Question

How do you divide $\dfrac{{{x}^{2}}+x-1}{2x+4}$ using polynomial long division?

Answer 1

Hint: We are given two polynomials ${{x}^{2}}+x-1$ and $2x+4$, we are asked to divide ${{x}^{2}}+x-1$ by $2x+4$ using the polynomial long division method. To solve this question, we will learn about the long division method, we will come across the degree of the polynomial and that only higher degree polynomial is divisible by lower degree polynomial, we will learn the steps to perform the polynomial long division of polynomials.

Complete step by step answer:
We are given, $\dfrac{{{x}^{2}}+x-1}{2x+4}$. In the numerator we have ${{x}^{2}}+x-1$ and in the denominator we have $2x+4$. We are asked to solve $\dfrac{{{x}^{2}}+x-1}{2x+4}$. 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 $\dfrac{4+{{x}^{2}}+{{x}^{3}}}{{{x}^{2}}+2x+4}$. Here, 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 ${{x}^{3}}+{{x}^{2}}+4$. Now, let us consider our question, so we have $\dfrac{{{x}^{2}}+x-1}{2x+4}$, 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.
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 $\dfrac{{{x}^{2}}+x-1}{2x+4}$. Here, the first term of the numerator is ${{x}^{2}}$ and that of the denominator is $2x$. So, on dividing them, we get $\dfrac{{{x}^{2}}}{2x}=\dfrac{x}{2}$.
Now, we multiply the denominator by $\dfrac{x}{2}$ and put it below the numerator.
\[2x+4\overline{\left){\begin{align}
  & {{x}^{2}}+x-1 \\
 & {{x}^{2}}+2x \\
 & \overline{-x-1} \\
\end{align}}\right.}\left( \dfrac{x}{2} \right.\]
Now, the new polynomial is $-x-1$. We will now divide –x by 2x, so we get,
$\dfrac{-x}{2x}=\dfrac{-1}{2}$ So, we will multiply the denominator by $\dfrac{-1}{2}$, and put it below the numerator, so we get,
\[2x+4\overline{\left){\begin{align}
  & {{x}^{2}}+x-1 \\
 & {{x}^{2}}+2x \\
 & \overline{\begin{align}
  & -x-1 \\
 & -x-2 \\
 & \overline{-1} \\
\end{align}} \\
\end{align}}\right.}\left( \dfrac{-1}{2} \right.\]
Now, we are left with -1, whose power is 0, while the denominator has the power 1, so it cannot be divided further. Hence, this is our answer.

Note:
We need to be careful when we subtract two values, the sign of the terms will get changed, so if we use the wrong sign at some point in the division, then our solution will become wrong. We can divide higher power by lower power terms, if we are left with the denominator with higher degree than the numerator, then division will stop.