Question

Find the zeros of the following

Answer 1

Question 1

$x^3 + x^2 - 4x-4=0\\\\x^2 (x+1)-4(x+1)=0\\\\(x^2 -4)(x+1)=0\\\\(x-2)(x+2)(x+1)=0\\\\x=-2, -1, 2$

Question 2

By inspection, we know that $x=2$ is a root.

$\frac{2x^3 - 11x^2 + 17x-6}{x-2}=2x^2 - 7x+3$.

So, we can factor the equation as

$(x-2)(2x^2 - 7x+3)=0\\\\(x-2)(x-3)(2x-1)=0\\\\x=\frac{1}{2}, 2, 3$