Question

How do you long divide $\dfrac{{{x^3} - 27}}{{x - 3}}?$

Answer 1

Hint: In this question, we have $\dfrac{{{x^3} - 27}}{{x - 3}}$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,
\[\begin{array}{*{20}{c}}
  {\,\,\,\,{x^2}} \\
  {x - 3)\overline {{x^3} + 0{x^2} + 0x - 27} } \\
  {\,\underline {{x^3} - 3{x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,3{x^2} + 0x}
\end{array}\]
Putting the divisor and dividend, we can now divide the numbers and subtract to find remainders,
\[\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,{x^2} + 3x} \\
  {x - 3)\overline {{x^3} + 0{x^2} + 0x - 27} } \\
  {\,\underline {{x^3} - 3{x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,3{x^2} + 0x}
\end{array}\]
Now we will write the product of this second term of the quotient and the divisor under the remainder, subtract it and bring down the next term of the dividend alongside it.
\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 3x} \\
  {x - 3)\overline {{x^3} + 0{x^2} + 0x - 27} } \\
  {\,\underline {{x^3} - 3{x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,3{x^2} + 0x}
\end{array}} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\underline {3{x^2} - 9x} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x - 27}
\end{array}\]
Next, we choose the next term of the quotient so when multiplied by the divisor matches the leading term 9x of our running remainder.
\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 3x + 9} \\
  {x - 3)\overline {{x^3} + 0{x^2} + 0x - 27} } \\
  {\,\underline {{x^3} - 3{x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,3{x^2} + 0x}
\end{array}} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\underline {3{x^2} - 9x} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x - 27}
\end{array}\]

We will then write the product of this third term of the quotient and the divisor under the remainder and subtract it.
\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 3x} \\
  {x - 3)\overline {{x^3} + 0{x^2} + 0x - 27} } \\
  {\,\underline {{x^3} - 3{x^2}\,\,\,\,\,\,\,\,\,} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,3{x^2} + 0x}
\end{array}} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\underline {3{x^2} - 9x} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x - 27}
\end{array}} \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {9x - 27} } \\
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0}
\end{array}\]
There is no remainder after the final division, since subtracting the last two equations will equal up to zero. In this example, there is no remainder. The division is exact. If we did get a remainder with degree less than the divisor then we would stop here anyway.

Therefore, $\dfrac{{{x^3} - 27}}{{x - 3}} = {x^2} + 3x + 9$ 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.