Question

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

Answer 1

As the the coefficient of highest power $x^3$ is1, if three roots are $\alpha ,\beta and\gamma$, we have

$f\left(x\right)=x^3-3x^2+x+1$

$=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\gamma \beta +\gamma \alpha )x+\alpha \beta \gamma$

Now let us compare coefficients of similar powers on each side.

First comparing the sum of roots from the coefficient of $x^2$, we have

$\alpha +\beta +\gamma =\frac{-b}{a}$

$a-b+a+a+b=-\left(\frac{-3}{1}\right)$

i.e. $3a=3$

i.e.$a=1.$

Also, coefficients of x give us the sum of the product of zeros

$\alpha \beta +\gamma \beta +\gamma \alpha =\frac{c}{a}=1$

$a(a-b)+a(a+b)+(a+b)(a-b)=1$

i.e. $a(a-b+a+b)+a^2-b^2=1$

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

and as
$a=1,$
$b^2=3\times (1{)}^2-1=2$ and

$b=\pm \sqrt{2}$

Further, products of roots is $(a-b)a(a+b)=1.$

Hence, $a=1$ and $b=\pm \sqrt{2}$

Note that the two values of b give the same set of roots.