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14 (a) Factorise fully \(4p^2 - 36\) (b) Show that \((m + 4)(2m - 5)(3m + 1)\) can be written in the form \(am^3 + bm^2 + cm + d\) where \(a, b, c\) and \(d\) are integers. - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 3
Question 15
14 (a) Factorise fully \(4p^2 - 36\) (b) Show that \((m + 4)(2m - 5)(3m + 1)\) can be written in the form \(am^3 + bm^2 + cm + d\) where \(a, b, c\) and \(d\) are i... show full transcript
Worked Solution & Example Answer:14 (a) Factorise fully \(4p^2 - 36\) (b) Show that \((m + 4)(2m - 5)(3m + 1)\) can be written in the form \(am^3 + bm^2 + cm + d\) where \(a, b, c\) and \(d\) are integers. - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 3
Step 1
Factorise fully \(4p^2 - 36\)
Answer
To factorise the expression (4p^2 - 36), we first recognize that it is a difference of squares:
We can apply the difference of squares formula, which states that (a^2 - b^2 = (a - b)(a + b)) to get:
Next, we can simplify this further by factoring out common factors. Notice that (2p - 6) can be factored as (2(p - 3)) and (2p + 6) can be factored as (2(p + 3)):
Thus, the full factorization is:
Step 2
Show that \((m + 4)(2m - 5)(3m + 1)\) can be written in the form \(am^3 + bm^2 + cm + d\)
Answer
To express ((m + 4)(2m - 5)(3m + 1)) in the standard polynomial form, we start by expanding the first two factors:
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Expand ((m + 4)(2m - 5)):
[(m + 4)(2m - 5) = m(2m - 5) + 4(2m - 5) = 2m^2 - 5m + 8m - 20 = 2m^2 + 3m - 20]
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Next, multiply the result by ((3m + 1)):
[(2m^2 + 3m - 20)(3m + 1) = 2m^2(3m) + 2m^2(1) + 3m(3m) + 3m(1) - 20(3m) - 20(1)]
This expands to:
[6m^3 + 2m^2 + 9m^2 + 3m - 60m - 20 = 6m^3 + 11m^2 - 57m - 20]
Finally, we can write it in the form (am^3 + bm^2 + cm + d) where:
- (a = 6)
- (b = 11)
- (c = -57)
- (d = -20)