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I am a problem solver Not a problem maker I am an answer giver not an answer taker I can do problems I haven’t seen I know that strict doesn’t equal mean.

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Presentation on theme: "I am a problem solver Not a problem maker I am an answer giver not an answer taker I can do problems I haven’t seen I know that strict doesn’t equal mean."— Presentation transcript:

1 I am a problem solver Not a problem maker I am an answer giver not an answer taker I can do problems I haven’t seen I know that strict doesn’t equal mean In this room we do our best Taking notes -practice –quiz- test The more I try the better I feel My favorite teacher is Mr. Meal…..ey

2 3 rd block 1 st block 4th block

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5 Take a closer look at the original equation and our roots: x 3 – 5x 2 – 2x + 24 = 0 The roots therefore are: -2, 3, 4 What do you notice? -2, 3, and 4 all go into the last term, 24!

6 Spooky! Let’s look at another 24x 3 – 22x 2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= The roots therefore are: -1/2, 2/3, 3/4

7 Take a closer look at the original equation and our roots: 24x 3 – 22x 2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= The roots therefore are: -1, 2, What do you notice? The numerators 1, 2, and 3 all go into the last term, 6! The denominators (2, 3, and 4) all go into the first term, 24!

8 This leads us to the Rational Root Theorem For a polynomial, If p/q is a root of the polynomial, then p is a factor of a n and q is a factor of a o

9 1. For polynomial Possible roots are ___________________________________ Here p = -3 and q = 1 Factors of -3 Factors of 1  ±3, ±1 ±1 2. For polynomial Possible roots are ______________________________________________ Here p = 12 and q = 3 Factors of 12 Factors of 3  ±12, ±6, ±3, ± 2, ±1 ±4 ±1, ±3 Or ±12, ±4, ±6, ±2, ±3, ±1, ± 2/3, ±1/3, ±4/3 Or 3,-3, 1, -1 Wait a second... Where did all of these come from???

10 Let’s look at our solutions ±12, ±6, ±3, ± 2, ±1, ±4 ±1, ±3 Note that + 2 is listed twice; we only consider it as one answer Note that + 1 is listed twice; we only consider it as one answer That is where our 9 possible answers come from! Note that + 4 is listed twice; we only consider it as one answer

11 Let’s Try One Find the POSSIBLE roots of 5x 3 -24x 2 +41x-20=0

12 Let’s Try One 5x 3 -24x 2 +41x-20=0

13 Find all the possible rational roots

14 That’s a lot of answers! Obviously 5x 3 -24x 2 +41x-20=0 does not have all of those roots as answers. Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.

15 Step 1 – find p and q p = -3 q = 1 Step 2 – by RRT, the only rational root is of the form… Factors of p Factors of q

16 Step 3 – factors Factors of -3 = ±3, ±1 Factors of 1 = ± 1 Step 4 – possible roots -3, 3, 1, and -1

17 Step 5 – Test each root X X³ + X² – 3x – (-3)³ + (-3)² – 3(-3) – 3 = -12 (3)³ + (3)² – 3(3) – 3 = 24 (1)³ + (1)² – 3(1) – 3 = -4 (-1)³ + (-1)² – 3(-1) – 3 = 0 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION =0 1

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19 Ex. 1: Factor a You can use this rule to factor trinomials that can be written in the form a 2 – b 2. a 2 – 64 = (a) 2 – (8) 2 = (a – 8)(a + 8)

20 Ex. 2: Factor 9x 2 – 100y 2 You can use this rule to factor trinomials that can be written in the form a 2 – b 2. 9x 2 – 100y 2 = (3x) 2 – (10y) 2 = (3x – 10y)(3x + 10y)

21 a 2 – 64 = (a – 8)(a + 8) 9x 2 – 100y 2 = (3x – 10y)(3x + 10y)

22 The sum or difference of two cubes will factor into a binomial  trinomial. same sign always opposite always + always opposite always + same sign

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25 Homework!!!

26 CONSTRUCTING THE TRIANGLE 1 ROW ROW ROW ROW R0W ROW ROW ROW ROW ROW 9 2

27 Pascal’s Triangle and the Binomial Theorem (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 +1 y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y x 2 y 3 + 5xy 4 + 1y 5 (x + y) 6 = 1x 6 + 6x 5 y x 4 y x 3 y x 2 y 4 + 6xy 5 + 1y

28 Expand the following. a) (3x + 2) 4 = 4 C 0 (3x) 4 (2) C 1 (3x) 3 (2) C 2 (3x) 2 (2) C 3 (3x) 1 (2) C 4 (3x) 0 (2) 4 n = 4 a = 3x b = 2 = 1(81x 4 )+ 4(27x 3 )(2)+ 6(9x 2 )(4)+ 4(3x)(8) + 1(16) = 81x x x x +16 b) (2x - 3y) 4 = 4 C 0 (2x) 4 (-3y) C 3 (2x) 1 (-3y) 3 = 1(16x 4 ) = 16x x 3 y + 216x 2 y xy y 4 n = 4 a = 2x b = -3y Binomial Expansion - Practice + 4 C 1 (2x) 3 (-3y) C 2 (2x) 2 (-3y) C 4 (2x) 0 (-3y) 4 + 4(8x 3 )(-3y)+ 6(4x 2 )(9y 2 )+ 4(2x)(-27y 3 )+ 81y 4

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