A student attempts to write $kx2−5x(x+1)(x−2)2\displaystyle \frac{kx^2 - 5x}{(x+1)(x-2)^2}$ in partial fractions with constant numerators.

Their incorrect attempt is shown below:

Step 1 $kx2−5x(x+1)(x−2)2≡A(x+1)+B(x−2)2\displaystyle \frac{kx^2 - 5x}{(x+1)(x-2)^2} \equiv \frac{A}{(x+1)} + \frac{B}{(x-2)^2}$

Step 2 $kx2−5x≡A(x−2)2+B(x+1)kx^2 - 5x \equiv A(x-2)^2 + B(x+1)$

Step 3 Let $x=2⇒B=4k−103\displaystyle x = 2 \Rightarrow B = \frac{4k-10}{3}$ Let $x=−1⇒A=k+59\displaystyle x = -1 \Rightarrow A = \frac{k+5}{9}$

Answer $k+59(x+1)+4k−103(x−2)2\displaystyle \frac{k+5}{9(x+1)} + \frac{4k-10}{3(x-2)^2}$

Explain the mistake that the student has made in step 1 and write the correct partial fraction decomposition in terms of $k$ .

A student attempts to write $kx2−5x(x+1)(x−2)2\displaystyle \frac{kx^2 - 5x}{(x+1)(x-2)^2}$ in partial fractions with constant numerators.

Their incorrect attempt is shown below:

Step 1 $kx2−5x(x+1)(x−2)2≡A(x+1)+B(x−2)2\displaystyle \frac{kx^2 - 5x}{(x+1)(x-2)^2} \equiv \frac{A}{(x+1)} + \frac{B}{(x-2)^2}$

Step 2 $kx2−5x≡A(x−2)2+B(x+1)kx^2 - 5x \equiv A(x-2)^2 + B(x+1)$

Step 3 Let $x=2⇒B=4k−103\displaystyle x = 2 \Rightarrow B = \frac{4k-10}{3}$ Let $x=−1⇒A=k+59\displaystyle x = -1 \Rightarrow A = \frac{k+5}{9}$

Answer $k+59(x+1)+4k−103(x−2)2\displaystyle \frac{k+5}{9(x+1)} + \frac{4k-10}{3(x-2)^2}$

Explain the mistake that the student has made in step 1 and write the correct partial fraction decomposition in terms of $k$ .

Algebraic Methods

Algebraic Methods Questions