Factorization
Solving quadratic equations by factorising
In this note
- 2.5Solve \(ax^{2}+bx+c=0\) by factorising.
- 2.5Rearrange to \(\ldots=0\) first, including \(x^{2}=7x\) and \(x(x+4)=21\).
- 2.5Form a quadratic from a problem, solve it and reject an impossible answer.
Before you start: Factorising quadratics. Other methods: completing the square and the quadratic formula.
Formula bank
If a product is zero ...
If two numbers multiply to 0, one of them must be 0. So \((x-2)(x-3)=0\) means \(x-2=0\) or \(x-3=0\): \(x=2\) or \(x=3\). This only works for 0 - lots of pairs multiply to 6 - so always make one side equal to 0 first.
General methodHow to solve a quadratic by factorising
- Expand any brackets and move everything to one side so it reads \(\ldots=0\).
- Factorise.
- Set each bracket equal to 0 and solve each one.
- Check in the original equation. In a real-life problem, reject an answer that is impossible (a negative length, for example).
Never divide both sides by \(x\): you lose the answer \(x=0\).
Example Solve \(x(x+4)=21\).
Expand and make it \(=0\), then factorise.
- \(x^{2}+4x-21=0\)
- \((x+7)(x-3)=0\)
- \(x+7=0\) or \(x-3=0\)
Answer \(x=-7\) or \(x=3\)
Example 2 Solve \(x^{2}=7x\).
Move \(7x\) across and take out \(x\) (don't divide by \(x\)).
- \(x^{2}-7x=0\)
- \(x(x-7)=0\)
Answer \(x=0\) or \(x=7\)
Try it yourself
Show answer
Answer 2, 7
Show answer
Answer 0, −6
Hint
Show answer
Answer \(2,\ -\tfrac32\)
Ways they test it
Type 1 Factorise, hence solve
"(a) Factorise \(x^{2}+x-20\). (b) Hence solve \(x^{2}+x-20=0\)." (2 + 1 marks)
Spot it 'hence' after a factorise part.
- Use your brackets from (a).
- Each bracket = 0.
For example \((x+5)(x-4)=0\)
Answer \(x=-5\) or \(x=4\)
Type 2 Not equal to zero yet
"Solve \(x(2x-1)=15\)." (3 marks)
Spot it a number on the right, or brackets to expand.
- Expand, subtract to make 0.
- Factorise and solve.
For example \(2x^{2}-x-15=0\)
- \((2x+5)(x-3)=0\)
Answer \(x=-\tfrac52\) or \(x=3\)
Type 3 Missing a term
"Solve \(3x^{2}=12x\)" or "Solve \(4x^{2}-9=0\)"
Spot it no number term, or no x term.
- No number: take out x (two answers, one is 0).
- No x term: DOTS, or \(x^{2}=\ldots\) then ± the square root.
For example \(4x^{2}-9=0\)
- \((2x+3)(2x-3)=0\)
Answer \(x=\pm\tfrac32\)
Type 4 Form it, solve it
"A rectangle is (x + 5) cm by (x − 2) cm with area 60 cm². Show that x² + 3x − 70 = 0 and find the length." (4-5 marks)
Spot it words, a shape, 'show that'.
- Write the equation (area = length × width).
- Expand and rearrange to the given form.
- Solve and reject the impossible value.
For example \((x+10)(x-7)=0\)
- \(x=7\) (−10 gives a negative width)
Answer 12 cm by 5 cm
Mistakes to avoid
- ✗ \(x^{2}=7x\Rightarrow x=7\)✓ \(x=0\) or \(x=7\)Dividing by \(x\) loses \(x=0\).
- ✗ \(x(x+4)=21\Rightarrow x=21\) or \(x+4=21\)✓ make it \(=0\) firstThe rule only works for 0.
- ✗ \((x+5)=0\Rightarrow x=5\)✓ \(x=-5\)Solve each bracket carefully.
- ✗ \(4x^{2}=9\Rightarrow x=\tfrac32\)✓ \(x=\pm\tfrac32\)A square root has two answers.
Summary
- Make it = 0, factorise, set each factor = 0.
- Never divide by \(x\).
- Usually two answers; in context reject the impossible one.
If you see …
- "hence solve"use the factorised form from the part before.
- a number on the right-hand sidesubtract it first.
Practice
Practice: easy, medium, hard
Easy
Show answer
Answer 2, 3
Show answer
Answer 4, −4
Hint
Show answer
Answer 0, 4
Medium
Guide me
What is it asking? x
What do we know? \(ac=6\): −6 and −1
Plan split and group
Show working
- \(3x^{2}-6x-x+2=0\)
- \((3x-1)(x-2)=0\)
- \(x=\tfrac13\) or \(x=2\)
Answer \(\tfrac13,\ 2\)
Guide me
What is it asking? x
What do we know? not = 0 yet
Plan \(x^{2}-3x-10=0\)
Show working
- \((x-5)(x+2)=0\)
Answer 5, −2
Guide me
What is it asking? x
What do we know? expand first
Plan \(2x^{2}-x-15=0\)
Show working
- \((2x+5)(x-3)=0\)
Answer \(3,\ -\tfrac52\)
Hard
Guide me
What is it asking? a length
What do we know? length × width = 60
Plan \(x^{2}+3x-70=0\), reject the negative
Show working
- \((x+5)(x-2)=60\)
- \(x^{2}+3x-70=0\)
- \((x+10)(x-7)=0\)
- \(x=7\), length \(=12\)
Answer 12 cm
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x^{2}+3x-70=0\) | M1A1 | ||
| \((x+10)(x-7)=0\) | M1 | ||
| 12 cm | A1 | ||
| Total: 4 marks | |||
Guide me
What is it asking? the smaller number
What do we know? \(x(x+3)=40\)
Plan rearrange, factorise, take the positive
Show working
- \(x^{2}+3x-40=0\)
- \((x+8)(x-5)=0\)
- \(x=5\) (−8 is not positive)
Answer 5
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x(x+3)=40\) | M1 | ||
| \((x+8)(x-5)=0\) | M1 | ||
| 5 | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? four values
What do we know? first: take out 5, DOTS
Plan second: \((x-7)(x+5)\)
Show working
- \(5(x^{2}-4)=0\Rightarrow x=\pm2\)
- \((x-7)(x+5)=0\Rightarrow x=7,\ -5\)
Answer ±2, 7, −5
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(\pm2\) | B1 | ||
| \((x-7)(x+5)=0\) | M1 | ||
| 7, −5 | A1 | ||
| Total: 3 marks | |||
Can I do this?
- I can solve quadratics by factorising.
- I can rearrange to = 0 first and never divide by x.
- I can form and solve a quadratic in context and reject impossible answers.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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