Divide $$$x^{2} - 7$$$ by $$$x - 4$$$
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Your Input
Find $$$\frac{x^{2} - 7}{x - 4}$$$ using long division.
Solution
Write the problem in the special format (missed terms are written with zero coefficients):
$$$\begin{array}{r|r}\hline\\x-4&x^{2}+0 x-7\end{array}$$$
Step 1
Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{x^{2}}{x} = x$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$x \left(x-4\right) = x^{2}- 4 x$$$.
Subtract the dividend from the obtained result: $$$\left(x^{2}-7\right) - \left(x^{2}- 4 x\right) = 4 x-7$$$.
$$\begin{array}{r|rrr:c}&{\color{DarkBlue}x}&&&\\\hline\\{\color{Magenta}x}-4&{\color{DarkBlue}x^{2}}&+0 x&-7&\frac{{\color{DarkBlue}x^{2}}}{{\color{Magenta}x}} = {\color{DarkBlue}x}\\&-\phantom{x^{2}}&&&\\&x^{2}&- 4 x&&{\color{DarkBlue}x} \left(x-4\right) = x^{2}- 4 x\\\hline\\&&4 x&-7&\end{array}$$Step 2
Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{4 x}{x} = 4$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$4 \left(x-4\right) = 4 x-16$$$.
Subtract the remainder from the obtained result: $$$\left(4 x-7\right) - \left(4 x-16\right) = 9$$$.
$$\begin{array}{r|rrr:c}&x&{\color{Brown}+4}&&\\\hline\\{\color{Magenta}x}-4&x^{2}&+0 x&-7&\\&-\phantom{x^{2}}&&&\\&x^{2}&- 4 x&&\\\hline\\&&{\color{Brown}4 x}&-7&\frac{{\color{Brown}4 x}}{{\color{Magenta}x}} = {\color{Brown}4}\\&&-\phantom{4 x}&&\\&&4 x&-16&{\color{Brown}4} \left(x-4\right) = 4 x-16\\\hline\\&&&9&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrr:c}&{\color{DarkBlue}x}&{\color{Brown}+4}&&\text{Hints}\\\hline\\{\color{Magenta}x}-4&{\color{DarkBlue}x^{2}}&+0 x&-7&\frac{{\color{DarkBlue}x^{2}}}{{\color{Magenta}x}} = {\color{DarkBlue}x}\\&-\phantom{x^{2}}&&&\\&x^{2}&- 4 x&&{\color{DarkBlue}x} \left(x-4\right) = x^{2}- 4 x\\\hline\\&&{\color{Brown}4 x}&-7&\frac{{\color{Brown}4 x}}{{\color{Magenta}x}} = {\color{Brown}4}\\&&-\phantom{4 x}&&\\&&4 x&-16&{\color{Brown}4} \left(x-4\right) = 4 x-16\\\hline\\&&&9&\end{array}$$Therefore, $$$\frac{x^{2} - 7}{x - 4} = \left(x + 4\right) + \frac{9}{x - 4}$$$.
Answer
$$$\frac{x^{2} - 7}{x - 4} = \left(x + 4\right) + \frac{9}{x - 4}$$$A