Question

Let $f(x)=x^3+x+1$, suppose $p(x)$ be another cubic polynomial such that $p(0)=-1$ & zero’s of $p(x)$ are the square of the roots of $f(x)=0$, then value of $p(9)$ equals

• A. $1331$
• B. $1024$
• C. $961$
• D. $899$

Answer 1

Answer: (D)

$f(x)=x^3+x+1=(x-\alpha )(x-\beta )(x-\gamma )\dots (A)$
where $\alpha ,\beta ,\gamma$ are roots of $f(x)=0$
$\therefore \alpha \beta \gamma =-1$
Now, roots of $p(x)=0$, are square of the roots of
$f(x)=0$ i.e., roots of $p(x)=0$ are ${\alpha }^2,{\beta }^2,{\gamma }^2$
$\therefore p(x)=\lambda (x-{\alpha }^2)(x-{\beta }^2)(x-{\gamma }^2)\dots (\ast )$
$\therefore p(0)=\lambda (-{\alpha }^2{\beta }^2{\gamma }^2)$
$\Rightarrow 1=\lambda$ (as $\alpha \beta \gamma =-1$ & $p(0)=-1)$
Now $p(x)=(x-{\alpha }^2)(x-{\beta }^2)(x-{\gamma }^2)$ (using (*))
$\therefore p(x^2)=(x^2-{\alpha }^2)(x^2-{\beta }^2)(x^2-{\gamma }^2)$
$=(x-\alpha )(x-\beta )(x-\gamma )(x+\alpha )(x+\beta )(x+\gamma )$
$=-f(x)f(-x)$ (using (A))
$p(x^2)=(x^3+x+1)(x^3+x-1)$
$\therefore p(3^2)=(27+3+1)(27+3-1)$ (putting x = 3)
$=(30+1)(30-1)$
$\therefore p(9)=899$