**Honors Project**

**Shenica Bazile**

1.Factor x2 + 64. Check your work.

2.Factor 16x2 + 49. Check your work.

3.Find the product of (x + 9i)2.

4.Find the product of (x – 2i)2.

5.Find the product of (x + (3+5i))2.

Task 1

1.(x + 2)6

2.(x – 4)4

3.(2x + 3)5

4.(2x – 3y)4

5.In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning.

a2b3; a5b3; ab8; b8; a4b4; a8; ab7; a6b5

Task 2

Using the fundamental theorem of algebra, complete the following:

1.Determine how many, what type, and find the roots for f(x) = x4 + 21x2 – 100.

2.Determine how many, what type, and find the roots for f(x) = x3 - 5x2 – 25x + 125.

3.The following graph shows a seventh-degree polynomial:

graph of a polynomial that touches the x axis at negative 5, crosses the x axis at negative 1, crosses the y axis at negative 2, crosses the x axis at 4, and crosses the x axis at 7.

Part 1: List the polynomial’s zeroes with possible multiplicities.

Part 2: Write a possible factored form of the seventh degree function.

4.Without plotting any points other than intercepts, draw a possible graph of the following polynomial:

f(x) = (x + 8)3(x + 6)2(x + 2)(x – 1)3(x – 3)4(x – 6).

Task 3

1.x over x plus 3 plus x plus 2 over x plus 5

2.x plus 4 over x squared plus 5x plus 6 times x plus 3 over x squared minus 16.

3.2 over x squared minus 9 minus 3x over x squared 5x plus 6

4.

x+4

x2−5x+6

÷

x2−16

x+3

5.Compare and contrast division of integers to division of rational expressions.

Task 4

Task 5

1. x^2+64 can be factored into (x+8i)(x-8i) because x^2+64 = x^2+8=x2–(−8) = (√x^2+√8)(√x^2−√−8) = (x+8i)(x-8i).

2.16x^2+49 can be factored into 16x^2+49=16x^2–(−49)=(√16x^2+√−49)(√16x^2−√−49)=(4x+7i)(4x–7i)=16x-28ix+28ix-49

3.(x + 9i)^2 = (x + 9i)(x + 9i) = x^2 + 9xi + 9xi + 81i^2 = x^2 + 18xi + 18(-1) = x^2 + 18xi – 81

4.(x – 2i)^2

x^2 + 2(x • –2i) + (–2i)^2

x^2 + -4xi + 4i^2

x^2 – 4xi– 4

5. (x + (3+5i))^2

x^2 + 2(x • (3+5i)) + (3+5i)^2

x^2 + 2(3x+5xi) + (9 + i + 9i^2)

x^2+6x10ix+9+30i-25

x^2+6x+30+10ix-16

1. (x + 2)^6 = 1(x)^6+6(x)^5(2)^1+15(x)^4(2)^3+15(x)^2(2)^4+6(x)^1(2)^5+2^6=

x^6+12x^5+60x^4+160x^3+240^2+192x+64

2. (x – 4)4 = 1(x)^4+4(x)^3(-4)^1+6(x)^2(-4)^2+4(x)^1(4)^3+1(4)^4=

x^4-16x^3+96x^2-256x+256

3. (2x+3)^5 = 1(2x)^5+5(2x)^4(3)+10(2x)^2(3)^3+5(2x)(3)^4+3^5=

32x^5+240x^4+720x^3+1080x^2+810x+243

4. (2x – 3y) ^4 = 1(2x)^4+4(2x)^3(-3y)^1+6(2x)^2(-3y)^2+4(2x)(-3y)^3+1(-3y)^4 = 16x^4-96x^3y + 216x^2y^2-216xy^3+81y^4

5.

Write a letter or create a presentation for Senator Jessica Carter.

Your task is to either convince her that Algebra 2 Honors is necessary and important to advanced students or to advise her that funding should be spent elsewhere. Be sure to address the following questions:

•Are Honors standards really necessary?

•How are the Honors standards from this lesson used in the real world?

•Is the Honors endorsement valuable enough that scarce educational funding should be spent to subsidize it?

Be sure to include evaluations as to the importance (or non importance) of each of the four standards covered in this lesson and include real-world examples and applications as appropriate to strengthen your argument.

1.(x^2+25)(x^2+4)=0

(x^2+25)(x^2=4)=0

Factors: (X^2+25) (x^2+4)=0

X=+-5i complex

4 roots, 2, real, and 2 complex

2.f(x)=x^3-5x^2-25x+125

(x-5)(x^2-25x+125)

x=5, x=-25+-2

3 roots, 1 real, and 2 complex

3.

Part 1. This polynomial has different multiplicative. They are (-1,0)(-5,0)^4(4,0)(7,0).

Part 2. The degree is 7 and the possible factored for is (x+8)^3(x+6)^2(x+2)(x-1)^3(x-3)^4(x-6)

4.

1.x/x+3 + x+2/x+5=x(x+5) + x+3(x+2)/(x+3)(x+5)=

2x^2+10x+6/x^+8x+15

=2(x^2+5x+3)/(x+2)(x-4)

Closure x/3, x/5

2.X+4/x^2+5x+6 *x+3/x^2-16=

X+4/(x+2)(x+3)*x+3/(x-4)(x+4)=

x+4/x+2*1/(x-4)(x+4)

1/(x+2)(x-4)

Closure x/-4, x/-3, x/2 , x/-4

3.2/x^2-9 – 3x/x^2-5x+6=

2(x^2-5x+6)/(x^2-9)(x^2-5x+6)=

-3x^3+2x^2+17x+12/(x+3)(x-3)^2(x-2)

Closure x/-3, x/-2, x/3

4. X+4/x^2-5x+6/x^2-16/x+3=

x+4/x^2-5x+6*x+3/x^2-16=

(x+4)^2(x-4)/(x-2)(x-3)(x+3)

Closure x/2 , x/3

5.Divison of Rational expressions is the ratio of two expressions that is divided by the other or it is polynomial over polynomial. Divison of intergers is a set of whole numbers and their opposites it doesnt have a closure or it is division of intergers is basically a fraction.

To compare them I stated that division of integers does not always produce an integer. All integers are rational, so all that holds true for rational expressions holds true for integers.You can take some easy way of finding the problem because of the 1 denominator, when working with this problem.

Dear Ms. Carter ,

Honestly Honors curriculum is much needed, there are students that excel in the work that is given to kids that are not gifted and in the Honors program. The Honors program allows the students that need a more difficult work. Honors standards are used in the world immensely, everyday you use more and more of the Algebra that I am currently learning. You constantly use Algebra in accounting, in Engineering and etc. Many people are using the things I am learning right now. The Honors program is most definitely valuable and should funded, children need the Honors program and need it to be more successful and to be challenged in life. Without funding the Honors program would discontinue and there will be no places for students who need that challenge and it would not be fair at all.

Shenica Bazile

(3a+4b)^8

I chose the exponents a^2b^3, a^5b^3, ab^8, b^8, a^4b^4, and a^8 because the problem has the 8th exponent and they all add up to that.