Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators. Her incorrect attempt is shown below. Step 1 $$\frac{2x^2 + x}{(x + 1)(x + 2)^2} = \frac{A}{(x + 1)} + \frac{B}{(x + 2)^2}$$ Step 2 $$2x^2 + x = A(x + 2)^2 + B(x + 1)$$ Step 3 Let $x = -1 \Rightarrow A = 1$ Let $x = -2 \Rightarrow B = -6$ Answer $$\frac{2x^2 + x}{(x + 1)(x + 2)^2} = \frac{1}{(x + 1)} + \frac{6}{(x + 2)}$$

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Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 1

Question 9

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Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators. Her incorrect attempt is shown below. Step 1 $$... show full transcript

Worked Solution & Example Answer:Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 1

Step 1

9 (a) (i) By using a counter example, show that the answer obtained by Chloe cannot be correct.

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Answer

To demonstrate that Chloe's answer cannot be correct, we can use a specific value for $x$ . Let's choose $x = -2$ .

Calculating the left-hand side (LHS): $LHS = \frac{2(-2)^2 + (-2)}{(-2 + 1)(-2 + 2)^2} = \frac{8 - 2}{(-1)(0)^2} = \frac{6}{0},$ which is undefined.

Now, considering the right-hand side (RHS) of Chloe's expression: $RHS = \frac{1}{(-2 + 1)} + \frac{6}{(-2 + 2)} = \frac{1}{-1} + \frac{6}{0} = -1 + \text{undefined},$ which leads to an inconsistency.

As we see, $LHS$ is undefined while $RHS$ also results in an undefined term. Therefore, since $LHS \neq RHS$ , we conclude that Chloe's expression cannot be correct.

Step 2

9 (a) (ii) Explain her mistake in Step 1.

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Answer

Chloe's mistake in Step 1 lies in failing to account for all terms when decomposing the fraction.

Specifically, she should have included an additional term involving $rac{C}{(x + 2)}$ in the denominator. This is crucial for ensuring the degrees of the numerator match the required form of the partial fractions, especially since $(x + 2)$ appears as a repeated factor in the denominator.

Step 3

9 (b) Write \frac{2x^2+x}{(x + 1)(x + 2)^2} as partial fractions, with constant numerators.

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Answer

To rewrite $\frac{2x^2 + x}{(x + 1)(x + 2)^2}$ as partial fractions, we express it in the form: $\frac{A}{(x + 1)} + \frac{B}{(x + 2)} + \frac{C}{(x + 2)^2}$

Setting up the equation: $2x^2 + x = A(x + 2)^2 + B(x + 1)(x + 2) + C(x + 1)$

Expanding the right side: $\Rightarrow 2x^2 + x = A(x^2 + 4x + 4) + B(x^2 + 3x + 2) + C(x + 1)$

Combining terms: $\Rightarrow (A + B)x^2 + (4A + 3B + C)x + (4A + 2B + C)$

Now, equating coefficients from both sides, we have:

$A + B = 2$
$4A + 3B + C = 1$
$4A + 2B + C = 0$

By solving this system of equations, we determine:

Starting with $A + B = 2$ , let $B = 2 - A$ .
Substitute into the other equations, ultimately resolving for $A$ , $B$ , and $C$ :

After calculations, we find: $A = 1, B = -4, C = 4.$

Thus, the partial fraction decomposition is: $\frac{2x^2 + x}{(x + 1)(x + 2)^2} = \frac{1}{(x + 1)} + \frac{-4}{(x + 2)} + \frac{4}{(x + 2)^2}.$

Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 1

Worked Solution & Example Answer:Chloe is attempting to write $$\frac{2x^2 + x}{(x + 1)(x + 2)^2}$$ as partial fractions, with constant numerators - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 1

9 (a) (i) By using a counter example, show that the answer obtained by Chloe cannot be correct.

9 (a) (ii) Explain her mistake in Step 1.

9 (b) Write \frac{2x^2+x}{(x + 1)(x + 2)^2} as partial fractions, with constant numerators.

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