Question

If x+2 and x-1 are factors of ${{x}^{3}}+10{{x}^{2}}+mx+n$, then the value of m and n respectively are
[a] 5 and -3
[b] 17 and -8
[c] 7 and -18
[d] 23 and -19

Answer 1

Hint: Recall factor theorem. According to factor theorem x-a is a factor of p(x) if p(a) = 0. Use the factor theorem and hence form two linear equations in m and n. Hence find the values of m and n such that x-1 and x+2 are the factors of p(x).

Complete step-by-step answer:
We have $p\left( x \right)={{x}^{3}}+10{{x}^{2}}+mx+n$
Since x+2 is a factor of p(x), we have
$p\left( x \right)=\left( x+2 \right)g\left( x \right)$, where g(x) is some polynomial in x.
Put x = -2, we get
$p\left( -2 \right)=\left( -2+2 \right)g\left( -2 \right)=0\times g\left( -2 \right)=0$
Hence x = -2 is a zero of the polynomial p(x).
Now, we have $p\left( x \right)={{x}^{3}}+10{{x}^{2}}+mx+n$
Substituting x=-2 in the expression of p(x), we get
$p\left( -2 \right)={{\left( -2 \right)}^{3}}+10{{\left( -2 \right)}^{2}}+m\left( -2 \right)+n=-8+40-2m+n=-2m+n+32$
Hence, p(-2) = -2m+n+32
But p(-2) = 0
Hence, we have
\[-2m+n+32\text{ }=0~~~~~\left( i \right)\]
Similarly, since x-1 is a factor of p(x), we have
p(1) = 0
We have $p\left( x \right)={{x}^{3}}+10{{x}^{2}}+mx+n$
Substituting x= 1 in the expression of p(x), we get
$p\left( 1 \right)={{\left( 1 \right)}^{3}}+10{{\left( 1 \right)}^{2}}+m\left( 1 \right)+n=1+10+m+n=m+n+11$
But since p(1) = 0, we have
$m+n+11=0\text{ }\left( ii \right)$
Multiplying equation (ii) by 2 and adding equation (i) and (ii), we get
$\begin{align}
  & -2m+2m+n+2n+32+22=0 \\
 & \Rightarrow 3n+54=0 \\
\end{align}$
Subtracting 54 from both sides, we get
$3n=-54$
Dividing by 3 on both sides, we get
$n=\dfrac{-54}{3}=-18$
Substituting the value of n in equation (ii), we get
$\begin{align}
  & m-18+11=0 \\
 & \Rightarrow m-7=0 \\
\end{align}$
Adding 7 on both sides, we get
$m=7$
Hence, we have
m=7 and n = -18
Hence option [c] is correct.

Note: Alternative solution:
We will use the property that if $\alpha ,\beta ,\gamma $ are the roots of the polynomial $p\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$, then we have
$\alpha +\beta +\gamma =\dfrac{-b}{a},\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a},\alpha \beta \gamma =\dfrac{-d}{a}$
Let the third root of the polynomial p(x) be a.
Hence, we have
$a+1+\left( -2 \right)=-10\Rightarrow a=-9$
Hence the third root is -9
Now, we have
$m=\left( 1 \right)\left( -2 \right)+\left( -9 \right)\left( -2 \right)+\left( 1 \right)\left( -9 \right)=-2+18-9=7$
Also, we have
$n=-\left( 1 \right)\left( -2 \right)\left( -9 \right)=-18$