Question

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

Answer 1

Interpret the given data.
We are given with the polynomial here,
$p\left(x\right)=x^3-3x^2+x+1$
And zeroes are given as $a-b,a,a+b$

Compare the given polynomial with general expression
Now, comparing the given polynomial with general expression we get :
$\therefore p{x}^3+q{x}^2+rx+s=x^3-3x^2+x+1$
$p=1,q=-3,r=1$ and $s=1$

Write down the relationship between the zeroes and the coefficients.
Sum of zeroes $=a-b+a+a+b=3a$
We know that,
$\text{Sum of zeroes =}\frac{\text{-coefficient of}{x}^2}{\text{coefficient of}{x}^3}=-\frac{q}{p}$
$-\frac{q}{p}=3a$ ------(1)
Product of zeroes $=(a-b)\times \left(a\right)\times (a+b)=(a^2-b^2)\times \left(a\right)$
$=a^3-a{b}^2$
We know that,
$\text{Product of zeroes=}\frac{\text{-constant term}}{\text{coefficient of}{x}^3}=-\frac{s}{p}$
$-\frac{s}{p}=a^3-a{b}^2$ -------(2)

Simplify the equation to find the value of a and b We know that,
$p=1,q=-3,r=1$ and $s=1$
Put the values of $p,q,r$ and $s$ in equations (1) and (2).
From equation (1), we get
$-\frac{q}{p}=3a$
$\therefore -\frac{-3}{1}=3a$
$a=1$
From equation (2) we get
$-\frac{s}{p}=1-b^2$
$-\frac{1}{1}=1-b^2$
$b^2=1+1=2$
$b=\pm \sqrt{2}$
hence, $1-\sqrt{2},1,1+\sqrt{2}$ are the zeroes of $x^3-3x^2+x+1.$