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Question 147

A student attempts to write 4x2−5x(x+1)(x−2)2\displaystyle \frac{4x^2 - 5x}{(x+1)(x-2)^2}(x+1)(x−2)24x2−5x​ in partial fractions with constant numerators.

Their incorrect attempt is shown below:

Step 1 4x2−5x(x+1)(x−2)2≡A(x+1)+B(x−2)2\displaystyle \frac{4x^2 - 5x}{(x+1)(x-2)^2} \equiv \frac{A}{(x+1)} + \frac{B}{(x-2)^2}(x+1)(x−2)24x2−5x​≡(x+1)A​+(x−2)2B​

Step 2 4x2−5x≡A(x−2)2+B(x+1)4x^2 - 5x \equiv A(x-2)^2 + B(x+1)4x2−5x≡A(x−2)2+B(x+1)

Step 3 Let x=2⇒B=2x = 2 \Rightarrow B = 2x=2⇒B=2 Let x=−1⇒A=1x = -1 \Rightarrow A = 1x=−1⇒A=1

Answer 1(x+1)+2(x−2)2\displaystyle \frac{1}{(x+1)} + \frac{2}{(x-2)^2}(x+1)1​+(x−2)22​

Explain the mistake that the student has made in step 1.

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