Question

How do you divide $\left( {{x}^{2}}+2x+15 \right)\div \left( x-3 \right)$ using synthetic division.

Answer 1

Hint: Now to solve the given division we will use synthetic division. Hence we will first consider the coefficient of the dividend and write it in synthetic division form. Now we will use the method and hence find the quotient and remainder.

Complete step-by-step answer:
Now let us first understand the concept of synthetic division. Synthetic division is a shortcut method to divide polynomials. The method is only defined for linear divisors. Linear divisors means the degree of divisors is 1. Now in synthetic division we just calculate using the coefficients of the dividend and constant of the divisor. Now we will write the coefficient and constant in the division form. Now consider the given expression. Using this we write it in synthetic division form as,
$\text{3 }\!\!|\!\!\text{ 1 2 15}$
Now write the first coefficient down without changing anything.
Hence we get,
$\begin{align}
  & \text{3 }\!\!|\!\!\text{ 1 2 15} \\
 & \text{ }\!\!|\!\!\text{ } \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ 1 }} \\
\end{align}$
Now multiply the first number with the number below first number and write it below the second number. Hence we will write $3\times 1=3$ below 2.
 $\begin{align}
  & \text{3 }\!\!|\!\!\text{ 1 2 15} \\
 & \text{ }\!\!|\!\!\text{ 3} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ 1 }} \\
\end{align}$
Now add the numbers in the second column and write the sum.
$\begin{align}
  & \text{3 }\!\!|\!\!\text{ 1 2 15} \\
 & \text{ }\!\!|\!\!\text{ 3} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ 1 5 }} \\
\end{align}$
Now again repeat the process. Hence we get,
$\begin{align}
  & \text{3 }\!\!|\!\!\text{ 1 2 15} \\
 & \text{ }\!\!|\!\!\text{ 3 15} \\
 & \text{ }\!\!|\!\!\text{ }\overline{\text{ 1 5 30 }} \\
\end{align}$
Now consider the last row. The last number in the row is 30 which is the remainder. Now the first two numbers are the coefficients of the quotient.
Hence we get, $\left( {{x}^{2}}+2x+15 \right)\div \left( x-3 \right)=\left( x+5 \right)\left( x-3 \right)+30$

Note: Now note that we can also solve the expression by factoring the term out. Now we will rewrite the expression such that we can take x – 3 common from the expression. Hence we will rewrite the dividend as $\left( {{x}^{2}}-3x+5x-15+15+15 \right)$ Now simplifying we get, $x\left( x-3 \right)+5\left( x-3 \right)+30$ which further gives us $\left( x+5 \right)\left( x-3 \right)+30$.