Further Pure Mathematics 4PM1 / Sketching Polynomials / 1.61.6: Past-paper question 34PM1/1/November/2023 — Question 3 · 8 marksMark as done · Save for laterg′(x)=mx2−10x−37wherem is an integerg'(x) = mx^2 - 10x - 37 \quad \mathrm{where} m \text{ is an integer}g′(x)=mx2−10x−37wherem is an integerThe curve y=g(x)\displaystyle y = g(x)y=g(x) passes through the point with coordinates (1,20)\displaystyle (1, 20)(1,20)Given that (x−5)\displaystyle (x-5)(x−5) is a factor of g(x)\displaystyle g(x)g(x)(a) show that g(x)=2x3−5x2−37x+60\displaystyle g(x) = 2x^3 - 5x^2 - 37x + 60g(x)=2x3−5x2−37x+60(5)(b) Hence, or otherwise, use algebra to solve the equation g(x)=0\displaystyle g(x) = 0g(x)=0(3)▸ Mark scheme← 1.5: 4PM1/1/June/2023 — Question 81.1: 4PM1/2R/June/2022 — Question 4 →
