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

$h=7$, $k=2$; $f(x)=(x+1)(3x+2)(2x-1)$; $\frac7{f(x)}=\frac7{3(x+1)}-\frac9{3x+2}+\frac4{3(2x-1)}$.

Polynomials, Cubic Equations and Partial Fractions - Cubic Expressions and Equations

Polynomials, Cubic Equations and Partial Fractions - Cubic Expressions and Equations

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