Here are the steps required for Synthetic Division of a Polynomial: Step 1 : 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. Instead of writing a positive 4, we write the negative of that. Several quick calculations can be fit on a single page, a convenience when doing large numbers of computations. Add again -7 + 10 and write the result 3 below on the third line.
But then we have our final answer. For example, 3x — 1 would become and 2x + 7 would become. It's a fast way of dividing polynomials, if you're dividing one polynomial by a linear expression like x+1 or x-3. 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. Let's take a look at a variety of examples: Let's see what happens if we use our regular synthetic division process, and ignore the fact that the leading coefficient of the divisor is 2 not 1.
The divisor number in brackets the 5 then multiplies this number. The last value in the bottom row is the remainder and is written as a fraction. There are a couple of points here. So I'll put a 0 as the coefficient for the x to the fourth term. Example 1 - Divide: Step 1: To set up the problem, first, set the denominator equal to zero to find the number to put in the division box.
But here it's going to be another just, let's go through the process of it just so that you get comfortable with it. Another Example Practice makes perfect! Not only are we getting a shortcut, but Synthetic Division can actually help us to evaluate polynomials! To illustrate the process, recall the example at the beginning of the section. Multiply the root value times this sum and add to the next coefficient. Well, with the power of the Remainder Theorem Synthetic Substitution we can skip all the mess and get the answer we need without out too much work. Since we are usually looking for factors of this form, this is not really a serious limitation.
Repeat steps 5 and 6, filling each column from left to right until you get to the end of the coefficients. Let's work out another example together. Synthetic division is a shorthand method of dividing polynomials where you divide the coefficients of the polynomials, removing the variables and exponents. If the leading coefficient is not a 1, then you must divide by the leading coefficient to turn the leading coefficient into a 1. The problem was that the leading coefficient in the divisor was not one.
This pattern is occurring because we are working with a divisor with a leading coefficient of one and a power of one! Example: What is the result when 4 x 4 -6 x 3 -12 x 2 - 10 x + 2 is divided by x - 3? To put the divisor, x + 2, into the form x — a, use the constant's negative. Bring down the leading coefficient 1 , multiply it with a 2 , and 3. Note: also watch out for a missing term in your original polynomial: if there is no x-squared term, or no x term, etc, then in order to get the right answer for synthetic division, you will need to make sure to put a zero number in your first row of coefficient numbers, as a placeholder for that missing term. If the last number you get is zero, in your final division numbers, then there is no remainder. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. And we essentially have our answer, even though it seems like voodoo. There's a slightly different process you would have to do if it was 3x or if was negative 1x or if it was 5x squared.
For the next step, we bring down the next term -7x 2 , so the new dividend term is 2x 3 - 7x 2. In this case, the first sum, 1, is placed next to an x to the second power one less than three. Solution Set up the division as before, with the slightly altered result: The quotient is now The original polynomial can now be rewritten Note that this is the same as Therefore, Which is just the value of the remainder! Then you multiply the positive 30 times the negative 4. You could think of it as a degree 0 term. The coefficient of the divisor variable, x, must be a one. Draw a horizontal line under the coefficients, leaving one empty row below the coefficients for work. Bring the first coefficient down below the horizontal line, which is the easiest step of all! Oh, I have to be careful here.
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. Since the dividend has degree 4, the quotient must have degree 3. So you multiply it times the negative 4. Now, the numbers in this column are added -3 + 5 and the result 2 is written below on the third line. The answer is x 3. Number of terms in remainder equals to number of divisor terms minus one. Use zeros for missing terms.
And synthetic division is going to seem like a little bit of voodoo in the context of this video. The best way to explain these steps is by way of an example. And I encourage you to verify it with traditional algebraic long division. The divisor must be a polynomial of degree one. This article has also been viewed 323,653 times.