# MTH 209 The University of Phoenix

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

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

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.

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.

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.

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**

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**

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.

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.

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.

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!

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

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

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

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

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**

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.

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**

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!

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)

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)

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**

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

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

Ex 3c y(y-3)-(y-3) Both have a (y-3) in them! Take it to the left

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!

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.

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**

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

Doing the Factor Factoring 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

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

Example 1 page 329 a) 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!

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)

Example 1c 4x2-9y2 Note 4x2=(2x)2 and 9y2=(3y)2 So  (2x+3y)(2x-3y)

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

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 + b2 We’ll be going backwards now we know what a and b are and get  (x+3)2

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. )

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 trinomial b) 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**

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

Ex 3 page 331 a) 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)2 b) 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**

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**

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!

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**

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.

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

Example 5a Grouping xy+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)

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!

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

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

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.

Example 1a pg 336 a) x2+5x+6 We’re looking for two numbers added that make 5 and when multiplied together make 6 x2+2x+3x+6 (x2+2x)+(3x+6) group again! x(x+2)+3(x+2) (x+3)(x+2)

Example 1b x2+8x+12 x2+2x+6x+12 (x2+2x)+(6x+12) (x+2)x+(x+2)6

Example 1c a2-9a+20 What 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**

Example 2 pg 337 Factoring the Illusive Trinomial
a) x2+5x I know 1+4=5 and 1·4=4 so I can just write down (x+4)(x+1)! done b) 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**

Example 3 page 338 More of the same!
a) 2x-8+x2 reorder x2 + 2x and 4-2=2 and 4·-2 = -8 (x+4)(x-2) b) t2-9t reorder to : t2 -9t -36 and with some though –12+3 = -9 and –12·3=-36 good so (t-12)(t+3) ** Ex 29-30**

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.

Ex 4b x2+0x+9 Product Sum 9=(3)(3) 3+3=6 9=(-3)(-3) -3+(-3)=-5
9=(3)(3) =6 9=(-3)(-3) (-3)=-5 9=(9)(1) =10 9=(-9)(-1) (-1)=-10 Again, none work. This IS done (prime) already. ** Ex 31-58**

The quick answer to the Sum of Two Squares
a2+b2 has no common factor other than 1 It is always a prime polynomial! Easy cheesy!

Yet another baby step, we’ll do the same now but mix in another variable Ex 5- Page 339
a) x2+2xy –8y2 We 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)

Polynomials with two variables
b) a2-7ab+10b2 We’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 –2b Writing it out (a-5b)(a-2b) ** Ex 59-66**

Just more factoring for the visuals… Ex 6 – pg 340
a) x3-6x2-16x factor out the GCF x x(x2-6x-16) then deal with the polynomial How about 2 and -8? Sum = -6, Product =-16 x(x-8)(x+2)

part b b) 4x3+4x2+4x factor out the GCF 4x
4x(x2+x+1) then deal with the polynomial How about 1 and 0? Sum = 1, Product =0 How about 1 and 1? Sum 2 , Product = 1 it’s done… it’s a prime polynomial

(bonus) part c 3wy2+18wy+27w factor out the GCF 3w
3w(y2+6y+9) then deal with the polynomial How about 3 and 3? Sum = 6, Product =9 3w(y+3)(y+3) 3w(y+3)2 ** Ex **

Section 5.3 Trial and Trial
Definitions Q1-6 Factor each trinomial Q7-20 Factor (if possible) each polynomial Q21-108 Spoken problems with words in them Q

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+c Since 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).

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

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 b 2. Replace bx by two terms using the two new numbers as coefficients 3. Factor the resulting four-term polynomial by grouping.

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)

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)

Example 1c 10x2+13x-3  ac = -30 What 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 6 10x2-2x+15x –3= (5x-1)2x+(5x-1)3= =(5x-1)(2x+3) ** Ex 5-42**

Example 2 pg 345 8x2-14xy+3y2 ac=24 we can look at:
-1 and –24, -3 and –8, -2 and –12, -4 and –6 Only –2 and –12 sum to –14 8x2-14xy+3y2=8x2-2xy-12xy+3y2= (4x-y)2x+(4x-y)(-3y)= (4x-y)(2x-3y) ** Ex 43-48**

Trial and Error 3x2+7x-6 The 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!

Example 3 page 346 Trial and Error
2x2+5x-3 Our pieces: 2x2 can only be 2x and x 1 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!

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 negative Trying the first gives –18x wrong Trying the second gives –11x GREAT! (3x-2)(x-3) Good. **49-68**

The cookbook 1. 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 difference x2+x-6 = (x+3)(x-2) and x2-x-6=(x-3)(x+2)

Why not add another variable into our trial and error method
(Do we HAVE to Mom?)

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**

Example 5 page 348 Factoring Completely (Doing what it takes)
Factor each polynomial completely: a) 4x3+14x2+6x take out a 2x  2x(2x2+7x+3) Then expand the trinomial 2x(2x+1)(x+3)

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 **

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 = -17x b) -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 **

You MUST practice these to get good! Section 5.4
Definitions Q1-4 Playing with the numbers Q5-10 Factor a trinomial using ac Q11-16 Factoring it out Q17-42 Complete the factoring Q43-48 Use trial and error Q49-72 Factor any polynomial thrown at ya Q73-104 Dialect Problems Q

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 · 13 Then you can break 77 up And 1001= 11·7·13

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

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**

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).

Doing the work once…

Factoring a Difference of Sum of Two Cubes
a3-b3 = (a-b)(a2+ab+b2) a3+b3 = (a+b)(a2-ab+b2) Always…!

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 **

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**

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 square If it’s not a perfect square, then use the ac method or trial and error If 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.

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!

Example 4 Potpourri c) –3x4-15x3+72x2 Take out –3x2 –3x2(x2+5x-24)
We want a product of –24 and a sum of 5 What numbers do that? 8 and –3 –3x2(x2+8x-3x-24) = –3x2(x-3)(x+8)

Example 4 Potpourri c) 60y3-85y2-25y Take out 5y 5y(12y2-17y-5)
We want a product of -60 and a sum of -17 What numbers do that? –20 and 3 5y(12y2-20y+3y-5) = 5y[4y(3y-5)+1(3y-5)]= 5y(3y-5)(4y+1) ** Ex **

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)

Uncle Sam wants YOU to factor. Section 5.5
Definitions Q1-6 Given one prime, factor the polynomial Q7-18 Factor each difference or sum of cubes Q19-34 Factoring a Difference of two fourth powers Q35-42 Factor anything thrown at you Q36-110 A smattering of applications Q More practice Q

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)

ax2+bx+c=0 If a,b, and c are real numbers and a isn’t 0 This is called a quadratic equation

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) .

You get to use all your factoring tools!
First you factor. Then you use the Zero Factor Property (ZFP?)

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=0 So either x=-3 or x=2

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 yes So BOTH are answers -3 and 2 ** Ex. 7-10**

Example 2 pg 360 Solve 3x2=-3x Coral EVERYTHING to the left… 3x2+3x=0
Factor : 3x(x+1)=0 Either 3x=0 or x+1 = 0 So either x=0 or x=-1 Check x=0  0=0 yes, x= =0, 0=0 yes Both x=0 and x=-1 are solutions (This is called a compound equation). ** Ex **

Caution: don’t play with your food
In the previous example 3x2=-3x We 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.

The Cookbook! Rewrite the equation with all stuff to the left.
Factor the left hand side completely Use the zero factor property to get simple linear equations Solve the linear equations Check the answers State the solution(s) to the original equation.

Example 3 pg 361 Solve (2x+1)(x-1)=14
We need to absorb the 14 into the stuff on the left 2x2-2x+1x-1= 14 2x2-x-15= 0 (2x+5)(x-3)=0 2x+5=0 or x-3=0 x=-5/2 or x=3 Plug them into the first equation and they both equal 14! So the solutions are –5/2 and 3. ** Ex **

Caution John? Why not just say (in the equation above: (2x+1)(x-1)=14 that 2x+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.

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=0 x=3 Even though x-3 is twice above, there is only ONE solution here. ** Ex **

Example 5 page 362 using grouping in your factoring
2x3-x2-8x+4=0 x2(2x-1)-4(2x-1)=0 (x2-4)(2x-1)=0 (x-2)(x+2)(2x-1)=0 x-2=0 or x+2=0 2x-1=0 x=2 or x=-2 or x= ½ So it has THREE solutions (note the highest order term!) ** Ex **

Example 6a pg 363 x+6=0 or x-4=0 x=-6 or x=4

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=0 2x=-1 x=-1/2 or x=10 **Ex **

Example 7 page 364 After the decoding of the word problem we get the expression x(2x+1)=55 Expand it and lump on the left 2x2+x-55 =0 (2x+11)(x-5)=0 2x+11=0 or x-5=0 2x=-11 or x=5 x=-11/2 or x=5 We can’t have a negative width so the solution we can use ifs 5 feet. ** Ex **

Enter the Pythagorean Theorem
If you have a triangle then a2+b2=c2

Example 8 pg 365 Using the Pythagorean Theorem.
x2+(x+1)2=52 x2+ x2+2x+1=25 2x2+2x-24=0 x2+x-12=0 (x-3)(x+4)=0 x-3=0 or x+4=0 x=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 **

Example 5.6 Definitions Q1-6 Solve pre-factored equations Q7-10
Solve messier equations Q11-Q32 Solve slightly messier equations Q33-70 Word problems Q71-104 End for this week!

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