 Download Presentation MTH 209 The University of Phoenix

# MTH 209 The University of Phoenix

Download Presentation ## MTH 209 The University of Phoenix

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1. MTH 209 The University of Phoenix Chapter 5 Factoring Polynomials Operations with Rational Expressions

2. Chapter 5 Section 1The Basics of Factors • Just the Fact-or Ma’am. • or • The O’Righty Factor • (I’m getting the puns out of my system).

3. Primes are your friend… • A prime number is a rock bottom, hard end, stop playing with the math, number. You can ONLY divide it by 1 and itself. • 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 etc. http://www.utm.edu/research/primes/lists/small/1000.txt

4. Prime Factorization of Integers-Pulling out the Primes(and puttin’ on the Ritz) 12 = 2 · 6 12 = 1 · 12 12 = 2 · 2 · 3 which is 22· 3 A Prime Number is a positive integer larger than 1 that has NO integral factors other than itself and one.

5. Beyond Primes? • All the REST of the numbers that are (basically) made of prime numbers multiplied together (and can be therefore factored) are called composite numbers.

6. Example 1 page 320 Prime Factorization • We write the number as two integers (anything you can ‘see’) . • 36 = 2 · 18 then break up the next thing you can • = 2 · 2 · 9 and keep breaking composite numbers apart • =2 · 2 · 3 · 3 and we are almost done… • = 22· 32 then covert them to exponents if possible ** Ex 7-12**

7. Le Example’ 2 page 321 • Find the prime factorization for 420 • We start by dividing by the smallest prime number we can see will go into it an integer number of times • How about 2? • Start Here  ** Ex 13-18**

8. And the winner is… • So the answer is 2 · 2 · 3 · 5 · 7 or (one more step) 22 · 3 · 5 · 7 It’s only in terms of prime numbers! Side note: For that first step, if the number is even, divide by 2, if it odd try 3. If it ends in a zero or 5, start with 5.

9. The GREATEST common factor • This adds to your toolkit when working with polynomials soon… • We want to know what the biggest number two numbers share as a factor.

10. It’s called the GCF by friends • Here’s how it works: 8 = 2 · 2 · 2 = 23 12= 2 · 2 · 3 = 22· 3 When you expand it, you see, magically, that there are two 2’s in each. 8 = 2 · 2 · 2 = 23 12= 2 · 2 · 3 = 22· 3

11. The cookbook • Following what we just did for those two numbers: • 1) Find the prime factorization of each integer (break it down all the way) • 2) Determine which primes are shared by both (including multiples like 22). • 3) Multiply the shared primes together and Voil’a!

12. Special Wonderfullness • If there are NO common factors, then the only common number is 1. • Nice when THAT happens… no?

13. Example 3 page 322 • a) 150, 225 So 150= 2·3 ·52 and 225 = 32·52 The GCF is 3 · 52 = 75

14. ex 3b • The numbers 216, 360, 504 • Using the same long division like trick, you can find • 216=23 · 33 · 3 and 360=23 · 32 ·5 and 504=23 ·32 ·7 • 216=23 ·32 · 3 and 360=23 ·32 ·5 and 504=23 ·32 ·7 • The answer is 23*32 = 72

15. 3c • Now for 55 and 168 • 55= 5 · 11 168=23 · 3 · 7 • There are NO primes in common, so the GCF = 1. ** Ex 19-28**

16. Example 4a page 323On the road to polynomials, one must start with monomials • What’s the GCF of 15x2 and 9x3 • 15x2 = 5 ·3 · x · x while 9x3 = 3 ·3 · x·x·x • They share 3·x·x so the GCF is 3x2 ** Ex 29-40**

17. The cookbook for monomials… • 1) Find the GCF for the coefficients of the monomials • 2) Form the GCF from the GCF of the coefficients, then the variables in the same manner.

18. pt b • 12x2y2 30x2yz 42x3y 12x2y2= 2·2·3·x·x·y·y 30x2yz = 2·3·5·x·x·y·z 42x3y= 2·3·7·x·x·x·y  6x2y ** Ex 29-40**

19. The why behind the madness • When you have a binomial you need to factor, you need to take out the biggest chunks you can. The GCF is the biggest chunk you can take!

20. Example 5 pg 323 GCF’s and Polynomials • a) 25a2+40a The GCF from 25 and 40 is 5. Pull out the 5. And we can pull out an ‘a’. 5a(5a+8)

21. 5b • 6x4-12x3+3x2 • We can pull out a 3. That’s the best we can do! • Then we can take out an x2 as well… • 3x2(2x2-4x+1)

22. 5c • x2y5+x6y3 • You can take out an x2 and a y3 from both. You don’t have any of those nasty coefficients to deal with. • x2 y3(y2+x4) • ** Ex 41-68**

23. Ex 6a page 324A binomial factor • (a+b)w+(a+b)6 • Why not take off an (a+b) from both terms? • (a+b) (w+6)

24. Ex 6b • x(x+2)+3(x+2) • Both have an (x+2) in them, so pull it out • (x+2)(x+3)

25. Ex 3c • y(y-3)-(y-3) • Both have a (y-3) in them! Take it to the left (y-3)(y-1) ** Ex 69-76**

26. The common danger… If you take the exact term out when factoring, you MUST leave a “1” in it’s place. ab+b take out the b! b(a+1) NOT b(a) Multiply it out to check!

27. Worrying about the Opposite of the GCF • Sometimes it’s best to take out the – of the GCF if it makes the end answer look better. • Starting with –4x+2xy you could take out the 2x and get 2x(-2+y) • You might also take out –2x and get -2x(2-y) You are forcing the y to become negative.

28. Let’s see it in action in Ex 7 pg 325 • We’ll do it BOTH ways… a) 3x-3y  3(x-y) or  -3(-x+y) b) a-b  1(a-b) sure you can factor out a 1 or  -1 (-a+b) • -x3+2x2-8x  x(-x2+2x-8) or  -x(x2-2x+8) ** Ex 77-92**

29. Danger! • Be sure you change the sign of EVERY term in the ( )’s when you pull out that negative sign!

30. Doing the FactorFactoring the Doing • Definitions Q1-Q6 • Find the prime factors Q7-18 • Find the GCF integers Q19-Q28 • Find the GCF monomials Q29-40 • Factor out the GCF in each expression Q41-76 • Factor out the GCF then the opposite Q77-92 • Word like problems Q 93-98

31. Section 5.2 Back to the FutureThe ‘Special Factors’ • Last time we looked at the special factors like (a+b)2 and (a-b)2 and (a+b)(a-b) • Last things first… well take (a+b)(a-b) • Remember : (a+b)(a-b) = a2-b2

32. Example 1 page 329 • a) y2-81 First, note the square root of 81 is 9, so you’re going to have 9 as the number in the factored polynomials. • So this is really y2-92 (right?) • Finally that comes from (y+9)(y-9) • done!

33. Ex 1b • 9m2=16 Next, move 16 over to the left… • So we have something familiar: 9m2-16=0 • Again, the number by itself is 16 which is 42. • Our squared variable has a 9 in front of it, and that is just 32 • This gives us 32m2-42 = (3m)2-42= (3m+4)(3m-4)

34. Example 1c • 4x2-9y2 Note 4x2=(2x)2 and 9y2=(3y)2 • So  (2x+3y)(2x-3y) ** Ex 7-20**

35. On to factoring Perfect Square Trinomials • Note we are just doing the polynomial section in reverse.

36. In 4.6 we saw… • (a+b)2= a2+2ab+b2 • So we saw: x2+6x+9 = x2 + 2 x 3 + 32 which is a2 + 2 a b + b2 We’ll be going backwards now we know what a and b are and get  (x+3)2

37. The trick to expose them… • Is it a perfect square trinomial? • 1) The first and last terms are a2 and b2 • 2) The middle term is 2ab or –2ab ( So, to see it quickly, take half of the middle coefficient and see if that squared is the b2ed term. )

38. Example 2 page 330 Identifying special products… • a) x2-14x+49 Let’s try x as the first term and ½(14) for the second. That is 7 and 72 IS 49. It works (So the middle term is –2 · x · 7). This is a perfect square trinomial • b) 4x2-81 Both terms are perfect squares (2x)2 and 92. So this is the difference of two squares. • c) 4a2+24a+25 The first term is (2a)2 and the last term is (5)2 . The middle term would be 2·2a·5 =20a BUT it’s really 24a so this is NOT a perfect square trinomial. • d) 9y2-24y-16 The first term is (3y)2 and the last term is (4)2. Our middle term would be 2 · 3y · 4 = 24y2 . But in a perfect square trinomial, the first and LAST terms are always positive. It’s not true and not one. • ** Ex 21-32**

39. Actually Factoring them… • Remember: • a2+2ab+b2 = (a+b)2 • a2 -2ab+b2 = (a-b)2 • Index cards…

40. Ex 3 page 331 • a) x2-4x+4 The first term is (x)2 and the last term is 22 . The middle term should be -2·x·2 or –4x. True! • So this is  (x-2)2 • b) a2 +16a+64 The first term is (a)2 while the last term is 82 . The middle should be 2·a·8 =16a True! • So this is  (a+8)2 ** Ex 33-50**

41. Ex 3c • c) 4x2-12x+9 • The first term is (2x)2 and the last is 32 . • The middle term is negative, does this work? -2 ·2x·3 = -12x Yes! • (2x-3)2 ** Ex 33-50**

42. The end of the road in factoring • Factoring Completely has happened when you can’t go any further. • If you can’t factor a polynomial it is called a prime or irreducible polynomial. • For example: 3x or w+1 or 4m-5 • When your factored polynomial is only made up of prime polynomials, you’re DONE!

43. Example 4 page 331 • a) Factor each completely: 2x3-50x = 2x(x2-25) = 2x(x+5)(x-5) • 8x2y-32xy+32y = 8y(x2-4x+4) = 8y(x-2)2 ** Ex 51-70**

44. Safety in Groups (or numbers) • Factoring by Grouping • remember when we had (x+a)(x+3)? • If you multiply these out, you get: • (x+a)x+(x+a)3 THIS is the important form for us, since it is the first step. • x2+ax+3x+3a messy! THIS is what your problems will usually look like.

45. Grouping Pt 2 • So if you are given a big long mess like • w2-bw+3w-3b • (w2-bw)+(3w-3b) The GROUP in group • w(w-b)+3(w-b) The factor in GROUPing • (w+3)(w-b) The second factoring in grouping

46. Example 5a Grouping • xy+2y+3x+6 we look at it and see we can pull a y out of the first two terms and a 3 out of the second two terms • (xy+2y) + (3x+6) • y(x+2)+3(x+2) • (y+3)(x+2)

47. 5b Grouping • 2x3-3x2-2x+3 • (2x3-3x2)+(-2x+3) just group • x2(2x-3)+1(-2x+3) pull out x2 and 1 • What is in the ( )’s is almost the same, except for –1 in the second term • x2(2x-3)-1(2x-3) factor out –1 ! • (x2-1)(2x-3)  (x-1)(x+1)(2x-3) Done!

48. Ex 5c • ax+3y-3x-ay = ax-3x-ay+3y rearrange the terms = (ax-3x)+(-ay+3y) = x(a-3)-y(a-3) = (x-y)(a-3) ** Ex 71-86**

49. Fun is Factoring Using Neurons • Definitions Q1-6 • Factor polynomials Q7-20 • What kind binomial is it? Q21-32 • Factor perfect square trinomials Q33-50 • Factor each polynomial Q51-70 • Factor each polynomial completely Q71-98 • Word problems Q89-94

50. Section 5.3 Sleuthing out ax2+bx+c where a=1 • This is the next step… beyond this (5.4) we’ll see a ≠1 (but not yet…) • Going backwards first: • (x+2)(x+3) = (x+2)x+(x+2)3 • x2+2x+3x+6 • x2+5x+6 This is what the problems will look like.