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

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Answer 1 · 2016-05-31T20:58:22

Factor by grouping to find zeros: #2, -2, -3#

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

So this cubic factors by grouping:

#x^3+3x^2-4x-12#

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

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

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

#=(x-2)(x+2)(x+3)#

Hence zeros: #2, -2, -3#

Answer 2 · 2018-07-15T16:35:53

#(x+2)(x-2)(x+3)#

We essentially have two parts to our function:

#color(steelblue)(x^3+3x^2)color(purple)(-4x-12)#

Let's factor a #x^2# out of the blue term, and a #-4# out of the purple term. We now have

#color(steelblue)(x^2(x+3))color(purple)(-4(x+3))#

Since both terms have an #x+3# in common, we can factor that out. We now have

#color(red)((x^2-4))(x+3)#

You might immediately recognize that what I have in red is a difference of squares, which can be factored as #(x+2)(x-2)#.

We now have

#(x+2)(x-2)(x+3)#

Hope this helps!