A student attempts to write 3x2−x(x+2)(x−3)2\displaystyle \frac{3x^2 - x}{(x+2)(x-3)^2}(x+2)(x−3)23x2−x in partial fractions with constant numerators.
Their incorrect attempt is shown below:
Step 1 3x2−x(x+2)(x−3)2≡Ax+2+B(x−3)2\displaystyle \frac{3x^2 - x}{(x+2)(x-3)^2} \equiv \frac{A}{x+2} + \frac{B}{(x-3)^2}(x+2)(x−3)23x2−x≡x+2A+(x−3)2B
Step 2 3x2−x≡A(x−3)2+B(x+2)3x^2 - x \equiv A(x-3)^2 + B(x+2)3x2−x≡A(x−3)2+B(x+2)
Step 3 Let x=3⇒B=4.8x = 3 \Rightarrow B = 4.8x=3⇒B=4.8 Let x=−2⇒A=0.56x = -2 \Rightarrow A = 0.56x=−2⇒A=0.56
Answer 0.56x+2+4.8(x−3)2\displaystyle \frac{0.56}{x+2} + \frac{4.8}{(x-3)^2}x+20.56+(x−3)24.8
Explain the mistake that the student has made in step 1.
