Question

​For polynomials p[x] = 6x³+11x²+kx-20 and q[x]=9x²+5x , x = 2 is a root of the polynomial
{p[x] + q[x]}.
​Then find the value of k ?

Answer 1

Dear Student,

$$\begin{gathered} p\left(x\right)=6x^3+11x^2+kx-20 \\ q\left(x\right)=9x^2+5x \\ \left\{p\left(x\right)+q\left(x\right)\right\}=6x^3+11x^2+kx-20+9x^2+5x \\ \left\{p\left(x\right)+q\left(x\right)\right\}=6x^3+20x^2+\left(k+5\right)x-20 \\ Givenx=2isarootofp\left(x\right)+q\left(x\right)i.e.,p\left(2\right)+q\left(2\right)=0 \\ \Rightarrow 6\times {\left(2\right)}^3+20\times {\left(2\right)}^2+\left(k+5\right)2-20=0 \\ \Rightarrow 48+80+2k+10-20=0 \\ \Rightarrow 118+2k=0 \\ \Rightarrow 2k=-118 \\ \Rightarrow k=\frac{-118}{2} \\ \Rightarrow \mathit{k}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{59} \end{gathered}$$

Regards