How do you long divide $\dfrac{{{x^2} + 4x - 39}}{{x + 8}}$?
Answer
621.9k+ views
Hint: In this question, we have a fraction $\dfrac{{{x^2} + 4x - 39}}{{x + 8}}$ and we have to divide it with the help of long division. Dividing the two just like we divide normal numbers with long division, with dividend under the bar and divisor on the left.
Complete step-by-step solution:
So, let us now put the dividend and divisor in their places, before we start dividing,
$ \Rightarrow x + 8)\overline {{x^2} + 4x - 39} $
Starting the division process,
We will now multiply it with one variable which will help us subtract the dividend inside the bracket,
\[\begin{array}{*{20}{c}}
{\,\,\,\,x} \\
{x + 8)\overline {{x^2} + 4x - 39} } \\
{\,\,\,\,\,\underline { - {x^2} - 8x} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x - 39}
\end{array}\]
Dividing the numbers even further until we find a remainder that cannot be divided,
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 4} \\
{x + 8)\overline {{x^2} + 4x - 39} } \\
{\,\,\,\,\,\underline { - {x^2} - 8x} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x - 39}
\end{array}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,4x + 32} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 7}
\end{array}$
Now, we have divided the above equation till we have found the remainder as $ - 7$, and we know that this particular function cannot be divided any further. If we did get a remainder with degree less than the divisor then we would stop here anyway, and since that is the case, we cannot divide it any further. Now, we treat in the normal way by dividing it by the divisor and adding it to the quotient:
$ \Rightarrow x - 4 + \dfrac{{ - 7}}{{x + 8}}$
Simplifying the equation even further in order to remove the complex signs, we get,
$ \Rightarrow x - 4 - \dfrac{7}{{x + 8}}$
So, now we have the final equation.
Therefore, for $\dfrac{{{x^2} + 4x - 39}}{{x + 8}}$ $ \Rightarrow x - 4 - \dfrac{7}{{x + 8}}$ is the solution of the long division.
Note: The long division method is usually used when dividing a long number with a shorter one, and when variables are involved, a similar process is followed with the dividend and divisor being on the right- and left-hand side of the parenthesis respectively. It gives a quotient and a remainder.
Complete step-by-step solution:
So, let us now put the dividend and divisor in their places, before we start dividing,
$ \Rightarrow x + 8)\overline {{x^2} + 4x - 39} $
Starting the division process,
We will now multiply it with one variable which will help us subtract the dividend inside the bracket,
\[\begin{array}{*{20}{c}}
{\,\,\,\,x} \\
{x + 8)\overline {{x^2} + 4x - 39} } \\
{\,\,\,\,\,\underline { - {x^2} - 8x} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x - 39}
\end{array}\]
Dividing the numbers even further until we find a remainder that cannot be divided,
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 4} \\
{x + 8)\overline {{x^2} + 4x - 39} } \\
{\,\,\,\,\,\underline { - {x^2} - 8x} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x - 39}
\end{array}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,4x + 32} } \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 7}
\end{array}$
Now, we have divided the above equation till we have found the remainder as $ - 7$, and we know that this particular function cannot be divided any further. If we did get a remainder with degree less than the divisor then we would stop here anyway, and since that is the case, we cannot divide it any further. Now, we treat in the normal way by dividing it by the divisor and adding it to the quotient:
$ \Rightarrow x - 4 + \dfrac{{ - 7}}{{x + 8}}$
Simplifying the equation even further in order to remove the complex signs, we get,
$ \Rightarrow x - 4 - \dfrac{7}{{x + 8}}$
So, now we have the final equation.
Therefore, for $\dfrac{{{x^2} + 4x - 39}}{{x + 8}}$ $ \Rightarrow x - 4 - \dfrac{7}{{x + 8}}$ is the solution of the long division.
Note: The long division method is usually used when dividing a long number with a shorter one, and when variables are involved, a similar process is followed with the dividend and divisor being on the right- and left-hand side of the parenthesis respectively. It gives a quotient and a remainder.
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