Question

find the value of k so that polynomial x3-3x2-4x+k is divisible by (x+2) and hence factorise the polynomial.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+\mathrm{k} \\ \mathrm{As}\mathrm{it}\mathrm{is}\mathrm{divisible}\mathrm{by}\left(\mathrm{x}+2\right) \\ \mathrm{so},\mathrm{x}+2=0 \\ \mathrm{x}=-2 \\ \mathrm{f}\left(-2\right)=0 \\ {\left(-2\right)}^3-3{\left(-2\right)}^2-4\left(-2\right)+\mathrm{k}=0 \\ -8-12+8+\mathrm{k}=0 \\ \mathrm{k}=12 \\ \mathrm{So},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+12 \\ ={\mathrm{x}}^2\left(\mathrm{x}-3\right)-4\left(\mathrm{x}-3\right) \\ =\left({\mathrm{x}}^2-4\right)\left(\mathrm{x}-3\right) \\ =\left(\mathrm{x}+2\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)\mathrm{ANS}...\left(\mathrm{Since}{\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right) \end{gathered}$$