SOLUTION: {{{(2x^4-3x^2+5x-2)/(x+2)}}} ----------- - Here's my work: {{{(x(2x^3-3x+5)-2)/(x+2)}}} {{{(2x^3-3x+5-1)/1}}} {{{(2x^3-3x+4)/1}}} {{{2x^3-3x+4}}} My textbook

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: {{{(2x^4-3x^2+5x-2)/(x+2)}}} ----------- - Here's my work: {{{(x(2x^3-3x+5)-2)/(x+2)}}} {{{(2x^3-3x+5-1)/1}}} {{{(2x^3-3x+4)/1}}} {{{2x^3-3x+4}}} My textbook      Log On


   

Question 636887: %282x%5E4-3x%5E2%2B5x-2%29%2F%28x%2B2%29
-------------
Here's my work:
%28x%282x%5E3-3x%2B5%29-2%29%2F%28x%2B2%29
%282x%5E3-3x%2B5-1%29%2F1
%282x%5E3-3x%2B4%29%2F1
2x%5E3-3x%2B4


My textbook says the answer is 2x%5E3-4x%5E2%2B5x-5%2B%288%2F%28x%2B2%29%29... I would appreciate it very much if someone would guide me through solving this problem, as I am just completely stumped here.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Your error is in going from
%28x%282x%5E3-3x%2B5%29-2%29%2F%28x%2B2%29
to
%282x%5E3-3x%2B5-1%29%2F1
When reducing fractions, only factors cancel!! The "x" in the numerator is only a factor of part of the numerator, not the whole numerator. And the "x" in the denominator is not a factor at all. And the 2's are not factors either. So neither is x nor the 2 cancels. In fact, the numerator is prime. It will not factor. So there no common factors to cancel. It is not possible to reduce this fraction.

But it is possible to divide the numerator by the denominator. Long division can be used:
      2x^3 - 4x^2 + 5x    -  5
     _________________________________
x+2 / 2x^4 + 0x^3 - 3x^2  +  5x  -2
      2x^4 + 4x^3
      -----------
            -4x^3 -3x^2
            -4x^3 -8x^2
           ------------
                   5x^2 +  5x
                   5x^2 + 10x
                   ---------
                          -5x -  2
                          -5x - 10
                          --------
                                 8

The remainder is 8. And as you may recall from when you first learned long division, you make a fraction out of the remainder and the divisor. So


With a divisor of x+2, synthetic division could also be used:
-2 |  2   0   -3    5   -2
         -4    8  -10   10
      --------------------
      2  -4    5   -5    8

As we should have expected, we get a remainder of 8 (in the lower right corner). And the quotient is represented in the rest of the bottom row. The "2 -4 5 -5" translates into 2x%5E3+-+4x%5E2+%2B+5x++++-++5. So with synthetic division we get the same answer:


P.S. Note that in both methods of division, it was important to use "0x^3" for the missing x^3 term. Any missing terms should be handled this way when dividing.