Question

find all the zeros of the polynomial- x^3+3x^2-4x-12. If one of its zeros is -3

Answer 1

One of the zeroes is -3.
So, x = -3

x + 3 = 0

Now on dividing the equation by (x + 3 =0) =>

${x}^{3} + 3 {x}^{2} - 4x - 12 \div x + 3$
We get the quotient
${x}^{ 2} - 4$
We can write it as
$(x + 2)(x - 2)$
So, the remaining two zeroes are=>

$x + 2 = 0 \\ x = - 2$
And,
$x - 2 = 0 \\ x = 2$

So the zeroes of the equation are=> -3, 2 and -2.