Question

If the polynomial $f\left(x\right)=2x^3+m{x}^2+nx-14$ has $(x-1)$ and $(x+2)$ as its factors, find the value of $\frac{m}{n}$

• A. $27$
• B. $\frac{1}{3}$
• C. $3$
• D. $\frac{1}{27}$

Answer 1

Answer: (C)

$3$

Given that,

$f\left(x\right)=2x^2+m{x}^2+nx-14$

$\left(x-1\right)and\left(x+2\right)$ Factor at $f\left(x\right)$

$x-1=0$

$\Rightarrow x=1$

$x+2=0$

$\Rightarrow x=-2$

$f\left(x\right)=2\times 1^3+m\times 1^2+1n-14$

$f(-2)=0$

$2{\left(-2\right)}^3+m{\left(-2\right)}^2+n\left(-2\right)-14=0$

$\Rightarrow 0=2+m+n-14$

$\Rightarrow m+n=12----\left(1\right)$

$-16+4m-2n=0$

$4m-2n-30=0$

$2\left(2m-n\right)=30$

$2m-n=15-----\left(2\right)$

on solving $\left(i\right)$ & $\left(ii\right)$ we get

$m+n=12$

$2m-n=15$

$\overline{3m=27}$

$m=\frac{27}{3}$

$m=9$

Now put $m=9$ in $eqn\left(i\right)$ we have

$\Rightarrow m+n=12$

$\Rightarrow 9+n=12$

$\Rightarrow n=3$

So,

$\frac{m}{n}=\frac{9}{3}=3$