SOLUTION: {{{(2x^5+3x^4+8x^3+26x^2+21x)/(2x^2+3x)}}}

Algebra.Com
Question 111643:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!



Factor x out of the top and 
factor x out of the bottom:

 

Cancel the x's



Now use long division:

               x³ + 0x² +  4x +  7 
       ———————————————————————————
2x + 3 )2x4 + 3x³ + 8x² + 26x + 21
        2x4 + 3x³ 
        —————————  
              0x³ + 8x²
              0x³ + 0x²
              ————————— 
                    8x² + 26x
                    8x² + 12x
                    —————————
                          14x + 21
                          14x + 21
                          ———————— 
                                 0

So the answer is x³ + 0x² + 4x + 7, or 
omitting the zero term:

                 x³ + 4x + 7

Edwin
RELATED QUESTIONS

Solve. (x – 3)^2 = 2 Solve. 2x^2 – 8x – 14 = 0 Solve. 3x^2 = –21x – 15 (answered by sofiyac)
This is a synthetic division. Please find the answer ASAP. Thank you! :)... (answered by stanbon)
2x^4+3x^3-8x^2+3x=0 (answered by MathLover1)
factor 2x^5-26x^3+72x compleately and also and solve... (answered by solver91311)
21x^2+2x-3= (answered by jim_thompson5910)
which expression is equivalent to 2(3x-4)-8x+3 2x+11 -2x-5 2x-5... (answered by jane maslow)
(6x^3-8x^2+3x-5)/2x^2 (answered by prateekagrawal)
i am trying to factor completely? 3x^2-26x+16... (answered by stanbon)
(9x^4+2x^3-x+5)+(9x^3-3x^2+8x+3)-(7x^4-x+12) (answered by CubeyThePenguin)