Question

How do you divide $\dfrac{3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7}{x-2}$ ?

Answer 1

Hint: This is polynomial division. In these kinds of questions, the degree of the numerator is generally greater than the degree of the denominator. If it is not the case, then we have to divide using partial fractions. So now the degree of the denominator is $1$ and the degree of the numerator is $4$. $x-2$ is the divisor. We just have to multiply a certain term to the divisor so that it is equal to the dividend which is $3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7$.

Complete step-by-step answer:
We just have to multiply a certain term to the divisor so that it is equal to the dividend which is $3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7$.
After doing so, we just have to subtract and get the remainder. We can only do it for two terms at a time since our divisor id a binomial term.
We will first multiply $3{{x}^{3}}$ to $x-2$ . It will become the following :
$\Rightarrow 3{{x}^{3}}\left( x-2 \right)=3{{x}^{4}}-6{{x}^{3}}$ .
We subtract this from $3{{x}^{4}}+22{{x}^{3}}$ .
$\left( x-2 \right)\overset{3{{x}^{3}}}{\overline{\left){\begin{align}
  & 3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7 \\
 & 3{{x}^{4}}-6{{x}^{3}} \\
 & \left( - \right)\text{ }\left( + \right) \\
 & \overline{0\text{ + 28}{{\text{x}}^{3}}-2{{x}^{2}}} \\
\end{align}}\right.}}$
Since we are subtracting, the (-) negative symbol changes to the positive.
Now we will multiply $28{{x}^{2}}$ to $x-2$ . It will become the following :
$\Rightarrow 28{{x}^{2}}\left( x-2 \right)=28{{x}^{3}}-56{{x}^{2}}$
We subtract this from $28{{x}^{3}}-2{{x}^{2}}$ .
\[\left( x-2 \right)\overset{3{{x}^{3}}+28{{x}^{2}}}{\overline{\left){\begin{align}
  & 3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7 \\
 & 3{{x}^{4}}-6{{x}^{3}} \\
 & \left( - \right)\text{ }\left( + \right) \\
 & \overline{\begin{align}
  & 0\text{ + 28}{{\text{x}}^{3}}-2{{x}^{2}} \\
 & \text{ + }28{{x}^{3}}-56{{x}^{2}} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
\end{align}} \\
 & \overline{\text{ 0+ 54}{{\text{x}}^{2}}\text{+2x }} \\
\end{align}}\right.}}\]
Now we will multiply $54x$ to $x-2$ . It will become the following :
$\Rightarrow 54x\left( x-2 \right)=54{{x}^{2}}-108x$
We subtract this from $54{{x}^{2}}+2x$.
\[\left( x-2 \right)\overset{3{{x}^{3}}+28{{x}^{2}}+54x}{\overline{\left){\begin{align}
  & 3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7 \\
 & 3{{x}^{4}}-6{{x}^{3}} \\
 & \left( - \right)\text{ }\left( + \right) \\
 & \overline{\begin{align}
  & 0\text{ + 28}{{\text{x}}^{3}}-2{{x}^{2}} \\
 & \text{ + }28{{x}^{3}}-56{{x}^{2}} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
\end{align}} \\
 & \overline{\begin{align}
  & \text{ 0 + 54}{{\text{x}}^{2}}\text{+2x} \\
 & \text{ +}54{{x}^{2}}-108x \\
\end{align}} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
 & \overline{\text{ 0+ 110x-7}} \\
\end{align}}\right.}}\]
Now we will multiply $110$ to $x-2$ . It will become the following :
$\Rightarrow 110\left( x-2 \right)=110x-220$
We subtract this from $110x-7$.
\[\left( x-2 \right)\overset{3{{x}^{3}}+28{{x}^{2}}+54x+110}{\overline{\left){\begin{align}
  & 3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7 \\
 & 3{{x}^{4}}-6{{x}^{3}} \\
 & \left( - \right)\text{ }\left( + \right) \\
 & \overline{\begin{align}
  & 0\text{ + 28}{{\text{x}}^{3}}-2{{x}^{2}} \\
 & \text{ +}28{{x}^{3}}-56{{x}^{2}} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
\end{align}} \\
 & \overline{\begin{align}
  & \text{ 0 + 54}{{\text{x}}^{2}}\text{+2x} \\
 & \text{ +}54{{x}^{2}}-108x \\
\end{align}} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
 & \overline{\text{ 0+ 110x-7}} \\
 & \text{ +110x-220} \\
 & \text{ }\left( - \right)\text{ }\left( + \right) \\
 & \overline{\text{ 0 + 213}\text{.}}\text{ } \\
\end{align}}\right.}}\]
This is how we divide.
We got a remainder of $213$.

Note: We should be very careful while doing these kinds of questions as there is a lot of scope for calculation mistakes. Just before dividing, we can equate the divisor to be zero and find the value of $x$. After finding, we have to substitute the value of $x$ in the function just to be sure whether our remainder would be zero or non-zero. If the function returns a value of zero upon substituting, then our remainder would zero too since this divisor is a factor of our function. If it doesn’t , then it is not a factor. Here $x=2$ is not a factor of $3{{x}^{4}}+22{{x}^{3}}-2{{x}^{2}}+2x-7$.