Question

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

Answer 1

Hint:In this problem, we will use the relationship between coefficients and zeroes of a given polynomial. If $\alpha ,\beta $ and $\gamma $ are zeroes (roots) of the polynomial $p{x^3} + q{x^2} + rx + s$ then the relationship between coefficients and zeros is given by $\alpha + \beta + \gamma = - \dfrac{q}{p},\;\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p}$ and $\alpha \beta \gamma = - \dfrac{s}{p}$ where $p,q,r,s$ are coefficients of the polynomial.

Complete step-by-step solution
In this problem, the given polynomial is ${x^3} - 3{x^2} + x + 1$. Let us compare this polynomial with $p{x^3} + q{x^2} + rx + s$. Therefore, we get $p = 1,q = - 3,r = 1$ and $s = 1$. Now we will use the relationship between coefficients and zeros of third degree polynomial.
If $\alpha ,\beta $ and $\gamma $ are zeroes (roots) of the polynomial $p{x^3} + q{x^2} + rx + s$ then the relationship between coefficients and zeros is given by
$\alpha + \beta + \gamma = - \dfrac{q}{p} \cdots \cdots \left( 1 \right)$
$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p} \cdots \cdots \left( 2 \right)$
$\alpha \beta \gamma = - \dfrac{s}{p} \cdots \cdots \left( 3 \right)$
In the problem, given that $a - b,a,a + b$ are zeros of the polynomial ${x^3} - 3{x^2} + x + 1$. Let us say $\alpha = a - b,\beta = a$ and $\gamma = a + b$. Therefore, from $\left( 1 \right)$ we can write
$
  a - b + a + a + b = - \left( {\dfrac{{ - 3}}{1}} \right)\quad \left[ {\because q = - 3,p = 1} \right] \\
   \Rightarrow 3a = 3 \\
   \Rightarrow a = \dfrac{3}{3} \\
   \Rightarrow a = 1 \\
 $
Therefore, we have $a = 1$.
Now from $\left( 3 \right)$, we can write
$
  \left( {a - b} \right)\left( a \right)\left( {a + b} \right) = - \dfrac{1}{1}\quad \left[ {\because s = 1,p = 1} \right] \\
   \Rightarrow a\left( {{a^2} - {b^2}} \right) = - 1 \cdots \cdots \left( 4 \right) \\
 $
Let us substitute $a = 1$ in equation $\left( 4 \right)$. Therefore, we get
$
  1\left( {{1^2} - {b^2}} \right) = - 1 \\
   \Rightarrow 1 - {b^2} = - 1 \\
   \Rightarrow {b^2} = 1 + 1 \\
   \Rightarrow {b^2} = 2 \\
   \Rightarrow b = \pm \sqrt 2 \\
 $
Therefore, we have $b = \pm \sqrt 2 $. Therefore, if $a - b,a,a + b$ are zeros of the polynomial ${x^3} - 3{x^2} + x + 1$ then $a = 1$ and $b = \pm \sqrt 2 $.

Note:The general form of third degree polynomial is $a{x^3} + b{x^2} + cx + d$ where $a \ne 0$. The polynomial of degree three is called cubic polynomial. For any value of $x$, if the polynomial $p\left( x \right)$ will be $0$ then we can say that that value of $x$ is zero (root) of the polynomial $p\left( x \right)$. Zeroes are also called roots of the polynomial. Every cubic polynomial has exactly three zeros. Every cubic polynomial has at least one real zero. If a cubic polynomial has exactly one real zero then the other two zeroes (roots) will be imaginary. Imaginary roots are always in pairs of the form $a \pm bi$ where $i$ is an imaginary number.