Question

Given that $2$ is a zero of the cubic polynomial $6x^3-2x^2-10x-20$, find its other two zeroes.

Answer 1

Given polynomial $=6x^3-2x^2-10x-20$

Also, $2$ is a zero of this polynomial.

That means, $(x-2)$ will be one of the factor or divisor for this polynomial.

We can, now divide this polynomial by $(x-2)$ to get the quotient and further find the roots of that quadratic polynomial.

By division method, the quotient is $6x^2+10x+10.$

That means, $6x^3-2x^2-10x-20$ can be factored as $6x^3-2x^2-10x-20=(x-2)(6x^2+10x+10).$

Now, $6x^2+10x+10$ can be factored further,

$6x^2+10x+10$

$\Rightarrow x=\frac{-10\pm \sqrt{{\left(10\right)}^2-4\left(6\right)\left(10\right)}}{2\left(6\right)}$

$\Rightarrow x=\frac{-10\pm \sqrt{100-240}}{12}=\frac{-10\pm \sqrt{-140}}{12}=\frac{-10\pm 2\sqrt{35}i}{12}$

$\Rightarrow x_1=\frac{-5}{6}+\frac{\sqrt{35}}{6}i$

$\Rightarrow x_2=\frac{-5}{6}-\frac{\sqrt{35}}{6}i$

so, the other zeroes are imaginary with values $x_1\&x_2.$

Hence, solve.