Question

Given that the zeroes of the cubic polynomial $x^3-6x^2+3x+10$ 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 zeroes of the given polynomial.

Answer 1

Step 1: Compare given polynomial $f\left(x\right)$ with standard polynomial of degree $3$ to obtain value of coefficients:

Given that

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

$\left(ii\right)a,a+b,a+2b$are zeroes/roots of polynomial $f\left(x\right)$

$\left(iii\right)a,b$ are real numbers

Standard polynomial in degree $3$ is given by $p{x}^3+q{x}^2+rx+s$

Comparing $f\left(x\right)$ with standard polynomial we get

$p=1,q=-6,r=3,s=10$

Step 2 : Use the formula for the sum of roots:

Sum of roots $=\frac{-q}{p}$

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

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

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

Step 3: Use the formula for the product of roots:

$\Rightarrow$ Product of roots $=\frac{-s}{p}$

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

$\Rightarrow$ $(2-b)\times 2\times \left(2-b+2b\right)=-10$ $...\left(from\right(i\left)\right)$

$\Rightarrow$ $4-b^2=-5$

$\Rightarrow$ $b^2-9=0$

$\Rightarrow$ $(b+3)(b-3)=0$

$\Rightarrow$ $b=3orb=-3$

Step 3 : Solve for $a$:

When $b=3,a=2-3\Rightarrow a=-1$

When $b=-3,a=2-(-3)\Rightarrow a=5$

$\therefore$ If $a=5,$ then $b=-3$

Or

If $a=-1,$ then $b=3$

Step 4 : Find the zeroes of the polynomial:

If $a=5,b=-3$ then zeroes of the polynomial are $a,a+b,a+2b\Rightarrow 5,5-3,5+2(-3)\Rightarrow 5,2,-1$

If $a=-1,b=3$ then zeroes of the polynomial are $a,a+b,a+2b\Rightarrow -1,-1+3,-1+2\left(3\right)\Rightarrow -1,2,5$

Hence the values of $a,b$ are $(5,-3)$ or $(-1,3)$ and their corresponding zeroes are $(5,2,-1)$ or $(-1,2,5)$