Question

Let $P(x)$ be the polynomial $x^3+a{x}^2+bx+c$, where $a,b,c\in R$. If $P(-3)=P(+2)=0$ and $P^{\prime }(-3)<0$, which of the following is a possible value of ' $c^{\prime }$ ?

• A. -27
• B. -18
• C. -6
• D. -3

Answer 1

Answer: (A)

$P(x)=x^3+a{x}^2+bx+c$
$P^{\prime }(x)=3x^2+2ax+b$
$P(-3)=0\Rightarrow -27+9a-3b+c=0$ ....(1)
$P(2)=0\Rightarrow 8+4a+2b+c=0$ ....(2)
hence $(1)-(2)$ gives
$5a-5b-35=0$
$a-b=7$ ....(3)
$P^{\prime }(-3)=27-6a+b<0\Rightarrow 27-6(a-b)-5b<0\Rightarrow 27-42-5b<0$
$\Rightarrow -15-5b<0\Rightarrow 3+b>0\Rightarrow b>-3$, hence $a-7=b>-3$
$a-7>-3[$ from $(3)]$
$a>4$....(4)
$\therefore$ from (2)
$8+16-6+c_{\max }=0$
$c_{\max }=-18\stackrel{\max }{\Rightarrow }c<-18$