Presentation on theme: "MTH 209 The University of Phoenix"— Presentation transcript:
1 MTH 209 The University of Phoenix Chapter 5Factoring PolynomialsOperations with Rational Expressions
2 Chapter 5 Section 1 The Basics of Factors Just the Fact-or Ma’am.orThe 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.etc.
4 Prime Factorization of Integers-Pulling out the Primes (and puttin’ on the Ritz) 12 = 2 · 612 = 1 · 1212 = 2 · 2 · 3 which is 22 · 3A 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 · 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 timesHow 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 · 7It’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 = 2312= 2 · 2 · 3 = 22 · 3When you expand it, you see, magically, that there are two 2’s in each.
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 3bThe numbers 216, 360, 504Using the same long division like trick, you can find216=23 · 33 · 3 and 360=23 · 32 · 5 and 504=23 · 32 · 7216=23 · 32 · 3 and 360=23 · 32 · 5 and 504=23 · 32 · 7The answer is 23*32 = 72
15 3cNow for 55 and 16855= 5 · =23 · 3 · 7There are NO primes in common, so the GCF = 1.** Ex 19-28**
16 Example 4a page 323 On the road to polynomials, one must start with monomials What’s the GCF of 15x2 and 9x315x2 = 5 · 3 · x · x while 9x3 = 3 · 3 · x·x·xThey 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 monomials2) 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+40aThe 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 5cx2y5+x6y3You 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 324 A binomial factor (a+b)w+(a+b)6Why 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
25 Ex 3c y(y-3)-(y-3) Both have a (y-3) in them! Take it to the left
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 1or -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 Factor Factoring the Doing Definitions Q1-Q6Find the prime factors Q7-18Find the GCF integers Q19-Q28Find the GCF monomials Q29-40Factor out the GCF in each expression Q41-76Factor out the GCF then the opposite Q77-92Word like problems Q 93-98
31 Section 5.2 Back to the Future The ‘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 329a) y 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 y (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=0Again, the number by itself is 16 which is 42.Our squared variable has a 9 in front of it, and that is just 32This 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)
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 a a b + b2We’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 b22) 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 Let’s try x as the first term and ½(14) for the second. That is 7 and 72 IS It works (So the middle term is –2 · x · 7). This is a perfect square trinomialb) 4x Both terms are perfect squares (2x)2 and 92. So this is the difference of two squares.c) 4a2+24a 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 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**
40 Ex 3 page 331a) x2-4x The first term is (x)2 and the last term is The middle term should be -2·x·2 or –4x. True!So this is (x-2)2b) a2 +16a The first term is (a)2 while the last term is The middle should be 2·a·8 =16a True!So this is (a+8) ** 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-5When 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 Groupingremember 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 groupw(w-b)+3(w-b) The factor in GROUPing(w+3)(w-b) The second factoring in grouping
46 Example 5a Groupingxy+2y+3x 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 1What is in the ( )’s is almost the same, except for –1 in the second termx2(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-y)(a-3)** Ex 71-86**
49 Fun is Factoring Using Neurons Definitions Q1-6Factor polynomials Q7-20What kind binomial is it? Q21-32Factor perfect square trinomials Q33-50Factor each polynomial Q51-70Factor each polynomial completely Q71-98Word 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)3x2+2x+3x+6x2+5x+6 This is what the problems will look like.
51 Example 1a pg 336a) x2+5x+6We’re looking for two numbers added that make 5 and when multiplied together make 6x2+2x+3x+6(x2+2x)+(3x+6) group again!x(x+2)+3(x+2)(x+3)(x+2)
52 Example 1b x2+8x+12 x2+2x+6x+12 (x2+2x)+(6x+12) (x+2)x+(x+2)6 You can use FOIL to check your answer!
53 Example 1ca2-9a+20What two numbers multiplied together make 20 but when added make –9?How about –4 and –5 ?a2-4a-5a+20(a2-4a)+(-5a+20)a(a-4)-5(a-4) I took out a –5 since I want the –4 inside to match the first a-4(a-5)(a-4) ** Ex 7-20**
54 Example 2 pg 337 Factoring the Illusive Trinomial a) x2+5x I know 1+4=5 and 1·4=4so I can just write down (x+4)(x+1)! doneb) y2+ 6y I know that 8-2 = 6 and 8·-2=-16 so it must be (y+8)(y-2)c) w2-5w I’ll try –8 +3 = -5 and –8·3=-24 good so (w-8)(w+3) ** Ex 21-28**
55 Example 3 page 338 More of the same! a) 2x-8+x2 reorderx2 + 2x and 4-2=2 and 4·-2 = -8(x+4)(x-2)b) t2-9t reorder to : t2 -9t -36and with some though –12+3 = -9 and –12·3=-36 good so (t-12)(t+3) ** Ex 29-30**
56 Prime Polynomials – you can’t get there from here Ex 4a pg 328 a) x2+7x are there any numbers that work? That add to 7 and multiply to become –6?Product Sum-6=(-1)(6) =5-6=(1)(-6) (-6)=-5-6=(2)(-3) (-3)=-1-6=(-2)(3) =1 None work. This IS done (prime) already.
57 Ex 4b x2+0x+9 Product Sum 9=(3)(3) 3+3=6 9=(-3)(-3) -3+(-3)=-5 9=(3)(3) =69=(-3)(-3) (-3)=-59=(9)(1) =109=(-9)(-1) (-1)=-10 Again, none work. This IS done (prime) already. ** Ex 31-58**
58 The quick answer to the Sum of Two Squares a2+b2 has no common factor other than 1It is always a prime polynomial!Easy cheesy!
59 Yet another baby step, we’ll do the same now but mix in another variable Ex 5- Page 339 a) x2+2xy –8y2We can guess that 4 and –2 will take care of our number parts (sum=2, product=-8)To get the 2y and –8y2 we need to add a y to to our 4 and –2 for 4y and –2y.So writing it we see (x+4y)(x-2y)
60 Polynomials with two variables b) a2-7ab+10b2We’ll need to guess that –5 and –2 will work (sum = -7 and product = 10)There is an extra b in the last term so we’ll use -5b and –2bWriting it out (a-5b)(a-2b) ** Ex 59-66**
61 Just more factoring for the visuals… Ex 6 – pg 340 a) x3-6x2-16x factor out the GCF xx(x2-6x-16) then deal with the polynomialHow about 2 and -8? Sum = -6, Product =-16x(x-8)(x+2)
62 part b b) 4x3+4x2+4x factor out the GCF 4x 4x(x2+x+1) then deal with the polynomialHow about 1 and 0? Sum = 1, Product =0How about 1 and 1? Sum 2 , Product = 1it’s done… it’s a prime polynomial
63 (bonus) part c 3wy2+18wy+27w factor out the GCF 3w 3w(y2+6y+9) then deal with the polynomialHow about 3 and 3? Sum = 6, Product =93w(y+3)(y+3)3w(y+3)2** Ex **
64 Section 5.3 Trial and Trial Definitions Q1-6Factor each trinomial Q7-20Factor (if possible) each polynomial Q21-108Spoken problems with words in them Q
65 Section 5.4 Factoring ax2+bx+c with a 0 The ac Method(not the ac/dc method – but rock on!)We really mean the a and c in ax2+bx+cSince a is some non-1 and non-0 number, we want to find the product of ac that sums to b. (We did this before, but a=1).
66 How it runs… 2x2+7x+6 ac=2·6=12 2x2+3x+4x+6 change 7x to 3x+4x 3·4=12 (2x+3)x+(2x+3)2 do the factor thing(2x+3)(x+2) Factor our 2x+3
67 The Strategy To factor the trinomial ax2+bx+c 1. Find two numbers that have a product equal to ac and a sum equal to b2. Replace bx by two terms using the two new numbers as coefficients3. Factor the resulting four-term polynomial by grouping.
68 Example 1 page 344 a) 2x2 + x –6 2·(-6) = -12 Possible pairs 1 and –12, -1 and 12, 2 and –6, -2 and 6, 3 and –4, -4 and 3 Which works?-3 and 4!2x2 –3x + 4x –6 = (2x2-3x)+(4x-6) =(2x-3)x+(2x-3)2 = (2x-3)(x+2)
69 Ex 1b b) 2x2+x-6 2·(-6) = -12 1 and –12, 2 and –6, 3 and –4 Change x to –3x and 4x= 2x2+x-6 = 2x2 –3x+4x-6=(2x-3)x+(2x-3)2= (2x-3)(x+2)
70 Example 1c10x2+13x-3 ac = -30What two numbers sum to 13 and have a product of –30?1 and –30, -1 and 30, 2 and –15, -2 and 15, 3 and –10, -3 and 10, 5 and –6, -5 and 610x2-2x+15x –3= (5x-1)2x+(5x-1)3==(5x-1)(2x+3)** Ex 5-42**
71 Example 2 pg 345 8x2-14xy+3y2 ac=24 we can look at: -1 and –24, -3 and –8, -2 and –12, -4 and –6Only –2 and –12 sum to –148x2-14xy+3y2=8x2-2xy-12xy+3y2=(4x-y)2x+(4x-y)(-3y)=(4x-y)(2x-3y) ** Ex 43-48**
72 Trial and Error3x2+7x-6The factors of 3x2 can only be 3x and x (right?)The factors of 6 can only be 6,3,2,or 1(3x 3)(x 2) or (3x 2)(x 3) or (3x 6)(x 1) or (3x 1)(x 6)The negative “c” means we have a + and – in the factors(3x + 3)(x - 2) This gives a middle term of –3x wrong(3x - 3)(x + 2) This gives a middle term of 3x wrong(3x + 2)(x - 3) And this gives a –7x(3x - 2)(x + 3) While this gives 7x We GOT IT!
73 Example 3 page 346 Trial and Error 2x2+5x-3Our pieces: 2x2 can only be 2x and x1 and 3 are the only ones for 3(2x 1)(x 3) and (2x 3)(x 1)The negative sign means we have a + and –(2x - 1)(x + 3) gives us 5x and the answer!
74 Ex 3b 3x2-11x+6 3x2 can only be 3x and x 6 can only be 6,1 and 2,3 (3x 1) (x 6) or (3x 2)(x 3) or (3x 6)(x 1) or (3x 3)(x 2)And with the c positive and b negative, then both signs must be negativeTrying the first gives –18x wrongTrying the second gives –11x GREAT!(3x-2)(x-3) Good. **49-68**
75 The cookbook1. If the signs of the terms of a trinomial are + + +, then both factors are sums.x2+5x+6 = (x+2)(x+3)2. If the signs are then both factors are differences.x2-5x+6 = (x-2)(x-3)3. If the signs are or then one factor is a sum, the other a differencex2+x-6 = (x+3)(x-2) and x2-x-6=(x-3)(x+2)
76 Why not add another variable into our trial and error method (Do we HAVE to Mom?)
77 Example 4 Factor 6x2-7xy+2y2 We can have (3x 2y)(2x y) or (3x y)(2x 2y) or (6x 2y)(x y) or (6x y)(x 2y)Both must be negative… try a few in your head… and bango… (3x-2y)(2x-y)** Ex 69-72**
78 Example 5 page 348 Factoring Completely (Doing what it takes) Factor each polynomial completely:a) 4x3+14x2+6xtake out a 2x 2x(2x2+7x+3)Then expand the trinomial2x(2x+1)(x+3)
79 Example 5b 12x2y+6xy+6y Take out the 6y 6y(2x2+x+1) Then use the ac method 2·1=2 and 1·2=2 but neither added = 1 so we are finished. That’s a prime polynomial.** Ex **
80 Example 6 Cleaning up… Factoring out the opposite of the GCF Best when a is negative.a) –18x2+51x-15x-3x(6x2-17x+5) = -3x(3x-1)(2x-5) the two numbers that multiply to 5 and with the 3 and 2 in front -2x-15x = -17xb) -3a2+2a+21-1(3a2-2a-21) = -1(3a+7)(a-3)the two numbers that multiply to become –21 and with the 3 and 1 in front add to 7a-9a = -2a.** Ex **
81 You MUST practice these to get good! Section 5.4 Definitions Q1-4Playing with the numbers Q5-10Factor a trinomial using ac Q11-16Factoring it out Q17-42Complete the factoring Q43-48Use trial and error Q49-72Factor any polynomial thrown at ya Q73-104Dialect Problems Q
82 Section 5.5 Linking the Division Stuff to Factoring: The Strategy If you are told to factor a big integer like 1001, you might guess or be told that a factor is 77.So 1001 = 77 · 13Then you can break 77 upAnd 1001= 11·7·13
83 Example 1 page 352 You are given one prime factor of a polynomial, find the rest… You are given x+1 is a prime factor of x3+2x2-5x-6
84 Ex 1 continued So we know (x+1) is one and the answer is the rest… (x+1) (x2+x-6)(x+1)(x+3)(x-2) is it!** Ex. 7-18**
85 Shortcuts starting with a3-b3 and a3+b3 And we are told that the first has (a-b) in it and the second has (a+b) in it.If we factor those primes out of the two binomials we can do the division (do it once and we don’t have to worry about it again).
87 Factoring a Difference of Sum of Two Cubes a3-b3 = (a-b)(a2+ab+b2)a3+b3 = (a+b)(a2-ab+b2)Always…!
88 Example 2 page 353 For these, just use the overall rules above and plug in the numbers… no more thinking than that!a) w note 23=8 so a=w, b=2 and use the difference formula= (w-2)(w2+2w+4)b) x3+1 note a=x and b=1 use the sum formula= (x+1)(x2-x+1)c) 8y note a=2y and b = 3 use the difference one and get: (2y-3)(4y2+6y+9) ** Ex **
89 Example 3 page 354 4th powers too much for us? x4-16 = (x2)2 –42 = (x2-4)(x2+4)= (x-2)(x+2)(x2+4)81m4-n4=(9m2)2-(n2)2= (9m2-n2)(9m2+n2)(3m-n)(3m+n)(9m2+n2)** Ex 35-42**
90 The overall winning Strategy for completely factoring polynomials If there are any common factors, take them out first.Next see if it is a difference of two squares or cubes, or a sum of two cubes. A sum of two squares does not factor.When factoring a trinomial, check to see if it is a perfect squareIf it’s not a perfect square, then use the ac method or trial and errorIf the polynomial has four terms, try grouping.Then check to see if any of the factors can be factored again… and on…and on until all are primes.
91 Example 4 Potpourri a) 2a2b-24ab+72b =2b(a2-12a+36) =2b(a-6)2 b) 3x3+6x2-75x take out a 3=3[x3+2x2-25x-50] group and factor=3[x2(x+2)-25(x+2)] factor again=3(x2-25)(x+2) expand the perfect square=3(x+5)(x-5)(x+2) done!
92 Example 4 Potpourri c) –3x4-15x3+72x2 Take out –3x2 –3x2(x2+5x-24) We want a product of –24 and a sum of 5What numbers do that? 8 and –3–3x2(x2+8x-3x-24) = –3x2(x-3)(x+8)
93 Example 4 Potpourri c) 60y3-85y2-25y Take out 5y 5y(12y2-17y-5) We want a product of -60 and a sum of -17What numbers do that? –20 and 35y(12y2-20y+3y-5) = 5y[4y(3y-5)+1(3y-5)]=5y(3y-5)(4y+1)** Ex **
94 Bonus Example a) Factor each completely ax3+ax = ax(x2+1) and that is it!b) by3+b= b(y3+1) = b(y+1)(y2-y+1)c) 8m3+22m2-6m= 2m(4m2+11m-3) = 2m(4m-1)(m+3)
95 Uncle Sam wants YOU to factor. Section 5.5 Definitions Q1-6Given one prime, factor the polynomial Q7-18Factor each difference or sum of cubes Q19-34Factoring a Difference of two fourth powers Q35-42Factor anything thrown at you Q36-110A smattering of applications QMore practice Q
96 Section 5.6 Enter the Quadratic Equation Don’t panic!It’s just a special polynomial!It looks like ax2+bx+c=0 no big deal! (you’ve already worked with ax2+bx+c stuff)(Bailiff, bar the doors)
97 The Quadratic Equation ax2+bx+c=0If a,b, and c are real numbers and a isn’t 0This is called a quadratic equation
98 Behind the scenes…To solve these uses the idea of the “Zero Factor Property”If the equation a * b = 0 then either a = 0, b= 0 (or both, of course) .
99 You get to use all your factoring tools! First you factor.Then you use the Zero Factor Property(ZFP?)
100 Example 1 pg 360 Use the ZFP to solve x2+x-6=0 x2+x-6=0 (x+3)(x-2)=0 So either x+3=0 or x-2=0So either x=-3 or x=2
101 Example 1 continued x2+x-6=0 So either x=-3 or x=2 is the solution… plug them in…9-3-6=0 yes or = 0 yesSo BOTH are answers -3 and 2** Ex. 7-10**
102 Example 2 pg 360 Solve 3x2=-3x Coral EVERYTHING to the left… 3x2+3x=0 Factor : 3x(x+1)=0Either 3x=0 or x+1 = 0So either x=0 or x=-1Check x=0 0=0 yes, x= =0, 0=0 yesBoth x=0 and x=-1 are solutions (This is called a compound equation). ** Ex **
103 Caution: don’t play with your food In the previous example 3x2=-3xWe DON’T divide by 3 or 3x before hand.Dividing by 3x would make it x=-1.We’d only get ONE of the two answers.
104 The Cookbook! Rewrite the equation with all stuff to the left. Factor the left hand side completelyUse the zero factor property to get simple linear equationsSolve the linear equationsCheck the answersState the solution(s) to the original equation.
105 Example 3 pg 361 Solve (2x+1)(x-1)=14 We need to absorb the 14 into the stuff on the left2x2-2x+1x-1= 142x2-x-15= 0(2x+5)(x-3)=02x+5=0 or x-3=0x=-5/2 or x=3Plug them into the first equation and they both equal 14! So the solutions are –5/2 and 3.** Ex **
106 CautionJohn? Why not just say (in the equation above: (2x+1)(x-1)=14 that2x+1=14 and x-1=14?Because it doesn’t have the zero factor property to help! Remember 6*1=6 as does 2*3=6 and 3*2=6 and 1*6=6. Which is which?Might it not be 7*2=14? so 2x+1=7 or x-1=2?Or the reverse? You see the problem.
107 Example 4 pg 360 Repeated Factor problem… 5x2-30x+45=0 5(x2-6x+9)=0 5(x-3)2=0 divide by 5(x-3)2=0 you can take the square root of both sides…x-3=0x=3 Even though x-3 is twice above, there is only ONE solution here.** Ex **
108 Example 5 page 362 using grouping in your factoring 2x3-x2-8x+4=0x2(2x-1)-4(2x-1)=0(x2-4)(2x-1)=0(x-2)(x+2)(2x-1)=0x-2=0 or x+2=0 2x-1=0x=2 or x=-2 or x= ½ So it has THREE solutions (note the highest order term!)** Ex **
110 Example 6b 0.02x2-0.19x=0.1 100(0.02x2-0.19x)=100(0.1) 2x2-19x=10 2x+1=0 or x-10=02x=-1x=-1/2 or x=10 **Ex **
111 Example 7 page 364After the decoding of the word problem we get the expression x(2x+1)=55Expand it and lump on the left2x2+x-55 =0(2x+11)(x-5)=02x+11=0 or x-5=02x=-11 or x=5x=-11/2 or x=5 We can’t have a negative width so the solution we can use ifs 5 feet.** Ex **
112 Enter the Pythagorean Theorem If you have a triangle then a2+b2=c2
113 Example 8 pg 365 Using the Pythagorean Theorem. x2+(x+1)2=52x2+ x2+2x+1=252x2+2x-24=0x2+x-12=0(x-3)(x+4)=0x-3=0 or x+4=0x=3 or x=-4 And since we are talking about lengths, we can’t have a negative length, so the working answer is 3.**Ex **
114 Example 5.6 Definitions Q1-6 Solve pre-factored equations Q7-10 Solve messier equations Q11-Q32Solve slightly messier equations Q33-70Word problems Q71-104End for this week!
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