EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Expanding and factorising
Algebra · Lesson 2 of 7
Last Lesson and Before
Answer each one, then check.
1. Simplify 3a + 5a − 2a.
6a
2. Work out −3 × −4.
12
3. What is the HCF of 12 and 18?
6
4. Last lesson: simplify x × x.
x²
5. Last lesson: simplify 2x × 5x².
10x³
Learning Objectives
1. Expand single brackets and simplify.
2. Factorise expressions by taking out the highest common factor.
3. Expand double brackets.
4. Factorise quadratic expressions of the form x² + bx + c.
5. Factorise the difference of two squares, x² − a².
The Words You Need
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Term One part of an expression: 3x, −5 and x² are terms. |
Expression Terms added or subtracted, with no equals sign: 3x² − 5x + 2. |
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Coefficient The number in front of a letter: the 3 in 3x². |
Like terms Same letters and powers: 4x and −7x, but not 4x and 4x². |
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Expand Multiply out the brackets. |
Factorise Put back into brackets - the opposite of expanding. |
Expanding Single Brackets
Multiply every term inside the bracket by the term outside.
▸ Numbers. 4(3x − 7) = 12x − 28.
▸ Letters. x(x + 5) = x² + 5x.
▸ Negative outside. −3(x − 4) = −3x + 12: a negative times a negative is positive.
▸ Expand and simplify. Expand every bracket, then collect like terms.
Expand and Simplify
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Expand and simplify 5(2x + 3) − 3(x − 4) |
1. Expand the first bracket
5(2x + 3) = 10x + 15
2. Expand the second: the −3 multiplies both terms
−3(x − 4) = −3x + 12
3. Write it all out
10x + 15 − 3x + 12
4. Collect like terms
10x − 3x = 7x and 15 + 12 = 27
Answer: 7x + 27
Factorising Fully
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Factorise fully 12x²y − 18xy² |
1. Find the HCF of the numbers
HCF of 12 and 18 = 6
2. Find the HCF of the letters
Both terms have x and y: xy
3. The HCF goes outside
6xy(…)
4. Divide each term by the HCF
12x²y ÷ 6xy = 2x and 18xy² ÷ 6xy = 3y
Answer: 6xy(2x − 3y)
PART ONE
Double Brackets
Every term in the first bracket multiplies every term in the second.
The Grid Method for (x + 4)(x - 3)
Multiply each row by each column, then collect like terms: x² − 3x + 4x − 12 = x² + x − 12.
|
× |
x |
−3 |
|---|---|---|
|
x |
x² |
−3x |
|
+4 |
+4x |
−12 |
Squaring a Bracket
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Expand and simplify (2x − 5)² |
1. Write it as two brackets
(2x − 5)(2x − 5)
2. First times first
2x × 2x = 4x²
3. Outer and inner
2x × (−5) = −10x and (−5) × 2x = −10x
4. Last times last
(−5) × (−5) = +25
5. Collect like terms
4x² − 20x + 25
Answer: 4x² − 20x + 25 (not 4x² + 25)
PART TWO
Factorising Quadratics
Double brackets in reverse.
Factorising x² + bx + c
Find two numbers that MULTIPLY to give c and ADD to give b.
▸ The brackets. x² + bx + c = (x + p)(x + q), where pq = c and p + q = b.
▸ c positive. Both numbers have the same sign as b: x² − 7x + 12 = (x − 3)(x − 4).
▸ c negative. The numbers have opposite signs: x² + x − 12 = (x + 4)(x − 3).
▸ Check. Expand your answer: it should give the expression you started with.
Factorising a Quadratic
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Factorise x² − 2x − 15 |
1. List pairs that multiply to −15
1 and −15, −1 and 15, 3 and −5, −3 and 5
2. Pick the pair that adds to −2
3 + (−5) = −2
3. Write the brackets
(x + 3)(x − 5)
4. Check by expanding
x² − 5x + 3x − 15 = x² − 2x − 15
Answer: (x + 3)(x − 5)
The Difference of Two Squares
One square number (or square term) minus another factorises straight away.
▸ The rule. x² − a² = (x + a)(x − a).
▸ Examples. x² − 49 = (x + 7)(x − 7). x² − 1 = (x + 1)(x − 1).
▸ Why it works. (x + 7)(x − 7) = x² − 7x + 7x − 49: the middle terms cancel.
▸ Spot it. Two terms, both square, with a MINUS between them. x² + 49 does not factorise.
Match the Expression to Its Factors
|
Expression |
Factorised |
|---|---|
|
x² + 5x + 6 (x + 2)(x + 3) |
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|
x² − 9 (x + 3)(x − 3) |
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x² + x − 6 (x + 3)(x − 2) |
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x² − 5x + 6 (x − 2)(x − 3) |
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4x + 12 4(x + 3) |
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x² + 3x x(x + 3) |
Key Terms
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Expand Multiply out brackets. |
Factorise Write an expression as a product of factors, using brackets. |
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Factorise fully Take out the highest common factor, so no common factor is left inside. |
Quadratic expression An expression whose highest power is x², such as x² + 5x + 6. |
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Difference of two squares An expression of the form x² − a², which factorises to (x + a)(x − a). |
Identity Two expressions that are equal for every value of x, written with ≡. |
Your Task: Expand, Then Undo
12 minutes
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Work in pairs. Partner A expands (a) (x + 5)(x + 2) (b) (x − 4)(x + 6) (c) (x − 3)². Partner B factorises (d) x² + 9x + 14 (e) x² − 4x − 21 (f) x² − 100. Then swap papers and check each other's answers by doing the opposite process. 1. Partner A: expand. 2. Partner B: factorise. 3. Swap and check by reversing. |
A good answer shows: (a) x² + 7x + 10 (b) x² + 2x − 24 (c) x² − 6x + 9 (d) (x + 2)(x + 7) (e) (x − 7)(x + 3) (f) (x + 10)(x − 10)
Can I...?
☐ Expand a single bracket.
☐ Expand and simplify two single brackets.
☐ Factorise by taking out the HCF.
☐ Factorise fully.
☐ Expand double brackets.
☐ Square a bracket.
☐ Factorise x² + bx + c.
☐ Factorise the difference of two squares.
Summary
✓ Expand: multiply every term inside by every term outside.
✓ Factorise: take out the HCF of the numbers AND the letters.
✓ Double brackets: four multiplications, then collect like terms.
✓ x² + bx + c: find two numbers that multiply to c and add to b.
✓ x² − a² = (x + a)(x − a).
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EXAM FOCUS Factorise x² + 9x + 20 (2 marks) Always check a factorisation by expanding it again. It takes ten seconds and catches sign errors that would cost the mark. |