How do you divide #(6x^2 + 7x - 2)# by #x+4#?

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Answer 1 · 2015-05-28T20:36:10

The process of synthetic division is somewhat like long division.

First choose a multiplier for #(x+4)# that will result in an expression whose highest order term matches the highest order term of #(6x^2+7x-2)#.

That multiplier is #6x#...

#6x*(x+4) = 6x^2+24x#

Subtract this from the original value to leave a remainder...

#(6x^2+7x-2) - (6x^2+24x)#

#=6x^2+7x-2-6x^2-24x#

#=-17x-2#

Now choose a multiplier for #(x+4)# that will result in an expression whose highest order term matches the highest order term of #(-17x-2)#

That multiplier is #-17#

#-17(x+4) = -17x-68#

Subtract this from the previous remainder to get a new remainder:

#(-17x-2)-(-17x-68)#

#=-17x-2+17x+68#

#=66#

So

#(6x^2+7x-2) = (x+4)(6x-17)+66#

or

#(6x^2+7x-2)/(x+4) = (6x-17)+66/(x+4)#