I am a problem solver Not a problem makerI am an answer giver not an answer takerI can do problems I haven’t seenI know that strict doesn’t equal meanIn this room we do our bestTaking notes -practice –quiz- testThe more I try the better I feelMy favorite teacher is Mr. Meal…..ey

3rd block

1st 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 2 3 4

• 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 2 3 4

The roots therefore are: -1, 2, 3 2 3 4

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)1. For polynomial 03x3xx 23

Possible roots are ___________________________________

Here p = -3 and q = 1

Factors of -3Factors of 1 ±3, ±1

±1

2. For polynomial 012x4x9x3 23

Possible roots are ______________________________________________

Here p = 12 and q = 3

Factors of 12Factors 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???

Let’s look at our solutions

41

4

11

1

21

2

31

3

61

6

121

12

±12, ±6 , ±3 , ± 2 , ±1, ±4 ±1 , ±3

3

4

3

43

1

3

13

2

3

2

13

3

23

6

43

12

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

Let’s Try One

Find the POSSIBLE roots of 5x3-24x2+41x-20=0

Let’s Try One5x3-24x2+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 pFactors of q

• Step 3 – factors

• Factors of -3 = ±3, ±1Factors of 1 = ± 1

• Step 4 – possible roots

• -3, 3, 1, and -1

• Step 5 – Test each root

X X³ + X² – 3x – 3

-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 LOOKINGFOR WHAT ROOTS WILL MAKE THE EQUATION =0

1

Ex. 1: Factor a2 - 64

• 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

• 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.

2233 babababa same sign

always oppositealways +

2233 babababa

always oppositealways +

same sign

2233 babababa

2233 babababa

2233 babababa

2233 babababa

Homework!!!

CONSTRUCTING THE TRIANGLE

1 ROW 0 1 1 ROW 1 1 2 1 ROW 2 1 3 3 1 ROW 3 1 4 6 4 1 R0W 4 1 5 10 10 5 1 ROW 5 1 6 15 20 15 6 1 ROW 6 1 7 21 35 35 21 7 1 ROW 7 1 8 28 56 70 56 28 8 1 ROW 8 1 9 36 84 126 126 84 36 9 1 ROW 9

2

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

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

n = 4a = 3x b = 2

= 1(81x4) + 4(27x3)(2) + 6(9x2)(4) + 4(3x)(8) + 1(16)

= 81x4 + 216x3 + 216x2 + 96x +16

b) (2x - 3y)4

= 4C0(2x)4(-3y)0 + 4C3(2x)1(-3y)3

= 1(16x4)

= 16x4 - 96x3y + 216x2y2 - 216xy3 + 81y4

n = 4a = 2x b = -3y

7.5.6

Binomial Expansion - Practice

+ 4C1(2x)3(-3y)1 + 4C2(2x)2(-3y)2 + 4C4(2x)0(-3y)4

+ 4(8x3)(-3y) + 6(4x2)(9y2) + 4(2x)(-27y3) + 81y4