How do you divide #(x^4 + 3x^3 + 28x + 15) /( x + 5)#?

1 Answer
George C.
Mar 17, 2016

#(x^4+3x^3+28x+15)/(x+5)=x^3-2x^2+10x-22+125/(x+5)#

Explanation:

You can separate out multiples of #(x+5)# from the numerator progressively as follows:

#(x^4+3x^3+28x+15)/(x+5)#

#=(x^4+5x^3-2x^3+28x+15)/(x+5)#

#=(x^3(x+5)-2x^3+28x+15)/(x+5)#

#=x^3+(-2x^3-10x^2+10x^2+28x+15)/(x+5)#

#=x^3-2x^2+(10x^2+28x+15)/(x+5)#

#=x^3-2x^2+(10x^2+50x-22x+15)/(x+5)#

#=x^3-2x^2+10x+(-22x+15)/(x+5)#

#=x^3-2x^2+10x+(-22x-110+125)/(x+5)#

#=x^3-2x^2+10x-22+125/(x+5)#

This is equivalent to long division of polynomials.

#color(white)()#
To check that the remainder is correct, substitute #x=-5# in the original numerator expression:

#x^4+3x^3+28x+15#

#=5^4-3*5^3-28*5+15=625-375-140+15 = 125#

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If you prefer (as I do), you can long divide the coefficients - not forgetting to include a zero for the 'missing' #x^2# term in the dividend...

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