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1 (a) Simplify $(x^y)^{y}$ (b) Expand and simplify $4(x + 3) + 7(4 - 2x)$ (c) Factorise fully $15x^2 + 3xy$ (Total for Question 1 is 5 marks) - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 2

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Question 1

1--(a)-Simplify--$(x^y)^{y}$---(b)-Expand-and-simplify--$4(x-+-3)-+-7(4---2x)$---(c)-Factorise-fully--$15x^2-+-3xy$----(Total-for-Question-1-is-5-marks)-Edexcel-GCSE Maths-Question 1-2022-Paper 2.png

1 (a) Simplify $(x^y)^{y}$ (b) Expand and simplify $4(x + 3) + 7(4 - 2x)$ (c) Factorise fully $15x^2 + 3xy$ (Total for Question 1 is 5 marks)

Worked Solution & Example Answer:1 (a) Simplify $(x^y)^{y}$ (b) Expand and simplify $4(x + 3) + 7(4 - 2x)$ (c) Factorise fully $15x^2 + 3xy$ (Total for Question 1 is 5 marks) - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 2

Step 1

Simplify $(x^y)^{y}$

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Answer

To simplify (xy)y(x^y)^{y}, we use the power of a power rule which states that (am)n=amn(a^m)^n = a^{m \cdot n}. Therefore:

(xy)y=xyy=xy2(x^y)^{y} = x^{y \cdot y} = x^{y^2}

Thus, the final simplification is:

xy2x^{y^2}

Step 2

Expand and simplify $4(x + 3) + 7(4 - 2x)$

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Answer

  1. Distribute the terms:

    • For the first part:
      4(x+3)=4x+124(x + 3) = 4x + 12
    • For the second part:
      7(42x)=2814x7(4 - 2x) = 28 - 14x
  2. Combine like terms:

    • Combine the results from above:
      4x+12+2814x=(4x14x)+(12+28)=10x+404x + 12 + 28 - 14x = (4x - 14x) + (12 + 28) = -10x + 40

Thus, the final expanded and simplified expression is:

10x+40-10x + 40

Step 3

Factorise fully $15x^2 + 3xy$

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Answer

To factorise the expression 15x2+3xy15x^2 + 3xy, we first look for the greatest common factor (GCF). In this case, the GCF of 15x215x^2 and 3xy3xy is 3x3x. We can factor this out:

  1. Factor out 3x3x: 15x2+3xy=3x(5x+y)15x^2 + 3xy = 3x(5x + y)

Therefore, the fully factored form is:

3x(5x+y)3x(5x + y)

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