Question

Let $P\left(x\right)$ be the polynomial $x^3+a{x}^2+bx+c,$ where $a,b,c\in R.$ If $P(–3)=P\left(2\right)=0$ and $P^{\prime }(–3)<0,$ then possible value of $c$ is

• A. $-27$
• B. $-18$
• C. $-6$
• D. $-3$

Answer 1

The correct option is A $-27$

$P\left(x\right)=x^3+a{x}^2+bx+c\cdots \left(1\right)$
$P(-3)=0=-27+9a-3b+c$
$\Rightarrow 9a-3b+c=27\cdots \left(2\right)$
and $P\left(2\right)=0=8+4a+2b+c$
$\Rightarrow 4a+2b+c=-8\cdots \left(3\right)$
From $\left(2\right)$ and $\left(3\right),$ we get
$a-b=7\cdots \left(4\right)$

Differentiating $\left(1\right)$ w.r.t. $x,$ we get
$P^{\prime }\left(x\right)=3x^2+2ax+b$
$P^{\prime }(-3)=27-6a+b<0$
$\Rightarrow 27-6(7+b)+b<0$ [From $\left(4\right)$]
$\Rightarrow 27-42-5b<0$
$\Rightarrow b>-3$
and $a>4$ [From $\left(4\right)$]
From $\left(3\right),$ we have
$4a+2(a-7)+c=-8$
$\Rightarrow 6a-14+c=-8$
$\Rightarrow \frac{6-c}{6}=a>4$
$\Rightarrow c<-18$