How do you factor #24x ^ { 3} + 60x ^ { 2} - 168x - 420#?

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Answer 1 · 2017-06-22T01:26:09

#(12x^2-84)(2x+5)#

Let's start with what we got:

#24x^3+60x^2-168x-420#

What we can do is break it up into:

#24x^3+60x^2# #&# #-168x-420#

We will factor this by grouping, find the GCF for each of them:

The GCF for #24x^3+60x^2# is #color(blue)(12x^2)#

The GCF for #-168x-420# is #color(blue)(-42)#

Now let's factor them:

#24x^3+60x^2->color(blue)(12x^2)(2x+5)#
#-168x-420->color(blue)(-84)(2x+5)#

We can see that #color(blue)(12x^2)color(blue)(-84)# is one factor the other one is #2x+5#

Our final answer is #(12x^2-84)(2x+5)#

Answer 2 · 2017-06-22T05:07:41

#2(2x + 5)(2x^2 - 7)#

#f(x) = 12(2x^3 + 5x^2 - 14x - 35) = 12y#
#Factor y by grouping:#
#y = 2x^2(x + 5) - 7(2x + 5) = (2x + 5)(2x^2 - 7)#
f#(x) = 12(2x + 5)(2x^2 - 7)#