SOLUTION: Can someone please help me divide the following. (6x^2 - 31x + 5) / (x - 5) and (x^3 + 2x^2

SOLUTION: Can someone please help me divide the following. (6x^2 - 31x + 5) / (x - 5) and (x^3 + 2x^2 - 3x + 2) / (x +1)

Question 254962: Can someone please help me divide the following. (6x^2 - 31x + 5) / (x - 5) and (x^3 + 2x^2 - 3x + 2) / (x +1)
Found 2 solutions by drk, palanisamy:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Lets take the first equation first. We can apply synthetic division. It is easier:
6x^2 - 31x + 5) / (x - 5)
It looks like this:
5 l . . . . 6 . . . . . -31 . . . . . . 5
. . . . . . . . . . . . .30 . . . . . . . -5
. . . . . . 6 . . . . . .-1 . . . . . . r = 0
So, we get 6x-1 with no remainder
--
We apply the same approach to the second one to get
(x^3 + 2x^2 - 3x + 2) / (x + 1)
(-1) l . . . . . . 1 . . . . . .2 . . . . . -3 . . . . . .2
. . . . . . . . . . . . . . . . .-1 . . . . . -1 . . . . .4
. . . . . . . . . .1 . . . . . .1 . . . . . -4 . . . . . .6
So, we get 1x^2 + 1x - 4 + 6/(x+1)
Answer by palanisamy(496) (Show Source): You can put this solution on YOUR website!
1)(6x^2 - 31x + 5) / (x - 5)
=(6x^2-30x-x+5)/(x-5)
=[6x(x-5)-1(x-5)]/(x-5)
=(x-5)(6x-1)/(x-5)
= (6x-1)
2)(x^3 + 2x^2 + 3x + 2) / (x +1)
=(x^3+x^2+x^2+x+2x+2)/(x+1)
=[x^2(x+1)+x(x+1)+2(x+1)/(x+1)
=(x+1)(x^2+x+2)/(x+1)
= x^2+x+2
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