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Question

How do you devided this problem by using long division? I try but keep on screwing up, please help

2x^3 + 4x^2 - 5 / x^2 + 1

Answer 1

A 2x +4
B ----------------------------------
C (x^2 +1) (2x^3 +4x^2 - 5)
D 2x^3 +2x
E -----------------------------------
F 0 +4x^2 -2x -5
G 4x^2 +4
H ------------------------------
I 0 -2x -9

answer is (2x +4) remainder -2x -9
or 2(x+2) remainder -2x -9
or 2(x+2) -(2x+9)/(x^2 +1)

2x^3 divided by x^2 is 2x
2x times x^2 is 2x^2, 2x times 1 is 2x (line D)
subtract line D from dividend in line C
divide 4x^2 by x^2 resulting to 4, bring down -5 (line F)
4 times x^2 is 4x^2, 4 time 1 is 4 (line G)
subtract G from F to get the remainder (line I)

remember:always add/subtract similar terms.

Answer 2

I get 2x+4 with a remainder of -(2x+9). Be sure that you are placing partial multiplication results in the right place; your cubic being divided and the divisor both have a missing term.

Answer 3

2x + 4 + [ (2x-9)/(x^2+1) ] <---answer
x^2 + 1 2x^3 + 4x^2 - 5
-(2x^3 + 2x)
4x^2 + 2x - 5
-(4x^2 + 4)
2x - 9

Answer 4

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Answer 5

2x^3+3x^2

all u have to do is simplify no working out is needed

(the way i got rid of the 1 on the denominator after i was working out is because anything over 1 is 1 ((( x/1 =x )))