Dividing Polynomials, The Remainder Theorem and Factor Theorem

Transcription

1 College Algebra - MAT 161 Page: 1 Copyright 2009 Killoran Dividing Polynomials, The Remainder Theorem and Factor Theorem 1 Long Division: a b D c C r b a is the Quotient b is the Divisor c is the Dividend r is the Remainder Example 1 Use long division to find the Dividend and Remainder: 2x 3 13 C 23x 8 2x 3 To perform Long Division of Polynomials, we must concentrate on the First Term in the Quotient. Here the question first becomes 1) What Times 2x is equal to 2x 3? The answer to this question will go on the top of the long division (Dividend), and then is Distributed to the Divisor and written under the Quotient..2x 3/ D 2x 3 3 2) This product goes under (like normal long division), and then is subtracted from above terms (be sure to match up like terms). ) 5x C 4 C 4 2x 3 When there are missing power.s/ of x, we MUST replace each with a zero to hold its place

2 College Algebra - MAT 161 Page: 2 Copyright 2009 Killoran Example 2 To perform long division for x4 2 5 x 4 C 0x 3 2 C 0x 5 and then: it will be necessary to rewrite as D x 3 C 6 continue pattern... What happens if the Remainder is Zero? Well that is called "dividing evenly". When numbers divide evenly it also tells us that the Product of the Divisor and the Dividend is equal to the Quotient. In other words we have Factored the Origional Polynomial. 1.1 Synthetic Division Looking at the patterns formed from the above long division, you might come to realize that there are only a few important numbers. The Divisor the first term doesn t effect the answer, just the Second term. For example in 3x3 8 C 90 when working with./ only the 2 gives us work. The x term cancels out upon subtraction. So we can condense all this down to the product and sum of numbers and just make the varialbes a "position" matrix. becomes The Left "box" comes from the binomial./ but represent this with just a positive 2. This will "take care of" the subtraction part of the long division so we don t have to constantly change signs. Multiplying this by the first coeficient we obtain the second line, but in the synthetic the line 3x 3 6 is represented by simply 6: Adding down we get the next value of 2 and we then just continue the patern. This cooresponds to the long division perfectly.

3 College Algebra - MAT 161 Page: 3 Copyright 2009 Killoran Reading a solution from Synthetic Division: For the Dividen, it is best to work from the Right to the Left starting with the constant, and then assending powers of x: Because the order of the numbers is important, it is also important to include missing powers of x. Example 3 Synthetically Divide: 15 C 2x 3 C 31x C 12 x C 1 Be sure to put all polynomials in Decending Order: 2x 3 C 15 C 31x C 12 1j j 6 D 2 C 13x C 18 6 x C 1 Example 4 Synthetically Divide: 6x4 7x 3 8x C 5 There is a missing power of x;specifically ; so we will need to put a zero in its place: 6x 4 7x 3 C0x {z } 2 8x C 5. The./ will give us our "box" number of.2/ 2j j 19 D 6x 3 C 5 C Theorem 1 Remainder Theorem: If the polynomial p.x/ is divided by.x c/ then the remainder will be p.c/ This is easy enough to show using an inductive Proof. Using a small polynomial p.x/ D 2 3x C 4 and a general divisor of.x c/ we can see that after performing synthetic division that the answer we get is precisely p.c/

4 College Algebra - MAT 161 Page: 4 Copyright 2009 Killoran This Gives us two powerful Tools.. 1) Given f.x/ D 4x 5 3x 3 C 61 we can find f. 3/ by synthetic division: 3j Thus f. 3/ D 930: This is useful if we do not have a calculator that does exponents. The only operations here are multiplying and adding, which can even be done BY HAND. Remember how to do that!!!! :) 2) If we are asked: Does./ divide 7 1 evenly? We can just calculate out f.1/ D.1/ 27 1 D 0 and since the remainder is zero (same as long division) we can conclude that Yes./ divides 7 1!! This leads us into our next Theorem: Theorem 2 Factor Theorem: If the polynomial p.x/ is divided by.x c/ and the remainder is 0; then.x c/ is a factor of p.x/ (and visa versa). Also the quotient is another Factor of p.x/ and usually has to be factored further! Proof. If p.x/ x c D q.x/ C 0 D q.x/ then: x c p.x/ x c D q.x/.x c/ p.x/ D.x x c c/ q.x/ p.x/ D.x c/ q.x/ Whish states that p.x/ can be factored into.x c/ and q.x/ Example 5 Use the fact that.x C 3/ is a factor of f.x/ D x 4 C 6x 3 C 9 42 to factor f.x/. Since.x C 3/ is a factor of f.x/ we know that it will divide evenly into f.x/ : This means that the remainder under Synthetic Division will be zero. If we use this fact, we can "create" a smaller polynomial to factor. 3j

5 College Algebra - MAT 161 Page: 5 Copyright 2009 Killoran Thus we have that f.x/ D x 4 C 6x 3 C 9 42 D.x C 3/ x 3 C Looking at q.x/ D x 3 C we can then use Factor by Grouping: q.x/ D.x C 1/ 1.x C 1/ q.x/ D./.x C 1/ 2 Finally we have that f.x/ D x 4 C 4x 3 C 2 4x 3 f.x/ D.x C 3/ q.x/ {z} f.x/ D.x C 3/./.x C 1/ 2 {z } so we have completely factored f.x/ : Pretty Cool!