The cubic polynomial $\mathrm{f}(x)=6{{x}^{3}}+h{{x}^{2}}-x-k$ is exactly divisible by $(x+1)$,

Question

The cubic polynomial $\mathrm{f}(x)=6{{x}^{3}}+h{{x}^{2}}-x-k$ is exactly divisible by $(x+1)$, and leaves a remainder of $72$ when it is divided by $(x-2)$. Calculate the values of $h$ and of $k$. Hence, factorise $\mathrm{f}(x)$ completely. Hence, express $\frac{7}{\mathrm{f}(x)}$ in partial fractions.

Accepted Answer

$h=7,\ k=2$ $f(x)=(x+1)(2x-1)(3x+2)$ $\frac{7}{f(x)}=\frac{7}{3(x+1)}+\frac{4}{3(2x-1)}-\frac{9}{3x+2}$

Polynomials, Cubic Equations and Partial Fractions - Partial Fractions

Polynomials, Cubic Equations and Partial Fractions - Partial Fractions

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