Synthetic Division Examples No Remainder

08 Mar, 2021

Identify the coefficients and constants. Synthetic division page 2 Common Mistakes to Avoid.

How To Do Synthetic Division

Write the problem in a division-like format.

Synthetic division examples no remainder. The remainder is zero so we have found that x 5 is a zero of the polynomial and a root of the equation. Solution We will use Synthetic Division to show that 2 is a zero. 4x x 4 so we put a 4 on top and perform the final division step.

As you can see the remainder is 68Since I started with a polynomial of degree 3 and then divided by x 3 that is by a polynomial of degree 1 I am left with a polynomial of degree 2Then the bottom line represents the polynomial 3x 2 7x 24 with a remainder of 68. Remember to add the terms inside the synthetic division process. In Section 25 we will discuss a trick for finding such a zero.

You can use synthetic division to help you with this type of problem. The Remainder Theorem states that f c the remainder. X 1 is a solution to p x 4x3 - 8x2 - 20x 24 0.

This is an example showing synthetic division with no remainder. For example you can use synthetic division to divide by x 3 or x 6 but you cannot use synthetic division to divide by x 2 2 or 3x 2 x 7. Example Show that 2 is a zero of fx4x35x27x2.

Finally construct a horizontal line just below the coefficients of the dividend. Whatever its product place it above the. Im going to again use synthetic division.

Use synthetic division to divide by x - 2. Multiply that number you drop by the number in the box. Drop the first coefficient below the horizontal line.

Remember that if you have a variable without a number in front of it. If the leading coefficient is not a 1 then you must divide by the leading coefficient to turn the leading coefficient into a 1. A monic linear binomial is simply a polynomial of the form x k.

X2 - 3x 2 div x 1. Setting the factors equal to zero I get that x 3 and x 2 are the zeroes of the quadratic. X 3 4.

For example suppose the dividend is fx 3x4 5x2 2. Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form. Since both the x3 and x terms are missing we would record the coecients as 3 0 5 0 2.

If synthetic division will not work then you must use long division. Take the constant term of the divisor with the opposite sign and write it to the left. Because the remainder is zero this means that x 3 is a factor and x 3 is a zero.

Write the coefficients of the dividend to the right. When you actually do these problems you will do a single calculation that looks like the last figure above. Learn how to perform synthetic division on polynomialsFor more help visit my website.

For example 3x 1 would become and 2x 7 would become. If there is no remainder then the is said to be a factor of the polynomial. By the Remainder Theorem f20 and so 2 is a zero.

The result or quoitient of such a division will either divide evenly or have a remainder. X 2 3 x 2 x 1. So if the remainder comes out to be 0 when you apply synthetic division then x - c is a factor of f x.

Factor fx completely and find all of its real zeros. Synthetic division can be used whenever you are dividing a polynomial by a monic linear binomial. X 4 x 3 x 2 x 1 x 0 5 1 5 7 34 1.

And it has a remainder of zero. So my answer is going to be 1x which is also just x plus 7. Demonstrates synthetic division by showing step-by-step solutions.

In the synthetic division I divided by x 3 and arrived at the same result of x 2 with a remainder of zero. If R is a root of p x the monomial x - R divides p x and there is no remainder. After youve divided p x with x - R and thus proven that R is a root you should have a quadratic equation which you can probably factor on your own.

If theres no remainder after polynomial division px x - n then you know that x - n is a factor of px and for that reason x n is a root of the polynomial. C - 2 c 2 inside the box. Examples of monic linear binomials are x 2 x2 x 2 x 4 x-4 x 4 and x 4 3.

Divide the polynomial x 4 5x 3 2x 2 28x 12 by the first degree binomial x 3. For example let p x. We saw this fx in Section 22.

Perform x 2 3 x 2 x 1. So x plus 7 is my answer. Use synthetic division to find your answer.

Do NOT forget to record a zero for any missing terms. For my last example Im going to divide the polynomial x squared plus 5x plus 4 by the polynomial x plus 3. Find x 4 5 x 3 7 x 2 34 x 1 x 5 using synthetic division.

C 2.

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