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14 (a) Factorise fully 4p² - 36 (b) Show that (m + 4)(2m - 5)(3m + 1) can be written in the form am² + bm + cm + d where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 3

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14-(a)-Factorise-fully-4p²---36--(b)-Show-that-(m-+-4)(2m---5)(3m-+-1)-can-be-written-in-the-form-am²-+-bm-+-cm-+-d-where-a,-b,-c-and-d-are-integers.-Edexcel-GCSE Maths-Question 15-2022-Paper 3.png

14 (a) Factorise fully 4p² - 36 (b) Show that (m + 4)(2m - 5)(3m + 1) can be written in the form am² + bm + cm + d where a, b, c and d are integers.

Worked Solution & Example Answer:14 (a) Factorise fully 4p² - 36 (b) Show that (m + 4)(2m - 5)(3m + 1) can be written in the form am² + bm + cm + d where a, b, c and d are integers. - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 3

Step 1

Factorise fully 4p² - 36

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Answer

To factorise the expression 4p² - 36, we start by recognizing that it is a difference of squares. We can express this as:

4p236=(2p)2624p^2 - 36 = (2p)^2 - 6^2

According to the difference of squares formula, a2b2=(a+b)(ab)a^2 - b^2 = (a + b)(a - b), we can rewrite the expression as:

(2p+6)(2p6)(2p + 6)(2p - 6)

Next, we can factor out a common factor from each of the terms:

2(p+3)(p3)2(p + 3)(p - 3)

So, the fully factored form of 4p² - 36 is:

2(p+3)(p3)2(p + 3)(p - 3)

Step 2

Show that (m + 4)(2m - 5)(3m + 1) can be written in the form am² + bm + cm + d

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Answer

To show that the expression (m + 4)(2m - 5)(3m + 1) can be written in the form am² + bm + cm + d, we will first expand the expression step by step.

  1. Start by expanding (m + 4)(2m - 5):

    (m+4)(2m5)=m(2m)+m(5)+4(2m)+4(5)(m + 4)(2m - 5) = m(2m) + m(-5) + 4(2m) + 4(-5)

    =2m25m+8m20= 2m^2 - 5m + 8m - 20

    =2m2+3m20= 2m^2 + 3m - 20

  2. Next, we will multiply this result by (3m + 1):

    (2m2+3m20)(3m+1)(2m^2 + 3m - 20)(3m + 1)

    Expanding gives:

    =2m2(3m)+2m2(1)+3m(3m)+3m(1)20(3m)20(1)= 2m^2(3m) + 2m^2(1) + 3m(3m) + 3m(1) - 20(3m) - 20(1)

    =6m3+2m2+9m2+3m60m20= 6m^3 + 2m^2 + 9m^2 + 3m - 60m - 20

    =6m3+(2m2+9m2)+(3m60m)20= 6m^3 + (2m^2 + 9m^2) + (3m - 60m) - 20

    =6m3+11m257m20= 6m^3 + 11m^2 - 57m - 20

Thus, we have expressed (m + 4)(2m - 5)(3m + 1) as:

6m3+11m257m206m^3 + 11m^2 - 57m - 20

This verifies that it can indeed be arranged in the desired form where all constants are integers, specifically where:

  • a = 6
  • b = 11
  • c = -57
  • d = -20.

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