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October 20, 2014

October 20, 2014

Posted by **katy** on Friday, September 23, 2011 at 3:55pm.

- algebra -
**Steve**, Friday, September 23, 2011 at 6:03pmThink of it as a long division problem, and do it the way you would if dividing a 2-digit number into a 3-digit number.

x goes into x^2 x times.

Multiply the divisor by x to get x^2 + 3x

Subtract that from x^2 + 5x and the x^2 terms go away.

Now you are left with x+3 into 2x+8. x goes into 2x 2 times.

Multiply the divisor by 2 to get 3x+6

Subtract that and the x terms go away, and you are left with a remainder of 2.

So, (x^2 + 5x + 8)/(x + 3) = x+2 remainder 2

or

x + 2 + 2/(x+3)

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