Question

How do you divide $6{{x}^{3}}+5{{x}^{2}}-4x+4$ by $2x+3$?

Answer 1

Hint: We are given two polynomials $6{{x}^{3}}+5{{x}^{2}}-4x+4$ and $2x+3$ and we are asked to divide them. To solve this question, we will learn about the numerator and denominator we have. We will see how the terms should be arranged in order to perform long division of the polynomials. Then we will learn how polynomials are divided using the long division method and we will also see what we get as the remainder.

Complete step-by-step answer:
We are given the polynomials $6{{x}^{3}}+5{{x}^{2}}-4x+4$ and $2x+3$ and we have to divide $6{{x}^{3}}+5{{x}^{2}}-4x+4$ by $2x+3$. So, here we have $6{{x}^{3}}+5{{x}^{2}}-4x+4$ as the numerator which is a polynomial of degree 3 and $2x+3$as the denominator which is a linear polynomial. In order to divide the polynomial, we should see whether the polynomials in the numerator and denominator are arranged in decreasing order of their power or not.
For example, let us consider $\dfrac{x+{{x}^{2}}+5{{x}^{3}}}{{{x}^{2}}+x+9}$. 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 $5{{x}^{3}}+{{x}^{2}}+x$.
Now, let us consider our question, so we have $\dfrac{6{{x}^{3}}+5{{x}^{2}}-4x+4}{2x+3}$. Here 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 $\dfrac{6{{x}^{3}}+5{{x}^{2}}-4x+4}{2x+3}$. The term with highest power in the numerator is $6{{x}^{3}}$ and that in the denominator is 2x. So, on dividing them, we will get $\dfrac{6{{x}^{3}}}{2x}=3{{x}^{2}}$. So, we will multiply the denominator by it and put it below the numerator and subtract.
\[2x+3\overline{\left){\begin{align}
  & 6{{x}^{3}}+5{{x}^{2}}-4x+4 \\
 & 6{{x}^{2}}+9{{x}^{2}} \\
 & \overline{-4{{x}^{2}}-4x+4} \\
\end{align}}\right.}\left( 3{{x}^{2}} \right.\]
Here, the higher power in the numerator is \[-4{{x}^{2}}\] and in 2x+3, we have 2x. So, we will divide \[-4{{x}^{2}}\] by 2x. So, we get, \[\dfrac{-4{{x}^{2}}}{2x}=-2x\]. So, we multiply (2x+3) by -2x and subtract that from the numerator, so we get,
\[2x+3\overline{\left){\begin{align}
  & 6{{x}^{3}}+5{{x}^{2}}-4x+4 \\
 & 6{{x}^{2}}+9{{x}^{2}} \\
 & \overline{\begin{align}
  & -4{{x}^{2}}-4x+4 \\
 & -4{{x}^{2}}-6x \\
 & \overline{\text{ }2x+4} \\
\end{align}} \\
\end{align}}\right.}\left( 3{{x}^{2}}-2x \right.\]
Now, we have the highest power is 2x in the numerator while in 2x+3, we have 2x. So, dividing 2x by 2x, we get \[\dfrac{2x}{2x}=1\]. So, now multiplying 1 with 2x+3 and then subtracting, we get,
\[2x+3\overline{\left){\begin{align}
  & 6{{x}^{3}}+5{{x}^{2}}-4x+4 \\
 & 6{{x}^{2}}+9{{x}^{2}} \\
 & \overline{\begin{align}
  & -4{{x}^{2}}-4x+4 \\
 & -4{{x}^{2}}-6x \\
 & \overline{\text{ }2x+4} \\
 & \text{ }2x+3 \\
 & \overline{\text{ }1} \\
\end{align}} \\
\end{align}}\right.}\left( 3{{x}^{2}}-2x+1 \right.\]
Now we are left with the term 1 in the numerator and its degree is less than that of that denominator, so we cannot divide them further. So, we get our solution as $3{{x}^{2}}-2x+1$ and the remainder as 1.

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