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Given that $x > 0$ and $x \neq 25$, fully simplify $$ \frac{10 + 5x - 2x^{2} - x^{3}}{5 - \sqrt{x}}\n$$ Fully justify your answer. - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 3

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Given-that-$x->-0$-and-$x-\neq-25$,-fully-simplify--$$-\frac{10-+-5x---2x^{2}---x^{3}}{5---\sqrt{x}}\n$$--Fully-justify-your-answer.-AQA-A-Level Maths Pure-Question 6-2021-Paper 3.png

Given that $x > 0$ and $x \neq 25$, fully simplify $$ \frac{10 + 5x - 2x^{2} - x^{3}}{5 - \sqrt{x}}\n$$ Fully justify your answer.

Worked Solution & Example Answer:Given that $x > 0$ and $x \neq 25$, fully simplify $$ \frac{10 + 5x - 2x^{2} - x^{3}}{5 - \sqrt{x}}\n$$ Fully justify your answer. - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 3

Step 1

Identify the expression to simplify

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Answer

We start with the expression to simplify, which is:

10+5x2x2x35x.\frac{10 + 5x - 2x^{2} - x^{3}}{5 - \sqrt{x}}.

Step 2

Factor the numerator

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Answer

Next, we need to factor the numerator. Let's rearrange and resize terms:

10+5x2x2x3=x32x2+5x+10.10 + 5x - 2x^{2} - x^{3} = -x^{3} - 2x^{2} + 5x + 10.

Factoring the numerator gives:

(x3+2x25x10)=(x+2)(x25).-(x^{3} + 2x^{2} - 5x - 10) = -(x + 2)(x^{2} - 5).

Step 3

Identify common factors

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Answer

Now, observe the denominator:

5x.5 - \sqrt{x}.

By substituting back, we schedule that no common factor is directly observable. However, multiplying the denominator by its conjugate we reformulate expressing it as:

(5x)(5+x)=25x.(5 - \sqrt{x})(5 + \sqrt{x}) = 25 - x.

Step 4

Complete the simplification

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Answer

Finally, simplify the entire expression:

(x+2)(x25)25x.\frac{-(x + 2)(x^{2} - 5)}{25 - x}.

We note that cancellation can occur, yielding:

(x+2).-(x + 2).

Thus, the fully simplified form is:

(x+2).-(x + 2).

This representation is valid under the condition that x25x \neq 25.

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