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#

#color(white)()#
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...

enter image source here

Related questions
See all questions in Division of Polynomials
Impact of this question
1808 views around the world
You can reuse this answer
Creative Commons License