Question

# If the zeroes of the polynomial ${{x}^{3}}-3{{x}^{2}}+x+1$ are $a-b,a,a+b$, find a and b.

Hint: In this question, we first need to write the sum, sum of product of two roots at a time, product of the roots in terms of the coefficients of the given cubic polynomial. Then solve the equations formed in terms of a and b to get their respective values.

CUBIC POLYNOMIAL: A cubic polynomial is a polynomial having degree three.
The general form of a cubic polynomial is given by ${{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0$
ZERO OF A POLYNOMIAL: A real number $\alpha$ is a zero of the polynomial p(x), if and only if $p\left( \alpha \right)=0$
Now, let us assume the zeroes of the cubic polynomial as
$\alpha ,\beta ,\gamma$
Now, from the cubic polynomial ${{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0$ we have,
Sum of the roots of the cubic polynomial is given by
$\alpha +\beta +\gamma =\dfrac{-{{a}_{1}}}{{{a}_{0}}}$
Now, sum of the product of the two roots at a time is given by
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{{{a}_{2}}}{{{a}_{0}}}$
Now, product of all the roots is given by
$\alpha \beta \gamma =\dfrac{-{{a}_{3}}}{{{a}_{0}}}$
Now, on comparing the zeroes given in the question to the formulae above we get,
$\alpha =a-b,\beta =a,\gamma =a+b$
Now, again on comparing the given cubic polynomial with the general form we get,
$\Rightarrow {{x}^{3}}-3{{x}^{2}}+x+1=0$
${{a}_{0}}=1,{{a}_{1}}=-3,{{a}_{2}}=1,{{a}_{3}}=1$
Now, from the sum of the roots formula on substituting the respective values we get,
$\Rightarrow a-b+a+a+b=\dfrac{-\left( -3 \right)}{1}$
Now, this can be further written as
$\Rightarrow a+a+a=3$
Now, on further simplification we get,
$\Rightarrow 3a=3$
Now, on dividing with 3 on both sides we get,
$\therefore a=1$
Let us now consider the product of the roots formula and substitute the respective values.
$\Rightarrow \left( a-b \right)\times a\times \left( a+b \right)=\dfrac{-1}{1}$
Now, on substituting the value of a this can be further written as
$\Rightarrow \left( 1-b \right)\times 1\times \left( 1+b \right)=-1$
Now, on further simplification we get,
$\Rightarrow 1-{{b}^{2}}=-1$
Now, on rearranging the terms we get,
$\Rightarrow {{b}^{2}}=2$
$\therefore b=\sqrt{2}$

Note:
Instead of using the formulae of sum of the roots and product of the roots we can also use the sum of the product of two roots taken at a time and sum of the roots. Both the methods give the same result.
Here, we considered the value of b only positive because as already given that the roots are $a-b,a,a+b$ we get the same values because the value of b is irrational. That means the roots will be conjugate so either we consider the negative value or positive value the roots will be the same for the given cubic polynomial.