Question

If f(-2)=0 what are all the factors of the function f(x)=x^3-2x^2-68x-120

Answer 1

Since $f(-2)=0$, it follows by the polynomial remainder theorem that $f(x)$ is perfectly divisible by $x+2$. Dividing, we have

$\dfrac{x^3-2x^2-68x-120}{x+2}=x^2-4x-60$

and factoring further, we have

$f(x)=(x+2)(x^2-4x-60)=(x+2)(x+6)(x-10)$

Answer 2

Answer:

All the factors of the function f(x) are:

         $(x+2),\ (x+6)\ ,(x-10)$

i.e.

         $f(x)=x^3-2x^2-68x-120=(x+2)\cdot (x+6)\cdot (x-10)$

Step-by-step explanation:

We are given a cubic function f(x) in terms of variable x as follows;

         $f(x)=x^3-2x^2-68x-120$

Also, we are given that:

$f(-2)=0$

This means that -2 is a zero of the function f(x)

Hence, f(x) could be written in the form of:

$f(x)=(x+2)q(x)$

where q(x) is a two degree polynomial.

i.e. we get:

$q(x)=\dfrac{f(x)}{x+2}\\\\i.e.\\\\q(x)=\dfrac{x^3-2x^2-68x-120}{x+2}\\\\i.e.\\\\q(x)=x^2-4x-60$

We can factorize q(x) further by using the method of splitting the middle term as follows:

$q(x)=x^2-10x+6x-60\\\\i.e.\\\\q(x)=x(x-10)+6(x-10)\\\\i.e.\\\\q(x)=(x+6)(x-10)$

i.e. The factors of f(x) are:

$(x+2),\ (x+6)\ ,(x-10)$