Question

How do you divide \[\left( 4{{x}^{3}}-2{{x}^{2}}+x-7 \right)\div \left( x-8 \right)\]?

Answer 1

Hint: This question is from the topic of algebra. In this question, we will use the long division method formula. In solving this question, we will divide the term \[\left( 4{{x}^{3}}-2{{x}^{2}}+x-7 \right)\] by the term \[\left( x-8 \right)\]. We will use the division formula to solve this question. We will find the quotient and remainder.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to divide the term \[4{{x}^{3}}-2{{x}^{2}}+x-7\] by the term \[x-8\]. Or, we can say that we have to solve \[\left( 4{{x}^{3}}-2{{x}^{2}}+x-7 \right)\div \left( x-8 \right)\].
We will divide the terms using the long division method. We can write
\[x-8\overline{\left){4{{x}^{3}}-2{{x}^{2}}+x-7}\right.}\]
From the above, we can see that \[4{{x}^{3}}-2{{x}^{2}}+x-7\] is dividend and \[x-8\] is divisor.
Now, we will write the first term of the quotient by dividing the first term of dividend with the term first term of divisor. We can write the above division as
\[x-8\overset{4{{x}^{2}}}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
\end{align}}\right.}}\]
Here, the below term \[4{{x}^{3}}-32{{x}^{2}}\] has been written by multiplying the quotient term \[4{{x}^{2}}\] with the divisor term \[x-8\]. Now, we will subtract the term \[4{{x}^{3}}-32{{x}^{2}}\] by the term \[4{{x}^{3}}-2{{x}^{2}}+x-7\] and do the further division. Remember that whenever we say that we subtract the terms, then we will change the signs of below terms and add them.
\[x-8\overset{4{{x}^{2}}}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
 & \overline{0+30{{x}^{2}}+x-7} \\
\end{align}}\right.}}\]
Now, we will add another quotient which will be equal to the first term of \[30{{x}^{2}}+x-7\] divided by the first term of \[x-8\]. And then, that added quotient will be multiplied by the term which is in the divisor that is \[x-8\] and subtract that multiplied term with the term \[30{{x}^{2}}+x-7\]. Similarly, we will do the same process as many times until we get a remainder as a constant.
\[x-8\overset{4{{x}^{2}}+30x}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
 & \overline{\begin{align}
  & \text{ }0+30{{x}^{2}}+x-7 \\
 & \text{ +}30{{x}^{2}}-240x \\
 & \text{ }\overline{\text{ }0+241x-7} \\
\end{align}} \\
\end{align}}\right.}}\]
Similarly, we can write
\[x-8\overset{4{{x}^{2}}+30x+241}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
 & \overline{\begin{align}
  & \text{ }0+30{{x}^{2}}+x-7 \\
 & \text{ +}30{{x}^{2}}-240x \\
 & \text{ }\overline{\begin{align}
  & \text{ }0+241x-7 \\
 & \text{ 241}x-1928 \\
 & \text{ }\overline{\text{ }0+1921} \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}\]
So, using the formula \[\text{Dividend}=\text{Divisor}\times \text{Quotient+Remainder}\], we can write
\[4{{x}^{3}}-2{{x}^{2}}+x-7=\left( 4{{x}^{2}}+30x+241 \right)\left( x-8 \right)+1921\]
Now, we can write the term \[\left( 4{{x}^{3}}-2{{x}^{2}}+x-7 \right)\div \left( x-8 \right)\] as
\[\dfrac{4{{x}^{3}}-2{{x}^{2}}+x-7}{x-8}=\dfrac{\left( 4{{x}^{2}}+30x+241 \right)\left( x-8 \right)+1921}{x-8}=\dfrac{\left( 4{{x}^{2}}+30x+241 \right)\left( x-8 \right)}{x-8}+\dfrac{1921}{x-8}\]
We can write the above as
\[\dfrac{4{{x}^{3}}-2{{x}^{2}}+x-7}{x-8}=4{{x}^{2}}+30x+241+\dfrac{1921}{x-8}\]
We have divided \[\left( 4{{x}^{3}}-2{{x}^{2}}+x-7 \right)\div \left( x-8 \right)\] and we get as \[4{{x}^{2}}+30x+241+\dfrac{1921}{x-8}\]

Note: We should have a better knowledge in the topic of algebra to solve this type of question easily. We should know how to divide the polynomials using the long division method. Remember the formula for division rule. The formula is:
\[\text{Dividend}=\text{Divisor}\times \text{Quotient+Remainder}\]
Let us understand this from the following example:
\[6\overset{32}{\overline{\left){\begin{align}
  & 195 \\
 & 18 \\
 & \overline{015} \\
 & \text{ }12 \\
 & \text{ }\overline{03} \\
\end{align}}\right.}}\]
Here, we can say that 195 is dividend, 32 is divisor, 6 is quotient, and 3 is remainder.
Don’t get confused in subtraction of below term:
\[x-8\overset{4{{x}^{2}}}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
 & \overline{0+30{{x}^{2}}+x-7} \\
\end{align}}\right.}}\]
We have first changed the signs and then added them. We can understand this from the following:
\[x-8\overset{4{{x}^{2}}}{\overline{\left){\begin{align}
  & 4{{x}^{3}}-2{{x}^{2}}+x-7 \\
 & 4{{x}^{3}}-32{{x}^{2}} \\
 & \overset{-+}{\mathop{\overline{0+30{{x}^{2}}+x-7}}}\, \\
\end{align}}\right.}}\]