How do you solve #v^3 - 2v^2 - 16v = -32#?

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Answer 1 · 2015-12-22T07:11:01

Rearrange and factor by grouping to find factors, hence solutions:

#v=4#, #v=-4#, #v=2#

We will use the difference of squares identity, which can be written:

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

First add #32# to both sides to get:

#v^3-2v^2-16v+32 = 0#

Factor the left hand side by grouping followed by the difference of squares identity:

#v^3-2v^2-16v+32#

#=(v^3-2v^2)-(16v-32)#

#=v^2(v-2)-16(v-2)#

#=(v^2-16)(v-2)#

#=(v^2-4^2)(v-2)#

#=(v-4)(v+4)(v-2)#

So our original equation becomes:

#(v-4)(v+4)(v-2) = 0#

which has solutions: #v=4#, #v=-4#, #v=2#