Question

If the zeros of the polynomial $f\left(x\right)=x^3-3x^2+x+1$ are $a-b,a,a+b$.Find $a$ and $b$.

Answer 1

Given,

The roots of the polynomial $f\left(x\right)=x^3-3x^2+x+1$ are $a-b,a,a+b$

We know that relation between roots$(x_1,x_2,x_3)$ of a cubic equation $p{x}^3+q{x}^2+rx+s=0$ and its coefficents$(p,q,r,s)$ is given by:

$x_1+x_2+x_3=-\frac{q}{p}....\left(1\right)$

$x_1\cdot x_2\cdot x_3=-\frac{s}{p}....\left(2\right)$

So we have

$x_1=a-b;x_2=a;x_3=a+b;$

and

$p=1;q=-3;r=1;s=1;$

Hence, from equation (1) we have:

$a-b+a+a+b=-\frac{-3}{1}$

$3a=3$

Hence, $a=1\dots ..\left(3\right)$

and from equation (2) we have:

$(a-b)(a+b)\left(a\right)=-\frac{1}{1}$

$(a^2-b^2)a=-1$

put value of $a=1$ from equation (3) we get

$(1-b^2)=-1$

$b^2=1+1=2$

Hence, $b=\sqrt{2}\dots \left(4\right)$