Question

How do you divide $\left( {{x}^{3}}-{{x}^{2}}-14x+25 \right)\div \left( x-3 \right)$ using synthetic division?

Answer 1

Hint: To divide the given polynomials by synthetic division, we first need to know what a synthetic division is. A synthetic division is the shortcut method for the polynomial division and the catch is that this is applicable only when we divide the given polynomial by a linear expression. In this method, we will use the coefficients of ${{x}^{3}},{{x}^{2}},x$ and constant and also we are going to put $x-3=0$ and then find the values of x and use this for finding the quotient and remainder for the division.

Complete step by step solution:
The polynomial given in the above problem which we have to divide is as follows:
$\left( {{x}^{3}}-{{x}^{2}}-14x+25 \right)\div \left( x-3 \right)$
We are asked to divide the above polynomial by synthetic division method.
Synthetic division method is a shortcut method for polynomial division. Now, using synthetic division method to find the quotient and remainder in the above division we get,
For that first of all, we will write the coefficients of ${{x}^{3}},{{x}^{2}},x$ and constant in the given polynomial $\left( {{x}^{3}}-{{x}^{2}}-14x+25 \right)$ and we get,
1, -1, -14 and 25 respectively
Now, we are going to write these numbers in one line in the following way:
$\left| \!{\underline {\,
  1\text{ }-1\text{ }-14\text{ }25 \,}} \right. $ ……………… (1)
The divisor polynomial given in the above problem is $x-3$ so equating this linear polynomial to 0 we get,
$\begin{align}
  & \Rightarrow x-3=0 \\
 & \Rightarrow x=3 \\
\end{align}$
Now, using this x equals 3 value in relation (1) we get,
$3\left| \!{\underline {\,
  1\text{ }-1\text{ }-14\text{ }25 \,}} \right. $
Carrying the number 1 below the division line symbol without changing it and we get,
$\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & 1\text{ }-1\text{ }-14\text{ }25 \\
 & \downarrow \\
\end{align} \,}} \right. \\
 & \text{ 1} \\
\end{align}$
Now, we will shift this number 1 below -1 and shift 3 below 1 and we get,
$\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & 1\text{ }-1\text{ }-14\text{ }25 \\
 & \downarrow \\
\end{align} \,}} \right. \\
 & \text{ 1} \\
\end{align}$
Multiplying 3 with 1 and then placing this multiplication of 3 with 1 and putting this 3 below -1. After than we will add -1 and 3 and we get,
\[\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & 1\text{ }-1\text{ }-14\text{ }25 \\
 & \begin{matrix}
   \downarrow & 3 \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \text{ }\begin{matrix}
   1 & 2 \\
\end{matrix} \\
\end{align}\]
Multiplying 3 by 2 and placing -14 below this multiplication. Then adding -14 and 6 we get,
\[\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & 1\text{ }-1\text{ }-14\text{ }25 \\
 & \begin{matrix}
   \downarrow & 3 & 6 \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \text{ }\begin{matrix}
   1 & 2 & -8 \\
\end{matrix} \\
\end{align}\]
Multiplying 3 by -8 and placing this result of multiplication below 25. After that adding 25 and -24 we get,
\[\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & 1\text{ }-1\text{ }-14\text{ }25 \\
 & \begin{matrix}
   \downarrow & 3 & 6 & -24 \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \text{ }\begin{matrix}
   1 & 2 & -8 & 1 \\
\end{matrix} \\
\end{align}\]
Now, in the last row of the above division, the last number written in this row is 1 and is the remainder. And the quotient is as follows:
${{x}^{2}}+2x-8$
Hence, we got the quotient as ${{x}^{2}}+2x-8$ and remainder as 1.

Note: The point to be noted here is that the synthetic division is only applicable when the divisor in the polynomial division is linear. If you have given the divisor which is not linear then synthetic division is not applicable so make sure the divisor is linear.