Question

Given that the zeroes of the cubic polynomial $x^3-6 x^2+3x+10$ are of the form $a, a+b, a+2 b$ for some real numbers $a$ and $b$, find the values of $a$ and $b$ as well as the zeroes of the given polynomial.

Answer 1

Let, $ \quad p(x)=x^3-6 x^2+3 x+10$

and $(a),(a+b)$ and $(a+2 b)$ are the zero of $p(x)$.

We know:

Sum of the zeroes $=-\left(\right.$ coefficient of $\left.x^2\right) \div$ coefficient of $x^3$

$\Rightarrow a+(a+b)+(a+2 b)=-(-6)$

$\Rightarrow 3 a+3 b=6$

$\Rightarrow a+b=2$

$\Rightarrow a=2-b$ ........(i)

Product of all the zeroes $=-($ constant term $)\div$ coefficient of $x^3$

$a(a+b)(a+2 b) =-10$

$(2-b)(2)(2+b) =-10$ [Using equation (i)]

$(2-b)(2+b) =-5$

$4-b^2 =-5$

$\Rightarrow \quad b^2 =9$

$\Rightarrow \quad b = \pm 3$

When $b=3, a=2-3=-1$ [Using equation (i)]

$\Rightarrow a=-1 \text { when } b=3$

$\text { When } b=-3, a=2-(-3)=5$

[Using equation (i)]

$\Rightarrow \quad a=5 \text { when } b=-3 .$

Case 1: when $a=-1$ and $b=3$

The zeroes of the polynomial are :

$a  =-1$

$a+b =-1+3=2$

$a+2 b =-1+2(3)=5$

$\Rightarrow-1,2$ and 5 are the zeroes.

Case 2: when $a=5$ and $b=-3$

The zeroes of the polynomial are :

$a =5$

$a+b=5-3=2 $

$a+2 b =5-2(3)=-1$

$\Rightarrow-1,2$ and $5$ are the zeroes.

In both the cases, the zeroes of the polynomial are $-1,2,5$.