|
|
|
|
|
|
|
|
|
|
|
|
Divide $$$x^{3}$$$ by $$$16 x^{2} + 1$$$
Related calculators: Synthetic Division Calculator, Long Division Calculator
Your Input
Find $$$\frac{x^{3}}{16 x^{2} + 1}$$$ using long division.
Solution
Write the problem in the special format (missed terms are written with zero coefficients):
$$$\begin{array}{r|r}\hline\\16 x^{2}+1&x^{3}+0 x^{2}+0 x+0\end{array}$$$
Step 1
Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{3}}{16 x^{2}} = \frac{x}{16}$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$\frac{x}{16} \left(16 x^{2}+1\right) = x^{3}+\frac{x}{16}$$$.
Subtract the dividend from the obtained result: $$$\left(x^{3}\right) - \left(x^{3}+\frac{x}{16}\right) = - \frac{x}{16}$$$.
$$\begin{array}{r|rrrr:c}&{\color{DarkCyan}\frac{x}{16}}&&&&\\\hline\\{\color{Magenta}16 x^{2}}+1&{\color{DarkCyan}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{DarkCyan}x^{3}}}{{\color{Magenta}16 x^{2}}} = {\color{DarkCyan}\frac{x}{16}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&+0 x^{2}&+\frac{x}{16}&&{\color{DarkCyan}\frac{x}{16}} \left(16 x^{2}+1\right) = x^{3}+\frac{x}{16}\\\hline\\&&&- \frac{x}{16}&+0&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrrr:c}&{\color{DarkCyan}\frac{x}{16}}&&&&\text{Hints}\\\hline\\{\color{Magenta}16 x^{2}}+1&{\color{DarkCyan}x^{3}}&+0 x^{2}&+0 x&+0&\frac{{\color{DarkCyan}x^{3}}}{{\color{Magenta}16 x^{2}}} = {\color{DarkCyan}\frac{x}{16}}\\&-\phantom{x^{3}}&&&&\\&x^{3}&+0 x^{2}&+\frac{x}{16}&&{\color{DarkCyan}\frac{x}{16}} \left(16 x^{2}+1\right) = x^{3}+\frac{x}{16}\\\hline\\&&&- \frac{x}{16}&+0&\end{array}$$Therefore, $$$\frac{x^{3}}{16 x^{2} + 1} = \frac{x}{16} + \frac{- \frac{x}{16}}{16 x^{2} + 1}$$$.
Answer
$$$\frac{x^{3}}{16 x^{2} + 1} = \frac{x}{16} + \frac{- \frac{x}{16}}{16 x^{2} + 1}$$$A