Thread: 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?

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

Hello, sologuitar!


Sorry, you messed up somewhere . . .


. .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .

Originally Posted by Soroban

Hello, sologuitar!


Sorry, you messed up somewhere . . .


. . &&&& \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}" alt="
&&&& \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}" />
. . . . . . . . . . . . . . . . . . . . . . . . . . . . &&&&&&\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}" alt="
&&&&&&\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.