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How do you divide #(x^2+7x+15)/(x-5)#?

Answer 1

#(x^2+7x+15)/(x-5) = x + 12 + 75/(x-5)#

EDIT: The formatting is screwed up for some reason, it looks fine in preview. Not really sure how to deal with it :/

#" "x^2+7x+15# #color(blue)(x)(x-5)->color(white)(....)ul(x^2-5x)" # #" "0color(white)(.)+12x+15# #color(blue)(12)(x-5)->" "color(white)(.)ul(12x-60)" # #" "color(red)("0+75 " larr " Remainder")#

The parts in blue denote the result from dividing in at each point, while the red deals with remainder. Combining these gives:

#(x^2+7x+15)/(x-5) = color(blue)(x + 12) + (color(red)(75))/(x-5)#

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Answer 2

Abigail Allen

To divide (x^2+7x+15) by (x-5), you can use long division or synthetic division. Here is the solution using long division:

     x + 12
_______________

x - 5 | x^2 + 7x + 15 - (x^2 - 5x) ____________ 12x + 15 - (12x - 60) ___________ 75

Therefore, (x^2+7x+15)/(x-5) simplifies to x + 12 with a remainder of 75.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.