Question

How do you divide \[\dfrac{{{x}^{4}}-81}{x-3}\] using synthetic division?

Answer 1

Hint: Here we have to do division of given terms using synthetic division method.
The synthetic division method is one of the easiest methods to perform division of polynomials.
In the method, the term is divided by the linear factor for which it must have more than one as a leading coefficient and degree also.
For performing the synthetic division method we have to follow some steps.
In general we can say that it can perform as “Bring down, multiply and add, multiply and add, multiply and add.”

Complete Step by Step solution:
Given that there is a term whose division we have to do by using a synthetic division method. The given expression is as follow:-
     \[\Rightarrow \dfrac{{{x}^{4}}-81}{x-3}\,\,\,\,\,\,\,\,.........\left( 1 \right)\]
In synthetic division methods there are some conditions, such as the denominator should not contain more than one as the degree of and leading coefficient of the first variable.
Otherwise it will become difficult to solve it. So, the given division term satisfies the conditions for synthetic division method.
Now, there are some steps which we have to follow to the given division in synthetic division method which is as follows.
Step-1:- Take denominator equal to zero for finding the divisor, also write the numerator variable a in descending order degree, if there is any term missing then put there zero. At the last write all co-efficient in numerator and denominator (as divisor) in descending order in the division form.
So, writing numerator variable term in descending order degree as
\[\Rightarrow \]\[{{x}^{4}}\,\,\,\,{{x}^{3}}\,\,\,\,\,{{x}^{2}}\,\,\,\,\,{{x}^{1}}\,\,\,\,\,\,{{x}^{0}}\]
Writing coefficient in descending order in as
\[\Rightarrow \]\[1\,\,\,\,0\,\,\,\,\,0\,\,\,\,\,0\,\,\,\,\,\,\,-81\]
Now, put denominator equals to zero ie.\[x-3=0\]
     \[\therefore \,\,x\,\,=\,3\]
\[\therefore \] \[3\] will be the divisor of the division.
\[\Rightarrow \] \[\left. {\underline {\,
 3 \,}}\! \right| \,\,1\,\,\,\,\,0\,\,\,\,\,\,0\,\,\,\,\,\,0\,\,\,\,\,\,\,\,-81\]
Step-2:- As the question is set in the division form, so now put down the first term or number straight down, as
\[\Rightarrow \] \[\begin{align}
  & \left. {\underline {\,
 3 \,}}\! \right| \,\,\underline{\underset{\downarrow }{\mathop{1}}\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\bullet \,\,\,\,\,\,\,\,-81} \\
 & \,\,\,\,\,1 \\
\end{align}\]
Step-3:- Now, multiply the number brought down with the divisor and put it in first column
     \[\begin{align}
  & \left. {\underline {\,
 3 \,}}\! \right| \,\,\underline{1\,\,\,\,\,\,\,\underset{3}{\mathop{0}}\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,-81} \\
 & \,\,\,\,\,1 \\
\end{align}\]
Step-4:- Now, add the two numbers and put the result at the bottom of the second column.
     \[\begin{align}
  & \left. {\underline {\,
 3 \,}}\! \right| \,\,\underline{1\,\,\,\,\,\,\,\underset{^{+}3}{\mathop{0}}\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,-81} \\
 & \,\,\,\,\,1\,\,\,\,\,\,\,3 \\
\end{align}\]
Step-5:- Now, again multiply the divisor with the number you just brought down and then add it with \[=\]in the third column. Repeat this steps until we reach at the end.
     So,
   \[\begin{align}
  & \left. {\underline {\,
 3 \,}}\! \right| \,\,\underline{\begin{align}
  & 1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\bullet \,\,\,\,\,\,\,\,\,-81\,\,\,\,\,\, \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,\,\,27\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,81 \\
\end{align}} \\
 & \,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,27\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\uparrow }{\mathop{0}}\,\,\,\,\,\,\,\,\,\, \\
\end{align}\]
Step-6:- Write the final answer we get. The last number in the answer is remainder which should written in the fraction. The variable should be start from one less than the numerator degree, and down with each term:
     Therefore the quotient is.
     \[1{{x}^{3}}\,\,+\,3{{x}^{2}}\,+\,9x\,+\,27\]
     And the remainder is zero \['0'\].
Therefore the final answer of given term using synthetic division method is,
     \[\dfrac{{{x}^{4}}\,\,-\,\,81}{x\,-\,3}\,\,=\,\,{{x}^{3}}+3{{x}^{2}}+9x+27+\dfrac{0}{x-3}\]
     \[={{x}^{3}}\,+\,3{{x}^{2}}\,+\,9x\,+\,27\]

Note:
In this numerical we have to perform division by using a synthetic division method.
The synthetic division is the method which is used to divide the one polynomial with another polynomial. It is the easiest way to perform the division of a polynomial.
The only condition for these methods is that the leading coefficient of the denominator should not have more than \[1\] as a term or number and it’s degree also should be \[1\]. This is also called a linear factor.
This method is not applicable if the divisor of the polynomial or the denominator of the polynomial is not a linear factor, it means if it carries more than \[1\] as a leading coefficient and degree.