# Synthetic Division

You’re not likely to get a CLEP question that requires you to do synthetic division, but you quite likely could get a question for which you need to divide polynomials. Synthetic division is an algorithm that allows you to divide polynomials faster than you can with regular old polynomial long division (pretty much because you don’t have to keep writing x, x, x . . . .). You can do it even if you never mastered polynomial long division.

Synthetic division works when you divide by a binomial: $x-c$, where $c$ is some real number.

Here’s how synthetic division works:

Say your problem is $\frac{x^3+8}{x+2}$. Then $x-c=2$ and $c=-2.$

• Set up something that looks like this:

• Write in $c$ where $-2$ is shown below. To the right of that write in the coefficients of the $x$ terms. The coefficient of $x^3$ is 1. There is no $x^2$ or $x$ term, but treat those as if they exist, with coefficients of 0. And the constant term is 8. The setup should look like this:

• From the list of coefficients, copy the one farthest to the left – that’s 1 – to directly below itself just below the horizontal line.

• Multiply that 1 by the -2 all the way over on the left. Write the result in just above the horizontal line, directly below the next term in the coefficient list (the first 0 in this case).

• Add that 0 to the number you just wrote under it, and write the sum below the horizontal line.

• Multiply the -2 all the way on the left by the -2 you just wrote under the horizontal line. Write the product under the next entry on the coefficient list, in this case the next 0. Add the number you just wrote to the number above it, and write the sum under the horizontal line. Keep going like that until you are out of numbers in the coefficient line.

The numbers below the line represent the answer. The expression is one degree lower than the one you divided. Since in this case the expression you divided is third degree, the answer is second degree. The first term is the coefficient of $x^2$, the second term is the coefficient of $x$, the third term is a constant term, and the fourth is the remainder.

The answer is $x^2-2x+4$.

To check, multiply back.

# Using the remainder theorem to find a remainder.

## Question

The polynomial ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue}&space;x+1}$. What is the remainder?

## Solution

Do you want to try this problem using polynomial long division? That could be a nightmare. How about synthetic division? I didn’t think so. You need the remainder theorem.

The remainder theorem states that when a polynomial (like, for example, the one in this problem) is divided by a binomial ${\color{Blue}&space;x-c}$, the remainder is equal to the value of the polynomial at point ${\color{Blue}&space;x=c}$.

Consider a simple example: When polynomial ${\color{Blue}&space;x^2-2x+2}$ is divided by ${\color{Blue}&space;x-1}$, the remainder is equal to the value of ${\color{Blue}&space;x^2-2x+2}$ at point ${\color{Blue}&space;x=1}$. Since 1 is an easy number at which to evaluate a function, it’s not too hard to see that at ${\color{Blue}&space;x=1}$, ${\color{Blue}&space;x^2-2x+2}=1.$  Thus when ${\color{Blue}&space;x^2-2x+2}$ is divided by ${\color{Blue}&space;x-1}$, the remainder is $1.$

You can confirm this result with polynomial long division or synthetic division.

Now try the question. To find the remainder when ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue}&space;x+1}$, evaluate ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$  at ${\color{Blue}&space;x=-1.}$ That’s

${\color{Blue}&space;(-1)^{55}-6&space;\cdot&space;(-1)^{42}-6&space;\cdot&space;(-1)^{29}-3&space;\cdot&space;(-1)^{23}&space;+&space;(-1)^4-2}$

Note that -1 taken to an even power is 1 and -1 taken to an even power is 1. Thus ${\color{Blue}&space;(-1)^{55}=-1}$, ${\color{Blue}&space;(-1)^{29}=-1}$, and ${\color{Blue}&space;(-1)^{23}=-1}$; while ${\color{Blue}&space;(-1)^{42}=1}$ and ${\color{Blue}&space;(-1)^{4}=1}$. The whole polynomial equals

${\color{Blue}&space;-1-\left&space;(6&space;\cdot&space;1&space;\right&space;)-\left&space;(6&space;\cdot&space;-1&space;\right&space;)-\left&space;(3&space;\cdot&space;-1&space;\right&space;)+1-2}$

${\color{Blue}&space;-1-6+6+3+1-2=1}$

The remainder is 1.

# Horizontal Transformation: Sliding a Function Sideways

## Question

This is the graph of $y=f\left&space;(&space;x&space;\right&space;)$

Sketch the graph of $y=f(x-1)$.

## Solution

The change from $y=f\left&space;(&space;x&space;\right&space;)$ to $y=f(x-1)$ is called a horizontal transformation; it slides the function sideways. The graph of $y=f(x-1)$ lies one unit to the right of the graph of $y=f\left&space;(&space;x&space;\right&space;)$. This may be counterintuitive, so let’s look at some numbers. At each x-value, the graph of $y=f(x-1)$ takes the y-value that the graph of $y=f\left&space;(&space;x&space;\right&space;)$ takes one unit earlier. Some values on the two graphs look like this:

The y-values from the original graph have all chugged along to correspond to higher x-values than they did before.

At any x-value on the graph, the function  operates on the  value 1 unit to the left of that x-value. That has the effect of moving the curve 1 unit to the right.

Similarly, $y=f(x+1)$ graphs one unit to the left of $y=f\left&space;(&space;x&space;\right&space;)$. Note that this is different from $y=f\left&space;(&space;x&space;\right&space;)+1,$ which graphs one unit up from $y=f\left&space;(&space;x&space;\right&space;)$ because the 1 that gets added to the function means 1 gets added to every y-value. Similarly, $y=f\left&space;(&space;x&space;\right&space;)-1$ graphs one unit below $f\left&space;(&space;x&space;\right&space;).$ If you don’t believe me on any of these, try playing around with values of a simple function to see what happens.