How do you find all zeros of #g(x)=x^3+3x^2-4x-12#?

1 Answer
Jan 23, 2017

Answer:

The zeros of #g(x)# are #2#, #-2# and #-3#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this with #a=x# and #b=2#.

Given:

#g(x) = x^3+3x^2-4x-12#

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

#x^3+3x^2-4x-12 = (x^3+3x^2)-(4x+12)#

#color(white)(x^3+3x^2-4x-12) = x^2(x+3)-4(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x^2-4)(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x^2-2^2)(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x-2)(x+2)(x+3)#

Hence zeros: #x=2#, #x=-2#, #x=-3#.