Polynomial Functions - Pre-Calculus

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We mulitply what's below the line by 1 and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with remainder

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficiens of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficient.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with reminder

This can be rewritten as:

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with remainder

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with remainder

This can be rewritten as

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

Remember to place a when there isn't a coefficient given.

We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

with remainder

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

To divide synthetically, we begin by drawing a box. On the inside separated by spaces, we write the coefficients of the terms of our polynomial being divided. On the outside, we write the root that would satisfy our binomial , namely . Leaving a space for another row of numbers, we then draw a line below our row of coefficients.

We then begin dividing by simply carrying our first coefficient (1) down below the line.

We then multiply this 1 by our divisor (3) and write the resulting product (3) below our next coefficient.

We then add the two numbers in that column and write the sum (5) below the line.

We then simply continue the process by multiplying this 5 by our divisor 3 and writing that product in the next column, adding it to the next coefficient, and continuing until we finish the columns.

We then need to translate our bottom row of numbers into the coefficients of our new quotient. Since the first column originally corresponded to our cubic term, it will now correspond to the quadratic term meaning that our 1 can be translated as . Similarly, our second column transitions from quadratic to linear, making our 5 become . Finally, our third column becomes the constant term, meaning 8 simply remains the constant 8. Finally, our former constant column becomes the column for our remainder. However, since we have a 0, we have no remainder and can disregard it.

Putting all of this together gives us a final answer of

First, set up the synthetic division problem by lining up the coefficients. There are a couple of different strategies - for this one, we will put a -7 in the top corner and add the columns.

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The first step is to bring down the first 1. Then multiply what is below the line by the -7 in the box, write it below the next coefficient, and then add the columns:

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We can interpret this answer as meaning

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 . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as it must be accounted for by a "place holder".

___ __ __ __

__ __ ___

where is the remainder.

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

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From here we see will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example

where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that with a remainder of .

And how do we check to know if we have the right answer? We multiply and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3)

=

now we just add the remainder which is 1 to yield which matches the original dividend and is therefore our answer!

To determine if is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the being divided in. Combine the result of that with the next coefficient , which is . Then, multiply that by . Combine that result with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that does not go in evenly to this function and is not a root.

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is . To put this in expanded form:

We begin by attempting to find any rational roots using the Rational Root Theorem, which states that the possible rational roots are the positive or negative versions of the possible fractional combinations formed by placing a factor of the constant term in the numerator and a factor of the leading coefficient in the denominator.

That was a lot of wordage in one sentence, so let's break that down. We begin with our polynomial.

The constant term is the term without a variable (just a plain number). In our case the constant is 60. What are the possible factors of 60?

The leading coefficient is the number in front of the largest power of the variable. When the terms are listed in descending order (highest to lowest power), the leading coefficient is always the first number. In our case the leading coefficient is hard to spot. Since there is no number in front of , the coefficient is 1 by default.

This is nice because the only factor of 1 is well ... 1.

We then create all the possible fractions with a factor of the constant in the numerator and a factor of the leading coefficient in the denominator. This actually isn't as bad as it could be since our only possible denominator is 1. Any fraction with a denominator of 1 is just the numerator. Therfore, our possible "fractions" are simply

However, we must consider the positive or negative versions of these, so our final list of possible rational roots is

Unfortunately, this is where the process (at least without the assitance of a graphing calculator) becomes less fun. Using synthetic division, we must simply try each possible root until we have success. There's really no consistent rule to tell us where to start. Generally starting with the smaller whole numbers is best because the synthetic division is easier. Therefore, we could begin with then proceed to , etc.

For the sake of keeping this explanation as short as possible, I am going to skip straight to 2, where we will first find success.

Therefore, 2 is a root. However, it is always important to check to see if a root is in fact a double root (it works twice). Therefore, let's try it one more time.

2 does in fact work twice and is thus a double root. Since we only have three terms remainng, we can convert from synthetic back to an algebraic expression.

We can then factor.

Writing our root of 2 as an algebraic expression gives . Since we have double root, we need two of these. Therfore, our final factored expression is.

Expansion and simplification give us which is properly expressed in standard form with the constant on the right. Below is a walkthrough of the expansion and simplification (including binomial and trinomial expansion techniques).

Remember to multiply each term in one polynomial with each term in the other polynomial.

To express this polynomial as a product of linear factors you have to find the zeros of the polynomial by the method of your choosing and then combine the linear expressions that yield those zeros.

Factoring will get you , but then you are left to sort through the thrid degree polynomial. We can quickly synthetically divide the polynomial by its potential roots factors of . So that's .

We know that is a zero and dividing the original polynomial by and gives us the polynomial . We can factor this or use the quadratic formula to give .

This leaves us with four linear expressions that compose the polynomial:

Use the grouping method to factorize common terms:

Use grouping method to factorize common terms:

First pull out the common factor of 2, and then factorize:

Use grouping to factorize the common terms:

Use the grouping method to factorize common terms: