Question

What must be subtracted from ${\mathrm{x}}^3-6{\mathrm{x}}^2-15\mathrm{x}+80$ so that the result is exactly divisible by ${\mathrm{x}}^2+\mathrm{x}-12$.

Answer 1

Finding remainder when ${\mathrm{x}}^3-6{\mathrm{x}}^2-15\mathrm{x}+80$ is divided by ${\mathrm{x}}^2+\mathrm{x}-12$:

On performing long division method, we get,

$$\begin{gathered} {\mathrm{x}}^2+\mathrm{x}-12)\overline{{\mathrm{x}}^3-6{\mathrm{x}}^2-15\mathrm{x}+80}(\mathrm{x}-7 \\ {\mathrm{x}}^3+{\mathrm{x}}^2-12\mathrm{x} \\ \overline{-7{\mathrm{x}}^2-3\mathrm{x}+80} \\ -7{\mathrm{x}}^2-7\mathrm{x}+84 \\ \overline{4\mathrm{x}-4} \end{gathered}$$

Hence, $4\mathrm{x}-4$must be subtracted from $({\mathrm{x}}^3-6{\mathrm{x}}^2-15\mathrm{x}+80)$.