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(a) Simplify $(xy)^y$ (b) Expand and simplify $4(x + 3) + 7(4 - 2x)$ (c) Factorise fully $15x^2 + 3xy$ - Edexcel - GCSE Maths - Question 4 - 2022 - Paper 2

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(a)-Simplify--$(xy)^y$--(b)-Expand-and-simplify--$4(x-+-3)-+-7(4---2x)$--(c)-Factorise-fully--$15x^2-+-3xy$-Edexcel-GCSE Maths-Question 4-2022-Paper 2.png

(a) Simplify $(xy)^y$ (b) Expand and simplify $4(x + 3) + 7(4 - 2x)$ (c) Factorise fully $15x^2 + 3xy$

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

Step 1

Simplify $(xy)^y$

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Answer

To simplify the expression (xy)y(xy)^y, we use the property of exponents that states ambn=(ab)m+na^m b^n = (ab)^{m+n}. Thus, we can express (xy)y(xy)^y as:

(xy)^y &= x^y y^y. ext{Therefore, the simplified form is:} \boxed{x^y y^y}. \end{align*}$$

Step 2

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

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Answer

First, we will expand both terms:

  1. For 4(x+3)4(x + 3): 4(x+3)=4x+12.4(x + 3) = 4x + 12.

  2. For 7(4−2x)7(4 - 2x): 7(4−2x)=28−14x.7(4 - 2x) = 28 - 14x.

Now, we combine these results:

4x+12+28−14x=(4x−14x)+(12+28)=−10x+40.4x + 12 + 28 - 14x = (4x - 14x) + (12 + 28) = -10x + 40.

So, the final simplified expression is: −10x+40.\boxed{-10x + 40}.

Step 3

Factorise fully $15x^2 + 3xy$

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Answer

To factorise the expression 15x2+3xy15x^2 + 3xy, we first identify the greatest common factor (GCF):

The GCF of 15x215x^2 and 3xy3xy is 3x3x.

Now, we factor out 3x3x:

15x2+3xy=3x(5x+y).15x^2 + 3xy = 3x(5x + y).

Thus, the fully factored form is: 3x(5x+y).\boxed{3x(5x + y)}.

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