Further Pure Mathematics 4PM1 / Sketching Polynomials / 1.71.7: Past-paper question 14PM1/2/June/2024 — Question 1 · 6 marksMark as done · Save for laterf(x)=6x3−13x2+ax−10wherea is a constantf(x) = 6x^3 - 13x^2 + ax - 10 \quad \mathrm{where} a \text{ is a constant}f(x)=6x3−13x2+ax−10wherea is a constantGiven that (3x−2)\displaystyle (3x - 2)(3x−2) is a factor of f(x)\displaystyle f(x)f(x)(a) show that a=21\displaystyle a = 21a=21(2)(b) Hence show algebraically that the curve y=f(x)\displaystyle y = f(x)y=f(x) has only one intersection with the x\displaystyle xx-axis.(4)▸ Mark scheme← 1.10: 4PM1/2R/June/2025 — Question 51.8: 4PM1/2/June/2024 — Question 4 →