Question:
If the zeros of the polynomial $f(x)=x^{3}-3 x^{2}+x+1$ are $(a-b), a$ and $(a+b)$, Find $a$ and $b$.
Solution:
By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes $=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficent of } x^{3}}$
$\therefore a-b+a+a+b=\frac{-(-3)}{1}$
$\Rightarrow 3 a=3$
$\Rightarrow a=1$
Now, Product of zeros $=\frac{-(\text { constant } t e r m)}{\text { coefficent of } x^{3}}$
$\therefore(a-b)(a)(a+b)=\frac{-1}{1}$
$\Rightarrow(1-b)(1)(1+b)=-1 \quad[\because a=1]$
$\Rightarrow 1-b^{2}=-1$
$\Rightarrow b^{2}=2$
$\Rightarrow b=\pm \sqrt{2}$