Expanding and factorising
Grade 8: expand a single bracket, factorise by taking out a common factor, collect like terms after expanding, substitute into a formula, rearrange a simple formula.
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Questions
Expand 3(x + 4)
- 3x + 12
- 3x + 4
- 3x + 7
- x + 12
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3x + 12Expand 5(2a − 3)
- 10a + 15
- 10a − 15
- 10a − 3
- 7a − 15
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10a − 15Expanding 4(y + 2) gives 4y + 8. (True or false?)
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TrueFactorise 6x + 9 by taking out the highest common factor.
- 2(3x + 9)
- 3(2x + 3)
- 3(2x + 9)
- 6(x + 9)
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3(2x + 3)Factorise 8a − 12
- 2(4a − 3)
- 4(2a − 12)
- 4(2a − 3)
- 8(a − 12)
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4(2a − 3)What is the highest common factor of 12x and 18? (Answer out loud)
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6Expand and simplify 2(x + 3) + 4x
- 2x + 18
- 6x + 3
- 6x + 6
- 8x + 6
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6x + 6Expand and simplify 3(x + 2) + 2(x − 1)
- 5x + 4
- 5x + 5
- 5x + 8
- 5x − 4
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5x + 4Factorising 5x² + 10x gives 5x(x + 2). (True or false?)
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TrueExpand and simplify 4(x + 5) − 2x. (Answer out loud)
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2x + 20Using P = 2l + 2w, find P when l = 7 and w = 3.
- 13
- 17
- 20
- 21
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20Using v = u + 5t, find v when u = 4 and t = 3.
- 12
- 15
- 19
- 27
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19If y = 3x − 2 and x = 4, then y = 14. (True or false?)
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FalseRearrange y = x + 7 to make x the subject.
- x = 7 − y
- x = 7y
- x = y + 7
- x = y − 7
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x = y − 7Make x the subject of y = 3x.
- x = 3 − y
- x = 3y
- x = y ÷ 3
- x = y − 3
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x = y ÷ 3Make x the subject of y = 2x + 5. (Answer out loud)
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x = (y − 5) ÷ 2