Unit+3+Virtual+Notebook


 * __Unit 3 Lesson 5__**

Solve the following problems on a separate piece of paper. Use your solutions to answer the following summary questions in your virtual notebook.

a.) Divide f(x) by g(x) using synthetic or polynomial division.
 * Example 1:**


 * f(x) = x^3 - x^2 + 2x -1 g(x) = x + 3**

b.) Find f(-3) when f(x) = x^3 - x^2 + 2x -1

a.) Divide f(x) by g(x) using synthetic or polynomial division.
 * Example 2:**


 * f(x) = 2x^3 - 3x^2 + 4x -7 g(x) = x - 2**

b.) Find f(2) when f(x) = 2x^3 - 3x^2 + 4x - 7

a.) Divide f(x) by g(x) using synthetic or polynomial division.
 * Example 3:**


 * f(x) = 4x^3 - 9x^2 + 3x -10 g(x) = x + 2**

b.) Find f(-2) when f(x) = 4x^3 - 9x^2 + 3x -10


 * __Summary Questions__**
 * 1. What do you notice about your solutions in part a and part b. In your explanation include what you got for solution to parts a and b to support your explanation.**
 * I noticed that my solution to part b was equal to my remainder of my solution in part a. The remainder in part a for example ` was -43, the same as f(-3). In example 2, the solutions were 5, and in example 3 the solutions were -84. **


 * 2. How can what you found be used as a short cut method to see if a number is a zero of a polynomial function or if a binomial is a factor before starting the synthetic division process? Explain.**
 * My results can be used to tell if a number is a zero of a polynomial function or if a binomial is a factor of the function. Simply plug a number into a function; if it equals zero, then that number is a zero of the function. If you enter a zero of the binomial divisor, and you are left with no remainder of the function, then that binomial is a factor of that function. **

**Remainder Theorem: If a polynomial is divided by a binomial x-k, then plugging in the k-value would yield the remainder. For example, in example 2, the k-value was 2. Plugging in 2 to the function, we were left with 5, which is the remainder of the solution.** **Factor Theorem: Binomial x-k is a factor of a polynomial only if the k-value plugged into that polynomial is 0. For example, in polynomial x^2-4x+4, binomial x-2 would be a factor of the polynomial because plugging in 2 to the polynomial would yield 0.**
 * 3. Look up the definition of the remainder theorem and factor theorem on page 215 and 216 of your text. Explain what these theorems mean in your own words using the examples above. Are there any restrictions to using the remainder theorem? Explain.**
 * Restrictions to using the remainder theorem are that you may only use it if the polynomial is to be divided by a binomial. **


 * 4. Explain when polynomial division is the appropriate method to use when dividing two polynomials. Explain when synthetic division is the most appropriate method to be used. Can you divide f(x) = 4x^3 - 8x^2 + 2x - 1 by g(x) = 2x + 1 using synthetic division? If you can explain what you would use as your k value.**
 * Polynomial division is appropriate to use when the divisor is not a binomial. Synthetic division can only be used when the divisor is a binomial. You can divide the above function by 2x+1 using synthetic division, because the divisor is a binomial. The k value of the divisor would be -(1/2), because that would be the zero of the divisor. **


 * __Unit 3 Lesson 9__**

1. The Fundamental Theorem of Algebra States: A polynomial function of a degree //n// has //n// zeros(real and non real). Some of these zeros may be repeated. Every polynomial of odd degree has at least one zero.


 * Explain what this statement means in your own words. In your description you should include an algebraic or graphical example to support your statement. You should also include the vocabulary of complex zeros, real zeros, and repeated zeros.**
 * This statement means that every polynomial function with a degree has a certain amount of zeroes that can be both real zeros and complex zeros. These may be repeated zeros. All polynomials with an odd degree have at least one real zero. For example, the polynomial x^3 + x^2 + 7 must have at least one real zero, since it has an odd degree, and the other two zeros may either be real of complex. **

2. Is it possible to find a polynomial with a degree of 3 with real number coefficients that has -2 as its only real zero? Explain. **Yes, it is possible; a polynomial with an odd degree must have at least one real zero, as this case does. The other two zeros may by either complex or real.**

3. The complex conjugate theorem states: Suppose that f(x) is a polynomial function with real coefficients. If a and b are real number with b not equal to zero and a + bi is a zero of f(x) then its complex conjugate a - bi is also a zero of f(x).


 * Explain what this statement means in your own words. You should include examples of complex conjugates when making your statement,.**
 * The complex conjugate theorem states that if f(x) is a polynomial and it has real coefficients, and if a and b are real, with b =/= 0, and a +bi is a zero, then the complex conjugate of a + bi, a - bi, is also a zero. For example, if 2 + i is a zero of f(x), and a and b are real, with b not equaling zero, then the complex conjugate, 2 - i, is also a zero of f(x). **

4. Is it possible to find a polynomial function of a degree of 4 with real coefficients that has zeros 1+3i and 1-i. Explain.
 * Yes, it is possible; if the function has real coefficients and a degree of 4, it means that it has 4 zeros, and the complex conjugates of the complex zeros are also zeros. In this case, we are given two complex zeros; this means the complex conjugates of these zeros, which are 1 - 3i, and 1 + i, are the other two zeros of the function; as such, it is possible to find a polynomial function with degree of 4 with 1 + 3i and 1 - i as zeros. **

5. Is it possible to find a polynomial function of a degree of 4 with real coefficients that has zeros -3, 1 + 2i, and 1 - i. Explain.
 * No, it is not possible; a polynomial with a degree of 4 has only 4 zeros, and since it has real coefficients, the complex conjugates of the complex zeros must also be zeros. In this case, we have two complex zeros, but also a real zero, -3. As such, if we were to add the complex conjugates as zeros, we would have 5 zeros, exceeding the limit of 4. Due to this, we can not find a polynomial function with a degree of 4 that has real coefficients and has zeros of -3, 1 + 2i, and 1 - i. **