1. ## Long Division

See attachment for question.

I got the answer: (3x^3/2) - (5x^2/8) - (7x/16) - (53/32) as the quotient.

I tried using synthetic division but the divisor is not in the form x - c. So, I could not apply synthetic in this case.

The remainder I got is -159/32.

Is any of this correct?

2. I've done this problem a few times now to make sure and got a quotient of

(3/2)x^3 - (13/4)x^2 + (35/8)x

and a remainder of

-121/8

Hope that helps

3. Hello, sologuitar!

Sorry, you messed up somewhere . . .

. . $\displaystyle \begin{array}{cccccccccccc} &&&& \frac{3}{2}x^3 & - & \frac{13}{4}x^2 & + & \frac{35}{8}x & - & \frac{121}{16} \\ & & -- & -- & -- & -- & -- & -- & -- & -- & -- \\ 2x+3 & ) & 3x^4 &-& 3x^3 &-& x^2 &-& 2x &+& 0 \\ & & 3x^4 &+& \frac{9}{2}x^3 \\ & & -- & -- & -- \\ &&&& -\frac{13}{2}x^3 &-& x^2 \\ &&&& -\frac{13}{2}x^3 &-& \frac{39}{4}x^2 \\ &&&& -- & -- & -- \end{array}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \begin{array}{cccccccccccc} &&&&&&\frac{35}{4}x^2 &-& 2x \\ &&&&&& \frac{35}{4}x^2 &+& \frac{105}{8}x \\ &&&&&& -- & -- & -- \\ &&&&&&&&-\frac{121}{8}x &+& 0 \\ &&&&&&&& -\frac{121}{8}x &-& \frac{363}{16} \\ &&&&&&&& -- & -- & -- \\ &&&&&&&&&& \frac{363}{16} \end{array}$

4. ## soroban...

Originally Posted by Soroban
Hello, sologuitar!

Sorry, you messed up somewhere . . .

. . $\displaystyle \begin{array}{cccccccccccc}$$\displaystyle &&&& \frac{3}{2}x^3 & - & \frac{13}{4}x^2 & + & \frac{35}{8}x & - & \frac{121}{16} \\ & & -- & -- & -- & -- & -- & -- & -- & -- & -- \\ 2x+3 & ) & 3x^4 &-& 3x^3 &-& x^2 &-& 2x &+& 0 \\ & & 3x^4 &+& \frac{9}{2}x^3 \\ & & -- & -- & -- \\ &&&& -\frac{13}{2}x^3 &-& x^2 \\ &&&& -\frac{13}{2}x^3 &-& \frac{39}{4}x^2 \\ &&&& -- & -- & -- \end{array} . . . . . . . . . . . . . . . . . . . . . . . . . . . .\displaystyle \begin{array}{cccccccccccc}$$\displaystyle &&&&&&\frac{35}{4}x^2 &-& 2x \\ &&&&&& \frac{35}{4}x^2 &+& \frac{105}{8}x \\ &&&&&& -- & -- & -- \\ &&&&&&&&-\frac{121}{8}x &+& 0 \\ &&&&&&&& -\frac{121}{8}x &-& \frac{363}{16} \\ &&&&&&&& -- & -- & -- \\ &&&&&&&&&& \frac{363}{16} \end{array}$
Thank you for pointing out that an error was made. This was a tough problem because of all the fractions in the quotient. Typically I solve long division problems where the divisor is in the form (x - c) meaning the coefficient of x is not > 1.