Question

If the zeros of the polynomial ${x^3} - 3{x^2} + x + 1$ are $a - b,a,a + b$. Then find the values of $a$ and $b$
A. $a = 1$ and $b = \pm \sqrt 2$
B. $a = \pm 1$ and $a = \pm \sqrt 2$
C. $a = 2$ and $b = \pm 1$
D. $a = - 1$ and $b = \pm \sqrt 2$

Answer 1

In this question, we will proceed by equating the values of sum of the roots or zeros and then product of the roots or zeroes to get the required values of $a$and $b$. So, use this concept to reach the solution of the given problem

Complete step-by-step answer :
The polynomial is ${x^3} - 3{x^2} + x + 1$ and their zeros or roots are $a - b,a,a + b$.
We know that for a cubic polynomial $a{x^3} + b{x^2} + cx + d$, we have
Sum of the roots $= \dfrac{{ - b}}{a}$
Sum of the product of two roots at a time $= \dfrac{c}{a}$
Product of the roots $= \dfrac{{ - d}}{a}$
So, for the given polynomial ${x^3} - 3{x^2} + x + 1$ we have
Sum of the roots is given by

$$\Rightarrow a - b + a + a + b = \dfrac{{ - \left( { - 3} \right)}}{1} \\\ \Rightarrow 3a = 3 \\\ \therefore a = 1 \\\$$

Product of the roots is given by

$$\Rightarrow \left( {a - b} \right)\left( a \right)\left( {a + b} \right) = \dfrac{{ - \left( { - 1} \right)}}{1} \\\ \Rightarrow \left( {1 - b} \right)\left( 1 \right)\left( {1 + b} \right) = - 1{\text{ }}\left[ {\because a = 1} \right] \\\ \Rightarrow \left( {1 - {b^2}} \right) = - 1{\text{ }}\left[ {\because \left( {x - y} \right)\left( {x + y} \right) = {x^2} - {y^2}} \right] \\\ \Rightarrow 1 + 1 = {b^2} \\\ \Rightarrow {b^2} = 2 \\\ \therefore b = \pm \sqrt 2 \\\$$

Therefore, we have $a = 1$ and $b = \pm \sqrt 2$
Thus, the correct answer is A. $a = 1$ and $b = \pm \sqrt 2$

Note : For a cubic polynomial $a{x^3} + b{x^2} + cx + d$, we have
Sum of the roots $= \dfrac{{ - b}}{a}$
Sum of the product of two roots at a time $= \dfrac{c}{a}$
Product of the roots $= \dfrac{{ - d}}{a}$
For a cubic polynomial the number of zeros or roots are equal to 3.