Question

How do you divide $\dfrac{{{x}^{2}}-2x+3}{x-1}$?

Answer 1

Hint: We are given $\dfrac{{{x}^{2}}-2x+3}{x-1}$ and we can see that the numerator is ${{x}^{2}}-2x+3$ and the denominator is $x-1$, we are asked to divide ${{x}^{2}}-2x+3$ by $x-1$. To solve this question, we will learn about the long division method, we will learn about how the terms should be defined in long division and we will learn at which step we should terminate the long division.

Complete step by step answer:
We are given, $\dfrac{{{x}^{2}}-2x+3}{x-1}$,here in the numerator we have a 2-degree polynomial, ${{x}^{2}}-2x+3$ and in the denominator we have $x-1$. In order to divide the polynomial, one should observe that the polynomial in the numerator and denominator should be arranged in decreasing order of their power.
For example, consider $\dfrac{2+{{x}^{2}}+5{{x}^{3}}}{{{x}^{2}}+2x+4}$. Here we can see that the numerator is not well defined, so we will arrange it in a descending order based on their power. So, we will get the numerator after arranging it as $5{{x}^{3}}+{{x}^{2}}+2$.
Now, let us consider our question, so we have $\dfrac{{{x}^{2}}-2x+3}{x-1}$. In our polynomial both the numerator and the denominator are in the correct order. Now, we will learn how to perform the long division of polynomials with the help of the following steps.
Step 1: Divide the first term of the numerator by the first term of the denominator and then put that in the quotient.
Step 2: Multiply the denominator by that value and then put that below the numerator.
Step 3: Now, subtract them to create the next polynomial.
Step 4: Repeat these steps again with the new polynomial.
Now, we have $\dfrac{{{x}^{2}}-2x+3}{x-1}$. Here, the first term of the numerator is ${{x}^{2}}$ and that of the denominator is x. So, on dividing them, we get $\dfrac{{{x}^{2}}}{x}=x$. So, we will put it in the quotient and multiply the denominator by it and put it below the numerator.
\[x-1\overline{\left){\begin{align}
  & {{x}^{2}}-2x+3 \\
 & {{x}^{2}}-x \\
 & \overline{\begin{align}
  & -x+3 \\
 & -x+1 \\
 & \overline{\text{ }2} \\
\end{align}} \\
\end{align}}\right.}\left( x-1 \right.\]
Now, the new polynomial is $-x+3$. We will now divide –x by x and multiply -1 by x-1 and will put it below –x+3.
Now, we are left with 2, whose power is 0, while the denominator has the power 1, so it cannot be divided further. Hence, this is our solution is x-1 and remainder is 2.

Note:
Students have to take care while subtracting two values, the sign of the terms will get changed, so if they use the wrong sign at some point in the division, then the solution will become wrong. Also, one can divide higher power by lower power terms, and if we are left with the denominator with higher degree than the numerator, then division will stop.