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

Last updated date: 20th Jun 2024
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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.

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.