Jamie is attempting to write 3x2+1(x−1)(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2}(x−1)(x+3)23x2+1 as partial fractions with constant numerators.
His incorrect attempt is shown below.
Step 1: 3x2+1(x−1)(x+3)2≡Ax−1+B(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2} \equiv \frac{A}{x - 1} + \frac{B}{(x + 3)^2}(x−1)(x+3)23x2+1≡x−1A+(x+3)2B
Step 2: 3x2+1≡A(x+3)2+B(x−1)3x^2 + 1 \equiv A(x + 3)^2 + B(x - 1)3x2+1≡A(x+3)2+B(x−1)
Step 3: Let x=1⇒4=16A⇒A=14x = 1 \Rightarrow 4 = 16A \Rightarrow A = \frac{1}{4}x=1⇒4=16A⇒A=41 Let x=−3⇒28=−4B⇒B=−7x = -3 \Rightarrow 28 = -4B \Rightarrow B = -7x=−3⇒28=−4B⇒B=−7
Answer: 3x2+1(x−1)(x+3)2≡14(x−1)−7(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2} \equiv \frac{1}{4(x - 1)} - \frac{7}{(x + 3)^2}(x−1)(x+3)23x2+1≡4(x−1)1−(x+3)27
(i) By using a counter-example, show that the answer obtained by Jamie cannot be correct.
(ii) Explain the mistake Jamie made in Step 1.
Write 3x2+1(x−1)(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2}(x−1)(x+3)23x2+1 as partial fractions, with constant numerators.
25 exam-style questions on Edexcel A Level Maths 1.4 Repeated Factors. Each one has a worked solution and a mark scheme showing where the marks go.