A student attempts to write 2x2+7x(x+4)(x−1)2\displaystyle \frac{2x^2 + 7x}{(x+4)(x-1)^2}(x+4)(x−1)22x2+7x in partial fractions with constant numerators.
Their incorrect attempt is shown below:
Step 1 2x2+7x(x+4)(x−1)2≡Ax+4+B(x−1)2\displaystyle \frac{2x^2 + 7x}{(x+4)(x-1)^2} \equiv \frac{A}{x+4} + \frac{B}{(x-1)^2}(x+4)(x−1)22x2+7x≡x+4A+(x−1)2B
Step 2 2x2+7x≡A(x−1)2+B(x+4)2x^2 + 7x \equiv A(x-1)^2 + B(x+4)2x2+7x≡A(x−1)2+B(x+4)
Step 3 Let x=1⇒B=1.8x = 1 \Rightarrow B = 1.8x=1⇒B=1.8 Let x=−4⇒A=0.16x = -4 \Rightarrow A = 0.16x=−4⇒A=0.16
Answer 0.16x+4+1.8(x−1)2\displaystyle \frac{0.16}{x+4} + \frac{1.8}{(x-1)^2}x+40.16+(x−1)21.8
Explain the mistake that the student has made in step 1.
