Factorising quadratics, HCF and difference of squares
What this lesson covers
- Factorising is expanding in reverse, and every factorisation can be checked by expanding it back.
- Taking out the highest common factor first makes every later step simpler.
Factorising is expanding in reverse
- Year 10: common factors, and x squared plus b x plus c.
- Every factorisation can be checked by expanding it back.
- This lesson adds four new kinds, with a reason for each step.
| Direction | Start | Gives |
|---|---|---|
| Expanding | 3(2x + 5) | 6x + 15 |
| Factorising | 6x + 15 | 3(2x + 5) |
| Factorising | x² + 7x + 12 | (x + 3)(x + 4) |
| The check | expand it back | same as you started |
A highest common factor with letters
- The number part first: the HCF of 6 and 9 is 3.
- Then each letter, at its lowest power: x and y.
- Divide each term by 3xy to see what is left inside.
| Part | Compare | Take out |
|---|---|---|
| 6x²y − 9xy² | numbers: 6 and 9 | HCF 3 |
| Letter x | x² and x | lowest power: x |
| Letter y | y and y² | lowest power: y |
| Take out 3xy | divide each term | 3xy(2x − 3y) |
| Check | 3xy × 2x, 3xy × 3y | 6x²y − 9xy² |
Common is not the same as highest
- Taking out 2x is correct, but not finished.
- The bracket still has a common factor of 2.
- Fully factorised: 4x, bracket, 3x plus 2.
| Step | Working | Result |
|---|---|---|
| 12x² + 8x | take out 2x | 2x(6x + 4) |
| Look inside | 6x + 4 shares 2 | not finished |
| Take out 4x | the highest | 4x(3x + 2) |
| Check | expands to 12x² + 8x | 3x + 2 is done |
Read the signs before you search
- c positive: both numbers share the sign of b.
- c negative: one number is positive, one negative.
- The one larger in size takes the sign of b.
| Signs | The two numbers | Example |
|---|---|---|
| c positive, b positive | both numbers + | x² + 7x + 12 |
| c positive, b negative | both numbers − | x² − 9x + 20 |
| c negative, b negative | larger size − | x² − 5x − 14 |
| c negative, b positive | larger size + | x² + 2x − 8 |
When b and c are both negative
- Multiply to minus 14, add to minus 5.
- Opposite signs, and the one larger in size is negative.
- Minus 7 and 2. Then expand back to check.
| Step | Reasoning | Result |
|---|---|---|
| x² − 5x − 14 | need × −14, + −5 | |
| c negative | opposite signs | larger size negative |
| Try 7 and 2 | −7 × 2 = −14 | −7 + 2 = −5 |
| So | (x − 7)(x + 2) | |
| Check | x² + 2x − 7x − 14 | x² − 5x − 14 |
A difference of two squares
- Two terms, both perfect squares, with a minus between.
- 4x squared is 2x all squared. 25 is 5 squared.
- Plus in one bracket, minus in the other. The middle cancels.
| Question | Answer | Note |
|---|---|---|
| 4x² − 25 | is 4x² a square? | yes, (2x)² |
| Is 25 a square? | yes, 5² | and a minus between |
| So | (2x + 5)(2x − 5) | |
| Check | 4x² − 10x + 10x − 25 | middle cancels |
Squares hiding behind a common factor
- 2 and 18 are not squares, so take the 2 out first.
- Inside is x squared minus 9: a difference of two squares.
- x² + 9, a sum of two squares, does not factorise into brackets.
| Step | Working | Note |
|---|---|---|
| 2x² − 18 | 2 and 18 not squares | look for HCF |
| Take out 2 | 2(x² − 9) | |
| Inside | x² − 3² | two squares |
| So | 2(x + 3)(x − 3) | |
| But x² + 9 | a sum, not a difference | does not factorise |
When a is not 1: the ac method
- Multiply a by c: 2 times 3 is 6.
- Find two numbers: multiply to 6, add to 7. So 6 and 1.
- Split the middle term, then group in pairs.
| Step | Working | Gives |
|---|---|---|
| 2x² + 7x + 3 | a × c = 2 × 3 | 6 |
| Need × 6, + 7 | 6 and 1 | |
| Split the 7x | 2x² + 6x + x + 3 | |
| Group in pairs | 2x(x + 3) + 1(x + 3) | |
| Common bracket | (2x + 1)(x + 3) |
The ac method with negatives
- a times c is 3 times minus 8: minus 24.
- Minus 12 and 2 multiply to minus 24 and add to minus 10.
- Split, group, take out the common bracket, check.
| Step | Working | Gives |
|---|---|---|
| 3x² − 10x − 8 | a × c = 3 × −8 | −24 |
| Need × −24, + −10 | −12 and 2 | |
| Split the −10x | 3x² − 12x + 2x − 8 | |
| Group in pairs | 3x(x − 4) + 2(x − 4) | |
| Check | (3x + 2)(x − 4) | expands back |
Why multiply a by c?
- Expand two brackets and look at the two middle pieces.
- They always multiply to first coefficient times last.
- So the numbers you want must multiply to ac.
| Look at | Working | Gives |
|---|---|---|
| (2x + 1)(x + 3) | middle pieces | 6x and 1x |
| Multiply them | 6 × 1 | 6 |
| First times last | 2 × 3 | 6 |
| Always | middle pair | multiplies to ac |
Which method? Ask in this order
- Always look for a common factor first.
- Then count the terms, and look at the number in front.
- And finish every one by expanding back.
| Ask | Then | |
|---|---|---|
| 1 | Common factor? | always take it out first |
| 2 | Two squares, a minus? | difference of squares |
| 3 | Three terms, a = 1? | product and sum |
| 4 | Three terms, a ≠ 1? | ac method |
| 5 | Finished? | expand back to check |
The order in action
- Three terms and a is 2, so it looks like the ac method.
- But step one first: every term shares a 2.
- Take it out, and what is left has a equal to 1.
| Step | Working | Gives |
|---|---|---|
| 2x² − 8x − 24 | common factor 2 | 2(x² − 4x − 12) |
| Inside: a = 1 | need × −12, + −4 | −6 and 2 |
| So | 2(x − 6)(x + 2) | |
| Check | expands to 2x² − 8x − 24 | matches |
When the factors are lengths
- A rectangular garden plot. Area is length times width.
- Factorise the area and one factor is the known width.
- Test with x equals 4: both give 98 square metres.
| Know | Working | Gives |
|---|---|---|
| Area 3x² + 11x + 6 | width x + 3 | length? |
| ac = 18 | need × 18, + 11 | 9 and 2 |
| Split and group | 3x(x + 3) + 2(x + 3) | |
| So | (3x + 2)(x + 3) | length 3x + 2 |
| Test x = 4 | area 98 | 14 × 7 = 98 |
What to take away
- Seven lines. The first and last are the same idea.
| Factorising | is expanding in reverse |
| HCF first | number, then lowest powers |
| c negative | opposite signs in the brackets |
| Square minus square | (a + b)(a − b) |
| a not 1 | find × ac, + b, then split |
| Which method? | HCF, then count the terms |
| Always | expand back to check |
Your turn: four questions
- Pause here and write all four out, with reasons.
- Questions 1 and 3 say fully: check the HCF.
- Question 4 has a front number of 2.
| Question | |
|---|---|
| 1 | Factorise 8a²b + 12ab² fully. |
| 2 | Factorise x² − 3x − 18. |
| 3 | Factorise 3x² − 75 fully. |
| 4 | Factorise 2x² − 5x − 3. |
Show answer
1. 4ab(2a + 3b). The HCF of 8 and 12 is 4, and the lowest powers are a and b. Check: 4ab × 2a = 8a²b and 4ab × 3b = 12ab². 2. (x − 6)(x + 3). The numbers multiply to −18 and add to −3, so they have opposite signs and the one larger in size is negative. 3. 3(x + 5)(x − 5). Take out 3 first, leaving x² − 25, a difference of two squares. 4. (2x + 1)(x − 3). ac is −6, and −6 and 1 add to −5. Split: 2x² − 6x + x − 3 = 2x(x − 3) + 1(x − 3).
The answers
- 1. 4ab(2a + 3b)
- 2. (x − 6)(x + 3)
- 3. 3(x + 5)(x − 5)
- 4. (2x + 1)(x − 3)
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