Unit+3

Unit 2 Lesson 5
 ** Example 1: ** a.) Divide f(x) by g(x) using synthetic or polynomial division.  ** 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

 ** Example 2: ** a.) Divide f(x) by g(x) using synthetic or polynomial division. b.) Find f(2) when f(x) = 2x^3 - 3x^2 + 4x - 7
 * f(x) = 2x^3 - 3x^2 + 4x -7 g(x) = x - 2 **

 ** Example 3: ** a.) Divide f(x) by g(x) using synthetic or polynomial division. b.) Find f(-2) when f(x) = 4x^3 - 9x^2 + 3x -10
 * f(x) = 4x^3 - 9x^2 + 3x -10 g(x) = x + 2 **

 ** 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. ** For every example, I noticed that the remainder in part a and solution to part b is the same. The remainder was the same as f(k), the h from x – k. For example 1, it was -43, for example 2 it was 5, and for example 3 it was -84.
 * __ Summary Questions __ **

 ** 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. **  From what I discovered, I can see if a number is a 0 of a polynomial function or if a binomial is a factor before starting the synthetic division process by finding f(k), from x – h. If f(k) = 0, then the remainder is 0 and h is also a 0 of the function. The binomial would be a factor. If it isn’t 0, no.

 ** 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. **  For the remainder theorem, it states that when polynomial function, such as those f(x)’s above, are divided by a form of x-h (the g(x)’s above), then the remainder would be equal to the function of the k term in x – k. As for the factor theorem, it is only usable when the function of k is equal to 0, making (x – k) a factor. To use the remainder theorem, the remainder can only equal f(k) if the polynomial function is divided by x – k/

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> ** 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. ** <span style="font-family: Arial,sans-serif; font-size: 10pt;">Polynomial division can be used for dividing any 2 polynomials, but is best used when the divisor has a leading coefficient with more than the power of 1. Synthetic division can only be used when the divisor has a leading coefficient has a power of 1 max. You can divide ** f(x) = 4x^3 - 8x^2 + 2x - 1 by g(x) = 2x + 1 using synthetic division. As the k value, you would have to use –1/2, which is found by isolating x. **

**__<span style="font-family: Arial,sans-serif;">Unit 2 Lesson 9 __** <span style="font-family: Arial,sans-serif; font-size: 10pt;">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.

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> ** 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. **

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> This statement means that the leading exponent a polynomial function has determines the number of zeros, whether real or unreal, and a repeating zero, like (x+2)^2, would count for 2 zeros and it would match the exponent of the expanded polynomial function x^2 + 4x + 4. There are complex zeros as well from those polynomial functions that don’t have a zero that can be seen on a graph, such as x^2 + 6, but since it has an exponent of 2, it will have 2 complex numbers that can be found using the quadratic formula. As for real zeros, they can be found using the quadratic formula as well, or by factoring after using, if necessary, synthetic division or polynomial division.

<span style="font-family: Arial,sans-serif; font-size: 10pt;">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.

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> Yes, it is. If a polynomial has a degree of three and you only want -2 as its only real zero, then you can do (x+2)^3. This is because then the only k in this would be -2.

<span style="font-family: Arial,sans-serif; font-size: 10pt;">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).

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> ** Explain what this statement means in your own words. You should include examples of complex conjugates when making your statement. **

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> This states that for a polynomial function that has real number coefficients, they might have a complex number. That complex number that is a zero of a function, has a + bi, where a and b which are real numbers but b cannot be zero because there would be no imaginary number, it would have a complex conjugate where the only different would be is that b has an opposite sign. This would also be a zero of f(x). Such as in 3 + 4i, the complex conjugate would be 3 – 4i. <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">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.

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> Yes, it is, because it can have up to four zeros, and the graph doesn’t have to touch the x-axis. There are 4 zeros in this case, with 1 + 3i, it’s complex conjugate 1 – 3i, and 1 – i, and it’s complex conjugate 1 – i.

<span style="font-family: Arial,sans-serif; font-size: 10pt;">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. <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">No, it can’t. Since this only has a degree of 4 and the zeros are -3, 1 + 2i, it’s complex conjugate 1 -2i, and 1 – i and it’s complex conjugate 1 + i, that would mean that there are 5 zeros total, and since it can only have 4 zeros, that can’t be.