GCSE · Maths · AQA · Spec 8300 · Foundation

Solving quadratics by factorising

Square a number, subtract the number itself, then subtract 6. You get zero. What was the number? There are two answers, and you will find both in three lines.

Where does this curve hit zero?

-4.5-20.535.5-8-161320xy = x² − x − 6(0.5, -6.25)

x: 0.5. y = x² − x − 6: -6.25

Drag the point along the curve and watch y, the second number. There are exactly two places where y is 0. Find both before you read on.

Exam line: Solving x² − x − 6 = 0 means finding the x-values where y = x² − x − 6 crosses the x-axis. Here they land exactly on 3 and −2. On a graph you draw yourself, or for a quadratic that will not factorise, a crossing usually falls between grid lines. Then a graph only gives approximate solutions, for example to 1 decimal place.
Watch out: The factorised form of this equation is (x − 3)(x + 2) = 0. The curve crosses at x = −2, not at +2. Each solution is the value of x that makes its bracket zero: x − 3 = 0 when x = 3, and x + 2 = 0 when x = −2. So the sign of each solution is the opposite of the sign in its bracket.

Finding the brackets

Solve x² + 2x − 15 = 0. First, build the grid backwards

The corners are x² and −15. The two middle cells must add up to +2x. Find two numbers that multiply to −15 and add to +2. Put the positive one down the side and the negative one along the top. Then fill the middle cells and collect the brackets.

Grid: each cell is its row times its column
x
xx²
−15

Type x² as x^2 if you cannot type ². Spaces do not matter.

Higher

Spot the slip

This answer looks tidy. It is wrong. Where?

Solve (x + 1)(x + 2) = 12

A student's working — which line goes wrong?
Higher

Maths · Algebra

Rearrange first, then solve

Step through it. Each line names the move being made.

GoalSolve x² = 5x + 14
1
x² = 5x + 14

Not equal to zero yet, so it is not ready to factorise.

2
3
4
5
6
7

Step 1 of 7

Not equal to zero yet, so it is not ready to factorise.

Choose your route

Which route does each equation need?

Pick an equation, then pick its route.

Still to sort

Already = 0: factorise now (0)

One side is 0 and two whole numbers fit the product-and-sum test.

Where the line is: Look at the right-hand side first. x² + 3x = 28 looks nearly the same, but it is not equal to 0 yet.

Rearrange to = 0 first (Higher) (0)

Neither side is 0 yet. Get everything onto one side, then factorise.

Where the line is: Brackets on the left do not make it ready. x(x − 1) = 20 has brackets, but the other side is 20, not 0.

Won't factorise: read approximate solutions from its graph (0)

No pair of whole numbers multiplies to the number term and adds to the x-coefficient.

Where the line is: Only choose this after you have tried every factor pair of the number term and none of them adds to the x-coefficient.

6 of 6 still to sort.

Before you solve anything, look at the equation and decide what to do first.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

If two things multiply to make zero, one of them has to be zero. That one fact turns a hard equation into two easy ones.

What you need to know

  • To solve by factorising, one side must be 0. Factorise the other side, then set each bracket equal to 0.
  • If two numbers multiply to make 0, at least one of them is 0. That is why each bracket gets its own small equation.
  • Each solution is the value that makes its bracket zero: (x − 3)(x + 2) = 0 gives x = 3 or x = −2.
  • Higher: if neither side is 0, rearrange first by doing the same thing to both sides.
  • The solutions are where y = 0: where the graph meets the x-axis, either crossing it or just touching it. Reading them from a graph gives approximate values.

The big picture

A quadratic equation can have two solutions, and factorising finds them. Make one side equal to 0, factorise the other side into two brackets, then set each bracket equal to 0. This works because two numbers can only multiply to make 0 if one of them is 0. The solutions are the x-values where y = 0, which is where the graph meets the x-axis. It crosses the axis there, or, if both brackets are the same, as in (x − 4)(x − 4) = 0, it just touches it. For a quadratic that will not factorise, you can read approximate solutions from its graph.

Key points

1To factorise x² + bx + c, find two numbers that multiply to c and add to b.
2If c is negative, the two numbers have opposite signs. If c is positive, they have the same sign as b.
3Two brackets give up to two different solutions. They can be the same number, as in (x − 4)(x − 4) = 0.
4Check each solution by substituting it into the original equation.
5At a point where the graph crosses or touches the x-axis, y = 0, so that x-value solves the equation.

Worked example

Problem

Solve x² − 2x − 24 = 0.

⚠ Watch out

Copying the signs from the brackets. (x + 7)(x − 2) = 0 does not give x = 7 and x = −2. Ask what makes each bracket zero: x + 7 = 0 when x = −7, and x − 2 = 0 when x = 2.

🧠

Memory hook

Make it zero. Make it brackets. Make each bracket zero.

✓

Check yourself

Solve x² + 9x + 14 = 0. Then say where the graph of y = x² + 9x + 14 crosses the x-axis, and check one of your answers by substituting it back in.

Flashcards

(9)
What must one side of a quadratic equation be before you factorise to solve it?
0. If it is not, rearrange first.
Why can you set each bracket equal to zero in (x − a)(x − b) = 0?
If two numbers multiply to make 0, at least one of them must be 0.
To factorise x² + bx + c, what two numbers do you look for?
Two numbers that multiply to c and add to b.
Solve (x + 8)(x − 3) = 0.
x = −8 or x = 3.
In x² + bx + c, c is negative. What do you know about the two numbers you need?
One is positive and one is negative, because their product is negative.
Where are the solutions of x² + bx + c = 0 on the graph of y = x² + bx + c?
At the x-values where the curve meets the x-axis, where y = 0. It crosses the axis there, or just touches it when both brackets are the same.
Why does reading solutions from a graph only give approximate answers?
You read the crossing by eye, and it can fall between grid lines, so you can only estimate it (for example, to 1 decimal place).
When would you read the solutions from a graph instead of factorising?
When the quadratic will not factorise: no pair of whole numbers multiplies to c and adds to b.
How do you check a solution of a quadratic equation?
Substitute it into the original equation and check both sides are equal.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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