Long Division Basics

To know how to divide polynomials, you need to know how to perform long division - a technique which we learned many many years ago. So if you forgot, you're already behind. But I won't get into that; I'll assume you know it. If not then here's a link:

http://argyll.epsb.ca/jreed/extras/longdiv/

Is it a good link? Who knows!? I Googled it and I was feeling lucky.

It's also important to know some of the vocabulary when dividing polynomials.

We can use the following example:

* 14x^4 + 0x^3 -3x^2 + 0x + 9 is the dividend

* x+3 is the divisor

* 14x^3 -42x^2 +123x -369 is the quotient

* 1116R is the remainder

Dividing Polynomials

To divide polynomials, you first have to identify the dividend and the divisor. Once they are identified you can now divide the dividend by the divisor using long division.

Dividing a polynomial is similar to dividing a number using long division... You check to see how many times a divisor can go into a dividend by dividing sections of the dividend by the divisor. For polynomials, you do this one term at a time, starting with the term with the largest exponent. Let's start with a tricky one:

1. This one's tricky because the question provided doesn't have an x^3-term or an x-term within the polynomial. In order to continue with this question "placeholders" are required.

Placeholders: Terms that are equivalent to 0 so that it doesn't have an effect on the polynomial itself, yet it allows the polynomial to be divided.

Another thing to watch out for are polynomials that are "out of order." Make sure the terms always go from highest to lowest exponents like so: (x^5 + x^4 + x^3 + x^2 + x + 9999)

2. Ask: How many times can the divisor (x+3) go into the dividend (14x^4)? The answer is 14x^3 times. Therefore, 14x^3 becomes the first term of the quotient and it's placed directly above 0x^3 (found in the dividend) because they are like-terms (this is important for proper communication).

Because (x+3)(14x^3) = 14x^4... you place 14x^4 below the first term of the dividend and subtract the two terms. The difference will always be 0... or else you did something wrong!

3. This is where you use the other part of the divisor, the 3 in this case. You multiply it by the same amount of times we multiplied the x-variable from step 2. SO... you multiply 3 by 14x^3 which = 42x^3. You place this directly below its like-term from the dividend (0x^3) and then you subtract the two terms.

4. Yay! We're done with the first term! Where to look next? The second term of the dividend? Nope. You use the answer which you got from step 3 (-42x^3) and divide that by you divisor (x+3). Solve: How many times does x+3 go into -42x^3?? Once you find your answer, it becomes the second term of your quotient... Simply repeat steps 2-3.

Continue to do this process until you reach the last term and eventually:

<= If there was a remainder of 0 then the divisor (x+3) would be known as an Even/Perfect Factor.

Division Statements

Once you're done all that dividing stuff, you'll need to summarize by using something known as a Division Statement.

Division Statement: Dividend = (Divisor)(Quotient) + Remainder.

So using the example that's been used the whole time above, the division statement would look like:

14x^4 - 3x^2 + 9 = (x+3)(14x^3 - 42x^2 +123x -369) + 1116.

Note: The remainder doesn't end with "R" so that we don't confuse it to be a variable.

This is not only useful for summarizing your work, but also can be used for word problems and for future calculations... (The Remainder Theorem!!!!)

Remainder Theorem

States that if you divide P(x) by (x-a), then the remainder will be the same as P(a).

This is assuming that a = the value which makes the divisor=0.

Therefore, if x=a then x-a = 0 (because a-a = 0 DUHH).

(Remember: (divisor) (quotient) + remainder... So if the quotient is 0, it automatically makes the DIVISOR 0 and you're left with the remainder, so you don't have to go through long division to figure out what the remainder is)

Let's make the divisor x+2

And the quotient P(x) = 2x^3 + 7x^2 -x + 1.

1. Think about what value of x would make the divisor = 0. In this case, that number would be -2.

2. Plug that number into the "function machine." P(-2) = 2(-2)^3 + 7(-2)^2 - (-2) + 1

3. Solve the function. The result is the remainder. P(-2) = 15R

So far in class, we've been using polynomials as the quotient and binomials as the divisor. As long as you have these 2 elements, the 3 steps above will definitely work.

Using Remainder Theorem to Solve Problems With 2 Variables

Polynomials with 2 variables may be intimidating at first, but are actually very simple to solve when you have:

- A divisor

- The remainder

Ex. Pg.92 #10a.

Determine the value of k such that when P(x) = kx^3 + 5x^2 - 2x + 3 is divided by x+1, the remainder is 7.

1. Think about what value of x would make the divisor = 0. In this case, that number would be -1.

2. Plug that number into the "function machine." You must also state that P(-1)=Remainder before you start to solve for "k". This would not only allow you to plug -1 into the function, but will also allow you to solve for the variable "k". P(-1) = 7.

3. Solve for "k". k=3

10b: Determine the remainder when P(x) is divided by x-3.

This is straight up Remainder Theorem because you solved for "k". Follow the 3 steps given to you in the "Remainder Theorem" section to solve this.

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FINALLY DONE! I apologize for my lateness, the post's length, and perhaps for a confusing scribe post. Today's lesson was very heavy content wise... So if you find any errors in this post please comment NICELY! :D

And I'll fix it when I have time. :)

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honestly the worst day ever to be scribe post :(

ReplyDeleteSTILL GOT ANOTHER 2 SUBJECTS TO DO FOR HOMEWORK!

YAYYYYyyaaayyaaayayayayyy *sob*