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(8x4 + 5x3 + 4x2 - x + 7)/(x + 1)

For this problem, where the leading coefficient of the divisor is 1, plain 'ol long division will suffice here.

x + 1  |8x4 + 5x3 + 4x2 - x + 7

So x + 1 goes into the dividend how many times? Ask yourself, what do I need to multiply x by in order to get back an 8x4?  8x3. So multiply 8x3 by the divisor like you would in regular long division, except when you're finished change the signs of those two terms (to account for the fact that we are really doing a subtraction) and sum together.

        8x3

x + 1 |8x4 + 5x3 + 4x2 - x + 7

     (-)8x4 (-) 8x3

          0x4 - 3x3 + 4x2

Bring down the 4x2 term and repeat. What do I need to multiply x by in order to get negative 3x3? -3x2. Multiply by the divisor and change the signs (again, shown in parentheses) and sum.

        8x3 - 3x2_+ 7x_- 8

x + 1 |8x4 + 5x3 + 4x2 - x + 7

     (-)8x4 (-)8x3

         0x4 - 3x3 + 4x2

             (+)3x3 (+)3x2

                 0x3  + 7x2  - x

                        (-)7x2 (-)7x

                          0x2 - 8x + 7

                              (+)8x (+)8

                                    0x + 15

This gives a remainder of 15, and like regular long division, can be written over the divisor as 15 / (x + 1)

So the answer is 8x3 - 3x2 + 7x - 8 + (15/(x+1))

the answer is b

not me

step-by-step explanation:

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