Question

How do you long divide \[\dfrac{{{x}^{2}}-5x+6}{x-2}\]?

Answer 1

Hint: This type of problem is based on the concept of division of polynomials. First, we need to place the polynomials in the long division table. Then, we should find a term which on multiplying with x, we get \[{{x}^{2}}\] as the first term. And as usual multiply x with x-2 and subtract the product with the polynomial \[{{x}^{2}}-5x+6\]. Then, find a term which on multiplying with x-2, we get -3x+6. Subtract the final term with the polynomial \[{{x}^{2}}-5x+6\]. Continue these steps again and again until we get 0 in the remainder. Once we get 0 in the remainder, the quotient is the required answer.

Complete step by step answer:
According to the question, we are asked to divide \[\dfrac{{{x}^{2}}-5x+6}{x-2}\].
We have been given the polynomials are \[{{x}^{2}}-5x+6\] and \[x-2\].
Let us represent the polynomials in the division table.
First, we need to find a term which on multiplying with x, we get \[{{x}^{2}}\] so that we can cancel \[{{x}^{2}}\].
We find that \[x\times x={{x}^{2}}\]. Then, we have to multiply x with x-2.
Therefore,
\[x-2\overset{x}{\overline{\left){\begin{align}
  & {{x}^{2}}-5x+6 \\
 & {{x}^{2}}-2x \\
\end{align}}\right.}}\]
Then, subtract the two polynomials \[{{x}^{2}}-5x+6\] and \[{{x}^{2}}-2x\].
On grouping the similar terms and subtracting, we get
\[{{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)={{x}^{2}}-5x+6-{{x}^{2}}+2x\]
On further simplification, we get
\[\Rightarrow {{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)=x\left( 2-5 \right)+6\]
\[\Rightarrow {{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)=-3x+6\]
Therefore,
\[x-2\overset{x}{\overline{\left){\begin{align}
  & {{x}^{2}}-5x+6 \\
 & \dfrac{{{x}^{2}}-2x}{-3x+6} \\
\end{align}}\right.}}\]
Now, we need to find a term which on multiplying with (x-2), we get -3x+6.
We know that -3(x-2)=-3x+6.
Therefore, we get
\[x-2\overset{x-3}{\overline{\left){\begin{align}
  & {{x}^{2}}-5x+6 \\
 & \dfrac{{{x}^{2}}-2x}{\begin{align}
  & -3x+6 \\
 & -3x+6 \\
\end{align}} \\
\end{align}}\right.}}\]
Now, we need to subtract -3x+6 from -3x+6.
We know that the same polynomials on subtracting cancel out.
Therefore, we get
\[x-2\overset{x-3}{\overline{\left){\begin{align}
  & {{x}^{2}}-5x+6 \\
 & \dfrac{{{x}^{2}}-2x}{\begin{align}
  & -3x+6 \\
 & \dfrac{-3x+6}{0} \\
\end{align}} \\
\end{align}}\right.}}\]
Since the remainder is zero, we can stop the division.
\[\therefore \dfrac{{{x}^{2}}-5x+6}{x-2}=x-3\]

Hence, the value of \[\dfrac{{{x}^{2}}-5x+6}{x-2}\] is x-3.

Note: we can check whether the obtained answer is correct or not.
Multiply the answer with the divisor that is x-2. If they get the final product the same as the dividend, that is \[{{x}^{2}}-5x+6\], then the final answer obtained is correct.
Let us check.
We need to find \[\left( x-2 \right)\left( x-3 \right)\].
\[\Rightarrow \left( x-2 \right)\left( x-3 \right)=x\times x-2x-3x-3\times -2\]
Let us now simplify the above equation.
\[\Rightarrow \left( x-2 \right)\left( x-3 \right)={{x}^{2}}-5x+6\]
We have got the dividend as the product of x-2 and x-3.
Therefore, the answer obtained is verified.