Question

Given that the zeroes of the cubic polynomial $f\left(x\right)=x^3-6x^2+3x+10$ are of the form $a,a+b,a+2b$ for some real numbers $a$ and $b$, find the values of $a$ and $b$ as well as the zeros of the given polynomial.

Answer 1

Given

Cubic polynomial is $f\left(x\right)=x^3-6x^2+3x+10$

And the roots are $a,a+b$ and $a+2b$

Using the relationship between coefficient and Zeroes of the polynomial we get,

$\text{Sum of roots}=-\frac{\text{coefficient of}x²}{\text{coefficient of}x³}$

$\Rightarrow \left(a\right)+(a+b)+(a+2b)=-\frac{(-6)}{1}$

$\Rightarrow 3(a+b)=6$

$\Rightarrow a+b=2...\left(i\right)$

And $\text{Product of roots}=-\frac{\text{constant}}{\text{coefficient of}x³}$

$\Rightarrow a(a+b)(a+2b)=-\frac{10}{1}$

$\Rightarrow a(a+b)(a+b+b)=-10$

$\Rightarrow a\left(2\right)(2+b)=-10[$Putting the value of $a+b$ from $\left(i\right)]$

$\Rightarrow 2a(2+2-a)=-10[$Putting the value of $b$ from $\left(i\right)]$

$\Rightarrow a(4-a)=-5$

$\Rightarrow 4a-a^2+5=0$

$\Rightarrow a^2+4a-5=0$

Now solving obtained quadratic equation we get,

$a=5,-1$

when $a=5,b=-3$ ; Zeroes are $5,(5+(-3\left)\right)=2,(5+2(-3\left)\right)=-1$

when $a=-1,b=3$ : Zeroes are $-1,(-1+3)=2,(-1+2(3\left)\right)=5$

Hence, $a=5\&b=-3$ or $a=-1\&b=3$ and Zeroes are $-1,2,5.$