Question

Let $P\left(x\right)=x^3-p{x}^2+qx-r$ be a polynomial. Then find the condition that must be satisfied by its coefficients when :

(i) its zeros are a-b, a and a+b.

(ii) the sum of its two zeros is zero.

Answer 1

$P\left(x\right)=x^3-p{x}^2+qx-r$ .....(1)

(i) Let $\alpha ,\beta ,\gamma$ are the zeros of $P\left(x\right)=x^3-p{x}^2+qx-r$,

where $\alpha =a-b,\beta =a,\gamma =a+b$.

Then,

$\alpha +\beta +\gamma =a-b+a+a+b=p$

$3a=p$

$a=\frac{p}{3}$

We know that $a=\beta$ is a zero of P(x). Thus, at $\beta =\frac{p}{3},P\left(x\right)=0$

Putting $x=\frac{p}{3}$ in equation 1, we get,

$(\frac{p}{3}{)}^3-p(\frac{p}{3}{)}^2+q\left(\frac{p}{3}\right)-r=0$

$\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r=0$

$p^3-3p^3+9pq-27r=0$

$-2p^3+9pq-27r=0$

$2p^3-9pq+27r=0$

Thus, this is the required equation.

(ii) Here, $\alpha ,\beta ,\gamma$ are the zeros of P(x) such that $\alpha +\beta =0$

We have, $\alpha +\beta +\gamma =p$

$\gamma =p$

Now, $\gamma =p$ is a zero of P(x). Thus,

At $\gamma =p,P\left(x\right)=0orP\left(\gamma \right)=0$

Putting $x=p$ in equation 1, we get,

$p^3-p^3+pq-r=0$

$pq-r=0$

$pq=r$

Thus, this is the required condition.