Divide (4x^2+8)/(2x+1)

Question

lgbasallote · Answer

____________
2x + 1 | 4x^2 + 8

first..divide 4x^2 by 2x..what do you have?

anonymous · Answer

2x

anonymous · Answer

So would the final,answer be 2x+8?

anonymous · Answer

2x + 8

lgbasallote · Answer

uhmm not really...since 4x^2 divided by 2x is 2x, we write 2x on top of the division i wrote...then we multiply 2x to both terms of 2x + 1

                2x
              ___________
2x + 1 | 4x^2 +  0x + 8
                4x^2 + 2x 
             -------------
                             -2x + 8

i subtracted that then brought down 8....now, divide the -2x by 2x then divide -1 by each term of 2x  + 1 again

               2x  - 1
              ___________
2x + 1 | 4x^2 +  0x + 8
                4x^2 + 2x 
             -------------
                             -2x + 8
                              -2x - 1
                          -----------
                                        9

that was subtraction again...now since you cannot divide 9 by 2x anymore, this is a remainder and we write it as

$\LARGE 2x - 1 + \frac{9}{2x+1}$

lgbasallote · Answer

got it? i have a tutorial on this stuff if you're interested in seeing it @Jtgrx