How do you find all zeros of g(x)=x3+3x2−4x−12?

1 Answer
George C.
Jan 23, 2017

The zeros of g(x) are 2, −2 and −3

Explanation:

The difference of squares identity can be written:

a2−b2=(a−b)(a+b)

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

Given:

g(x)=x3+3x2−4x−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:

x3+3x2−4x−12=(x3+3x2)−(4x+12)

x3+3x2−4x−12=x2(x+3)−4(x+3)

x3+3x2−4x−12=(x2−4)(x+3)

x3+3x2−4x−12=(x2−22)(x+3)

x3+3x2−4x−12=(x−2)(x+2)(x+3)

Hence zeros: x=2, x=−2, x=−3.

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