How do you find the quotient of #(2h^3+8h^2-3h-12)div(h+4)#?

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Answer 1 · 2017-08-20T08:55:45

#2h^2-3#

I'm going to rewrite this as:

#(2h^3+8h^2-3h-12)/(h+4)#

We can factor the numerator this way:

#((2h^3+8h^2)+(-3h-12))/(h+4)#

#(2h^2(h+4)+(-3)(h+4))/(h+4)#

#((2h^2-3)(h+4))/(h+4)#

We can now cancel the #(h+4)# on top and bottom:

#((2h^2-3)cancel((h+4)))/cancel((h+4))#

And be left with:

#2h^2-3#