# I am a problem solver Not a problem maker

## Presentation on theme: "I am a problem solver Not a problem maker"— Presentation transcript:

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

1st block 3rd block 4th block

Take a closer look at the original equation and our roots:
x3 – 5x2 – 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!

Spooky! Let’s look at another
24x3 – 22x2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= 0 The roots therefore are: -1/2, 2/3, 3/4

Take a closer look at the original equation and our roots:
24x3 – 22x2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= 0 The roots therefore are: -1, 2, 3 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!

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 an and q is a factor of ao

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

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 Note that + 4 is listed twice; we only consider it as one answer That is where our 9 possible answers come from!

Let’s Try One Find the POSSIBLE roots of 5x3-24x2+41x-20=0

Let’s Try One 5x3-24x2+41x-20=0

Find all the possible rational roots

That’s a lot of answers! Obviously 5x3-24x2+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.

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

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

Step 5 – Test each root X X³ + X² – 3x – 3 1 -3 3 1 -1
(-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

Ex. 1: Factor a2 - 64 a2 – 64 = (a)2 – (8)2 = (a – 8)(a + 8)
You can use this rule to factor trinomials that can be written in the form a2 – b2. a2 – 64 = (a)2 – (8)2 = (a – 8)(a + 8)

Ex. 2: Factor 9x2 – 100y2 9x2 – 100y2 = (3x)2 – (10y)2
You can use this rule to factor trinomials that can be written in the form a2 – b2. 9x2 – 100y2 = (3x)2 – (10y)2 = (3x – 10y)(3x + 10y)

a2 – 64 = (a – 8)(a + 8) 9x2 – 100y2 = (3x – 10y)(3x + 10y)

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

Homework!!!

CONSTRUCTING THE TRIANGLE
2 CONSTRUCTING THE TRIANGLE ROW 0 ROW 1 ROW 2 ROW 3 R0W 4 ROW 5 ROW 6 ROW 7 ROW 8 ROW 9

Pascal’s Triangle and the Binomial Theorem
(x + y)0 = 1 (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1 y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y1 + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6 7.5.2

Binomial Expansion - Practice
a = 3x b = 2 Expand the following. a) (3x + 2)4 = 4C0(3x)4(2)0 + 4C1(3x)3(2)1 + 4C2(3x)2(2)2 + 4C3(3x)1(2)3 + 4C4(3x)0(2)4 = 1(81x4) + 4(27x3)(2) + 6(9x2)(4) + 4(3x)(8) + 1(16) = 81x x x2 + 96x +16 n = 4 a = 2x b = -3y b) (2x - 3y)4 = 4C0(2x)4(-3y)0 + 4C1(2x)3(-3y)1 + 4C2(2x)2(-3y)2 + 4C3(2x)1(-3y)3 + 4C4(2x)0(-3y)4 = 1(16x4) + 4(8x3)(-3y) + 6(4x2)(9y2) + 4(2x)(-27y3) + 81y4 = 16x4 - 96x3y + 216x2y xy3 + 81y4 7.5.6

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