Question

How do find the quotient of \[\dfrac{{{y}^{3}}-125}{y-5}\].

Answer 1

Hint: In this problem, we are going to find the quotient of the polynomial division \[\dfrac{{{y}^{3}}-125}{y-5}\], we have two methods to solve this problem, one is the long polynomial division and the other is using difference of cubes formula and simplifying it. Here, we are going to use the long polynomial division in order to get the quotient of the given problem. Now we can start from step by step to find the quotient.

Complete step by step answer:
We know that the given polynomial division is,
\[\dfrac{{{y}^{3}}-125}{y-5}\]
Now we can set up the polynomials to be divided in long division, if there is any missing terms for every exponent, add one with a value of 0, we get
\[y-5\overset{{}}{\overline{\left){{{y}^{3}}+0{{y}^{2}}+0y-125}\right.}}\]
Now we can divide the highest order term in the dividend \[{{y}^{3}}\] by the highest order term in the divisor y, we get
\[y-5\overset{{{y}^{2}}}{\overline{\left){{{y}^{3}}+0{{y}^{2}}+0y-125}\right.}}\]
We can now multiply the quotient term to the divisor, we get
\[y-5\overset{{{y}^{2}}}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{{{y}^{3}}-5{{y}^{2}}} \\
\end{align}}\right.}}\]
We know that the expression is to be subtracted in the dividend, so we can change the sign in \[{{y}^{3}}-5{{y}^{2}}\], we get
\[y-5\overset{{{y}^{2}}}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{-{{y}^{3}}+5{{y}^{2}}} \\
 & \text{ +5}{{y}^{2}} \\
\end{align}}\right.}}\]
Now we can bring down the next term from the dividend to the current dividend,
\[y-5\overset{{{y}^{2}}}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{-{{y}^{3}}+5{{y}^{2}}} \\
 & \text{ +5}{{y}^{2}}+0y \\
\end{align}}\right.}}\]
Now we should divide the highest order term in the dividend \[5{{y}^{2}}\] by the divisor y, we get
\[y-5\overset{{{y}^{2}}+5y}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{-{{y}^{3}}+5{{y}^{2}}} \\
 & \text{ +5}{{y}^{2}}+0y \\
\end{align}}\right.}}\]
We can now multiply the new quotient to the divisor, then subtract those expression to get the next dividend value, we get
\[\begin{align}
  & y-5\overset{{{y}^{2}}+5y}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{-{{y}^{3}}+5{{y}^{2}}} \\
 & \text{ +5}{{y}^{2}}+0y \\
 & \text{ }\underline{-5{{y}^{2}}+25y} \\
 & \text{ }+25y \\
 & \text{ } \\
\end{align}}\right.}} \\
 & \\
\end{align}\]
Now we can bring down the next term to the current dividend, we can divide the highest order term in the dividend 25y by the divisor y and we can multiply the new quotient term by the divisor.
Subtract the expression to get the remainder
\[\begin{align}
  & y-5\overset{{{y}^{2}}+5y+25}{\overline{\left){\begin{align}
  & {{y}^{3}}+0{{y}^{2}}+0y-125 \\
 & \underline{-{{y}^{3}}+5{{y}^{2}}} \\
 & \text{ +5}{{y}^{2}}+0y \\
 & \text{ }\underline{-5{{y}^{2}}+25y} \\
 & \text{ }+25y-125 \\
 & \text{ }\underline{-25y+125} \\
 & \text{ }0 \\
 & \text{ } \\
\end{align}}\right.}} \\
 & \\
\end{align}\]
Since the remainder is 0, we got the final answer, the quotient is \[{{y}^{2}}+5y+25\].

Note:
We also have another method to solve this problem,
We know that the given polynomial division is
\[\dfrac{{{y}^{3}}-125}{y-5}\]……… (1)
We also know that the difference of cubes formula is,
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
We can write the quotient of the expression (1) as,
\[\begin{align}
  & \dfrac{{{y}^{3}}-{{5}^{3}}}{y-5} \\
 & \because {{5}^{3}}=125 \\
\end{align}\]
Now we can write the numerator in difference of cubes formula, we get
\[\Rightarrow \dfrac{\left( y-5 \right)\left( {{y}^{2}}+5y+{{5}^{2}} \right)}{y-5}\]
Now by cancelling similar terms, we get
\[{{y}^{2}}+5y+25\]
Therefore, the quotient of \[\dfrac{{{y}^{3}}-125}{y-5}\] is \[{{y}^{2}}+5y+25\].