. The equation $(x + 2) (x + 3) = 0$ asks: where is $y$ zero? That is where the curve meets the $x$ -axis. So the roots of the equation are where the graph crosses the axis.

Left: y = (x + 2)(x + 3) crosses the x-axis at x = −3 and x = −2, the two roots of (x + 2)(x + 3) = 0. Right: y = (x − 3)² only touches the x-axis, at x = 3, because both brackets are zero at the same place. That is a repeated root.

Each bracket $(x - r)$ is zero at exactly one place, $x = r$ . So the brackets and the crossing points match one for one. Choose two roots below and watch the equation and the graph that go with them.

Explore — from roots to equation to graph

Choose two roots, r and s. The equation (x − r)(x − s) = 0 has exactly these solutions. Watch the expanded equation and where the graph meets the x-axis. Then make r and s equal.

r: −2

s: −3

Four pairs of roots with their equations. In the last row the root is repeated, so the graph touches the x-axis instead of crossing it.

Roots	Factorised	Expanded
−2, −3	(x + 2)(x + 3) = 0	x² + 5x + 6 = 0
−2, 5	(x + 2)(x − 5) = 0	x² − 3x − 10 = 0
0, 7	x(x − 7) = 0	x² − 7x = 0
3 (twice)	(x − 3)² = 0	x² − 6x + 9 = 0

Notice the pattern in the expanded equations. For roots $r$ and $s$ , the equation is $x^{2} - (r + s) x + rs = 0$ . The middle number is minus the sum of the roots, and the last number is their product. That is exactly why the factorising method looks for a pair with a given sum and product.

Zero, one or two roots

The explorer always gives at least one root, because you chose the roots first. But some quadratic equations have none. How many roots can a quadratic equation have?

Stop and think

Solve $x^{2} + 4 = 0$ .

Show the answer

There is no real solution. $x^{2}$ is never negative, so $x^{2} + 4$ is always at least 4 and can never be 0. The difference-of-two-squares lecture showed the same thing from the other side: a sum of two squares does not factorise into brackets with real numbers. On a graph, $y = x^{2} + 4$ sits entirely above the $x$ -axis and never meets it.

So a quadratic equation can have two roots (the graph crosses the axis twice), one repeated root (it touches once), or no real roots (it misses the axis). It can never have three: a quadratic factorises into at most two brackets, and each bracket gives one root. Can you tell which case you are in before you start? Yes, and a later lecture in this unit, on the discriminant, shows how with a single calculation.

Worked examples

Worked Example 1 — Rearrange first

Solve $x^{2} = 3 x + 10$ .

The equation does not equal zero yet, so move every term to the left: $x^{2} - 3 x - 10 = 0$ . Now find two numbers with product −10 and sum −3. They are 2 and −5.

(x + 2) (x - 5) = 0

So $x = - 2$ or $x = 5$ . Check in the original equation. For $x = 5$ : $25 = 15 + 10$ . For $x = - 2$ : $4 = - 6 + 10$ . Both work.

Worked Example 2 — A number in front of x²

Solve $6 x^{2} - x - 2 = 0$ .

Use the ac method from the factorising lecture. Here $a c = 6 \times (- 2) = - 12$ . You need two numbers with product −12 and sum −1: they are 3 and −4. Split the middle term and group:

6 x^{2} + 3 x - 4 x - 2 3 x (2 x + 1) - 2 (2 x + 1) (3 x - 2) (2 x + 1) = 0 = 0 = 0

So $3 x - 2 = 0$ , giving $x = \frac{2}{3}$ , or $2 x + 1 = 0$ , giving $x = - \frac{1}{2}$ . When the $x^{2}$ coefficient is not 1, the roots are often fractions. Check $x = - \frac{1}{2}$ : $6 \times \frac{1}{4} + \frac{1}{2} - 2 = 1.5 + 0.5 - 2 = 0$ .

Worked Example 3 — A repeated root

Solve $x^{2} - 6 x + 9 = 0$ .

Two numbers with product 9 and sum −6 are −3 and −3. So $(x - 3) (x - 3) = 0$ , which is $(x - 3)^{2} = 0$ .

Both brackets give $x = 3$ . That is not two solutions that happen to be equal. There is only one solution, $x = 3$ , and it is called a repeated root (or double root). On the graph, the curve touches the $x$ -axis at 3 and turns back without crossing it, as in the figure above. You can also see it with completing the square: $x^{2} - 6 x + 9$ is already $(x - 3)^{2}$ , whose smallest value is 0.

Worked Example 4 — No constant term, and no x term

Solve (a) $x^{2} = 7 x$ and (b) $x^{2} - 49 = 0$ .

(a) Rearrange: $x^{2} - 7 x = 0$ . The common factor is $x$ : $x (x - 7) = 0$ . So $x = 0$ or $x = 7$ . Do not divide both sides by $x$ to get $x = 7$ . That throws away the root $x = 0$ , because dividing by $x$ is not allowed when $x$ could be 0.

(b) This is a difference of two squares: $(x + 7) (x - 7) = 0$ , so $x = 7$ or $x = - 7$ , written $x = \pm 7$ . Taking the square root of $x^{2} = 49$ gives the same answer, as long as you remember both signs.

Worked Example 5 — A rectangle

A rectangle is 3 cm longer than it is wide. Its area is 40 cm². Find its width.

Let the width be $x$ cm. The length is $x + 3$ , so $x (x + 3) = 40$ . Rearrange: $x^{2} + 3 x - 40 = 0$ . Two numbers with product −40 and sum 3 are 8 and −5:

(x + 8) (x - 5) = 0

So $x = - 8$ or $x = 5$ . A width cannot be negative, so the width is 5 cm and the length is 8 cm. Check: $5 \times 8 = 40$ . The algebra gives every solution of the equation; the context decides which ones make sense.

Now look at a student's working for another equation with no constant term.

Spot the error — solving 3x² = 12x

One line of this working is wrong. Tap the line you think is wrong. Every line has an explanation.

The working, with the mistake marked and corrected:

Line 1. 3x² = 12x. (The equation.)
Line 2. Divide both sides by 3x: x = 4. The mistake: dividing by x is not allowed when x might be 0, and it loses the root x = 0.
Line 3. Check: 3 × 16 = 48 and 12 × 4 = 48. (True, but it only checks the root that was found.)
Line 4. So the only solution is x = 4. (Wrong. The correct working is 3x² − 12x = 0, so 3x(x − 4) = 0, giving x = 0 or x = 4.)

Three classic mistakes

Trap 1 — Splitting brackets that do not equal zero

$(x + 2) (x + 3) = 6$ does not mean $x + 2 = 6$ or $x + 3 = 6$ . The zero product property works only for a product equal to 0. Rearrange so one side is 0 before you split anything.

Trap 2 — Dividing by x

From $x^{2} = 7 x$ , dividing by $x$ gives only $x = 7$ and loses $x = 0$ . Take $x$ out as a common factor instead: $x (x - 7) = 0$ .

Trap 3 — Wrong signs in the answers

$(x + 2) (x - 5) = 0$ gives $x = - 2$ and $x = 5$ , not $x = 2$ and $x = - 5$ . Each root is the value that makes its bracket zero, so the sign flips. Substituting each answer back catches this at once.

Try it yourself

Solve each equation and check every answer by substituting it back.

Solve $x^{2} - x - 12 = 0$ .

Hint 1
Find two numbers with product −12 and sum −1.

Hint 2
They are −4 and 3.

Full answer
$(x - 4) (x + 3) = 0$ , so $x = 4$ or $x = - 3$ . Check: $16 - 4 - 12 = 0$ and $9 + 3 - 12 = 0$ .
Solve $2 x^{2} + 7 x + 3 = 0$ .

Hint 1
$a c = 6$ . Find two numbers with product 6 and sum 7.

Hint 2
Split $7 x$ into $6 x + x$ and group.

Full answer
$2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3) = 0$ . So $x = - \frac{1}{2}$ or $x = - 3$ . Check $x = - 3$ : $18 - 21 + 3 = 0$ .
Solve (a) $3 x^{2} = 12 x$ and (b) $4 x^{2} - 9 = 0$ .

Hint 1
(a) Rearrange and take out the common factor $3 x$ .

Hint 2
(b) $4 x^{2} = (2 x)^{2}$ and $9 = 3^{2}$ .

Full answer
(a) $3 x^{2} - 12 x = 0$ , so $3 x (x - 4) = 0$ , giving $x = 0$ or $x = 4$ . (b) $(2 x + 3) (2 x - 3) = 0$ , so $x = - \frac{3}{2}$ or $x = \frac{3}{2}$ .
Work backwards. Find a quadratic equation with whole-number coefficients whose solutions are $x = \frac{2}{3}$ and $x = - 5$ .

Hint 1
$x = \frac{2}{3}$ comes from the bracket $3 x - 2 = 0$ . Which bracket gives $x = - 5$ ?

Hint 2
Expand $(3 x - 2) (x + 5)$ .

Full answer
$(3 x - 2) (x + 5) = 0$ , which expands to $3 x^{2} + 13 x - 10 = 0$ . Check $x = \frac{2}{3}$ : $3 \times \frac{4}{9} + \frac{26}{3} - 10 = \frac{4}{3} + \frac{26}{3} - 10 = 10 - 10 = 0$ . Any multiple, such as $6 x^{2} + 26 x - 20 = 0$ , has the same solutions.

The bigger idea

Zero is special: a product is zero only when one of its factors is zero, and no other number gives you that grip. That is why every method for solving equations in this unit, and many beyond it, works by getting a 0 on one side. The idea scales up: a cubic that factorises into three brackets has up to three roots, one from each bracket.

But factorising only works when the quadratic factorises neatly. The completing-the-square lecture showed that $x^{2} + 4 x - 1$ needs surds, and no whole-number pair will find them. The next lecture builds a formula that works for every quadratic, factorisable or not.

Key concepts in this lesson

The zero product property: if A × B = 0, then A = 0 or B = 0.
To solve by factorising: rearrange to = 0, factorise, set each factor to 0, solve, check.
The property works only when the product is 0; a product of 6 tells you nothing about either factor alone.
Never divide by x; take it out as a common factor so the root x = 0 is kept.
The roots are where the graph meets the x-axis; a repeated root, as in (x − 3)² = 0, is where it touches.