Question

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

Answer 1

Step 1. Basic structure of the cubic polynomial.

As we know that, the cubic polynomial is in the form ${\mathbf{ax}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{bx}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{cx}\mathbf{}\mathbf{+}\mathbf{}\mathbf{d}$ and the zeroes are $\mathrm{\alpha }$, $\mathrm{\beta }$, and $\text{γ}$.

We will find the value of $\mathrm{a}$ by using the formula of sum of roots and value of $\mathrm{b}$ by using the formula of product of roots.

The sum of roots are,

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

Where $\mathrm{b}$ is the coefficient of ${\mathrm{x}}^2$ and $\mathrm{a}$ is the coefficient of ${\mathrm{x}}^3$.

The products of roots are,

$\mathbf{\alpha }\mathbf{·}\mathbf{\beta }\mathbf{=}\frac{\mathbf{-}\mathbf{d}}{\mathbf{a}}$

Where $\mathrm{d}$ is the constant term and $\mathrm{a}$ is the coefficient of ${\mathrm{x}}^3$.

It is given that, the polynomial equation is ${\mathrm{x}}^3-3{\mathrm{x}}^2+\mathrm{x}+1$.

Step 2. Compare the given equation with standard form of cubic polynomial.

Comparing the given equation with standard form of cubic polynomial.

Here,

$\mathrm{a}=1$

$\mathrm{b}=-3$

$\mathrm{c}=1$

$\mathrm{d}=1$

The zeroes are, $\left(\mathrm{a}-\mathrm{b}\right),\left(\mathrm{a}\right)$ and $\left(\mathrm{a}+\mathrm{b}\right)$.

Step 3. Find the value of $\mathrm{a}$.

We will find the value of $\mathrm{a}$ by using the formula $\frac{-\mathrm{b}}{\mathrm{a}}$.

$\Rightarrow$ $\left(\mathrm{a}-\mathrm{b}\right)+\left(\mathrm{a}\right)+\left(\mathrm{a}+\mathrm{b}\right)=3$

Group like terms

$\Rightarrow$ $\mathrm{a}+\mathrm{a}+\mathrm{a}-\mathrm{b}+\mathrm{b}=3$

Add similar elements.

$\Rightarrow$ $3\mathrm{a}=3$

Divide both sides by $3$.

$\Rightarrow$ $\frac{3\mathrm{a}}{3}=\frac{3}{3}$

$\Rightarrow$ $a=1$

Step 4. Find the value of $\mathrm{b}$.

.We will find the value of $\mathrm{b}$ by using the formula $\frac{-\mathrm{d}}{\mathrm{a}}$.

$\Rightarrow$ $\left(\mathrm{a}-\mathrm{b}\right)\times \mathrm{a}+\left(\mathrm{a}\right)\times \left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{a}+\mathrm{b}\right)\times \left(\mathrm{a}-\mathrm{b}\right)=1$

$\Rightarrow$ ${\mathrm{a}}^2-\mathrm{ba}+{\mathrm{a}}^2+\mathrm{ab}+{\mathrm{a}}^2-{\mathrm{b}}^2=1$

$\Rightarrow$ ${\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2=1$

$\Rightarrow$ $3{\mathrm{a}}^2-{\mathrm{b}}^2=1$

Put the calculated value of $\mathrm{a}$ .

$\Rightarrow$ $3{\left(1\right)}^2-{\mathrm{b}}^2=1$

$\Rightarrow$ $3-{\mathrm{b}}^2=1$

$\Rightarrow$ ${\mathrm{b}}^2=2$

$\Rightarrow$ $\mathrm{b}=\sqrt{2}$

Hence, the values of $\mathrm{a}$ is $1$ and $\mathrm{b}$ is $\sqrt{2}$.