Question

For what value of m, is x cube -2mx square+19 divisible by x-2

Answer 1

Given,

$\longrightarrow\sf{p(x)=x^3-2mx^2+19}$

Factor theorem states that if a polynomial $\sf{p(x)}$ is divisible by $\sf{x-a}$ then $\sf{p(a)=0.}$

Hence, for the polynomial $\sf{p(x)}$ divisible by $\sf{x-2,}$

$\longrightarrow\sf{p(2)=0}$

$\longrightarrow\sf{2^3-2m\times2^2+19=0}$

$\longrightarrow\sf{8-2m\times4+19=0}$

$\longrightarrow\sf{8-8m+19=0}$

$\longrightarrow\sf{27-8m=0}$

$\longrightarrow\sf{8m=27}$

$\longrightarrow\sf{\underline{\underline{m=\dfrac{27}{8}}}}$

Therefore, $\sf{p(x)=x^3-2mx^2+19}$ is divisible by $\sf{x-2}$ for $\sf{m=\dfrac{27}{8}.}$