Chapter 9.3, Problem 53E

To determine

To find: the solution by dividing using synthetic division method.

Answer to Problem 53E

  $x^4+2x^3+4x^2+5x+10$ .

Explanation of Solution

Given:

  $x^5-3x^2-20$ by $x-2$ .

Concept used:

Steps required for synthetic division of a polynomial:

$\left(1\right):$ To set up the problem, first, set the denominator equal to zero to find the number to put in the division box. Next, make sure the numerator is written in descending order and if any terms are missing you must use a zero to fill in the missing term, finally list only the coefficient in the division problem.

$\left(2\right):$ Once the problem is set up correctly, bring the leading coefficient (first number) straight down.

$\left(3\right):$ Multiply the number in the division box with the number you brought down and put the result in the next column.

$\left(4\right):$ Add the two numbers together and write the result in the bottom of the row.

$\left(5\right):$ Repeat steps 3 and 4 until you reach the end of the problem.

$\left(6\right):$ Write the final answer. The final answer is made up of the numbers in the bottom row with the last number being the remainder and the remainder must be written as a fraction. The variables or x’s start off one power less than the original denominator and go down one with each term.

Calculation:

It has to divide $x^5-3x^2-20$ by $x-2$ using synthetic division.

For that rewrite the given expression so that all coefficients of variable $x$ in decreasing power are included, that is:

  $x^5+0x^4+0x^3-3x^2+0x-20$ .

And it can be express the divisor in the form of $x-a$ that is:

  $x-\left(+2\right)$ .

In synthetic division, it is done to following steps.

First drop the first coefficient just below the horizontal line.

Then multiply that number and drop by $a+2$ and place the product above the horizontal line just below the second coefficient.

Next add the column of numbers and put the sum directly below the horizontal line.

Repeat the steps $\left(i\right)\text{ and }\left(ii\right)$ until it run out of columns to add.

Now, for $x^5+0x^4+0x^3-3x^2+0x-20÷x-\left(+2\right)$ :

  $\begin{array}{l}\underset{\_}{\begin{array}{l}+2\text{}100-30-20\\ 2481020\end{array}}\\ 1245100\end{array}$

Thus, it has:

  $\begin{array}{l}{x}^5+0x+0x^3-3x^2+0x-20÷x-\left(+2\right).\\ \Rightarrow x^4+2x^3+4x^2+5x+10.\end{array}$

Hence, $x^4+2x^3+4x^2+5x+10$ .