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

Hint: 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.