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Simplify fully \[ \frac{(x + 3)(6 - 2x)}{(x - 3)(3 + x)} \] for \( x \neq -3 \) Circle your answer, −2 2 \[ \frac{(6 - 2x)}{(x - 3)} \] \[ \frac{(2x - 6)}{(x - 3)} \] - AQA - A-Level Maths Mechanics - Question 2 - 2021 - Paper 3

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Simplify-fully--\[-\frac{(x-+-3)(6---2x)}{(x---3)(3-+-x)}-\]-for-\(-x-\neq--3-\)--Circle-your-answer,--−2--2--\[-\frac{(6---2x)}{(x---3)}-\]--\[-\frac{(2x---6)}{(x---3)}-\]-AQA-A-Level Maths Mechanics-Question 2-2021-Paper 3.png

Simplify fully \[ \frac{(x + 3)(6 - 2x)}{(x - 3)(3 + x)} \] for \( x \neq -3 \) Circle your answer, −2 2 \[ \frac{(6 - 2x)}{(x - 3)} \] \[ \frac{(2x - 6)}{(x -... show full transcript

Worked Solution & Example Answer:Simplify fully \[ \frac{(x + 3)(6 - 2x)}{(x - 3)(3 + x)} \] for \( x \neq -3 \) Circle your answer, −2 2 \[ \frac{(6 - 2x)}{(x - 3)} \] \[ \frac{(2x - 6)}{(x - 3)} \] - AQA - A-Level Maths Mechanics - Question 2 - 2021 - Paper 3

Step 1

Step 1: Expand the numerator and denominator

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Answer

The numerator is obtained by expanding: [ (x + 3)(6 - 2x) = 6x + 18 - 2x^2 - 6x = -2x^2 + 18 ]

The denominator is expanded as follows: [ (x - 3)(3 + x) = 3x + x^2 - 9 = x^2 + 3x - 9 ]

Step 2

Step 2: Substitute in the expression

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Answer

Now, substituting the expanded forms back into the expression, we get:

[ \frac{-2x^2 + 18}{x^2 + 3x - 9} ]

Step 3

Step 3: Factor out the numerator and denominator

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Answer

Factor the numerator: [ -2(x^2 - 9) = -2(x - 3)(x + 3) ]

The denominator can be factored as: [ (x - 3)(x + 3) ]

Step 4

Step 4: Cancel common factors

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Answer

Cancelling out the common factors, we have: [ \frac{-2(x - 3)(x + 3)}{(x - 3)(x + 3)} = -2 ]

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