Warm-Up

1. Warm-Up • Simplify the following terms:

2. TEST • Our Ch. 9 Test will be on 5/29/14

3. Complex Number Operations

4. Learning Targets • Adding Complex Numbers • Multiplying Complex Numbers • Rules for Adding and Multiplying Conjugates

5. Addition of Complex Numbers • When adding imaginary numbers we combine like terms Ex:

6. Multiplication of Complex Numbers • When multiplying complex numbers we will distribute the factors throughout Ex:

7. Multiplying Notes • Be careful to notice that when multiplying we will often end with an imaginary term to the second power. • These terms will always simplify to their opposite value. • Ex: ***

8. You Try

9. You Try

10. Conjugate Operations • Complex Conjugate operations are needed in order to factor quadratics and determine their complex roots. • There are two main operations that we need to know about

11. Sum of Complex Conjugates • The sum of our conjugates will always result in twice the value of our real terms

12. Multiplication of Complex Conjugates • Multiplying the conjugates will always result in the sum of our a terms squared and b terms squared

13. Why is the Conjugate Important • The conjugate is important because our non real roots of polynomials always come in pairs Our pairs of complex numbers will always be conjugates

14. Conjugate cont. • So if we multiply our roots we should get our polynomial in standard form

15. Now we can begin to divide polynomials • In order to divide polynomials we have to be able to determine one of its factors • Once a factor is known we can begin to divide it throughout the standard form of the polynomial and simplify it • If the factor used is indeed a root our remainder will be zero

16. Division Cont. • We can then repeat the process until we are only left with all of the roots of the polynomial • This process allows us to transform a polynomial from Standard Form to Factored Form

17. Types of Division • There are two methods that we can use to divide polynomials • Long Division • Synthetic Division (preferred method)

18. Long Division Now multiply by the divisor and put the answer below. Bring down the next number or term 32 698 First divide 3 into 6 or x into x2 Now divide 3 into 5 or x into 11x So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way: If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps. Subtract (which changes the sign of each term in the polynomial) x + 11 2 1 Multiply and put below x - 3 x2 + 8x - 5 Remainder added here over divisor 64 x2 – 3x subtract 5 8 11x - 5 32 11x - 33 This is the remainder 26 28

19. Bring first number down below line Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Let's try a problem where we factor the polynomial completely given one of its factors. You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. - 2 4 8 -25 -50 - 8 Add these up 0 50 Add these up Add these up No remainder so x + 2 IS a factor because it divided in evenly 4 x2 + x 0 - 25 0 Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is the divisor times the quotient: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. You could check this by multiplying them out and getting original polynomial

20. Comparison Between Synthetic and Long Division • Why Synthetic Division Works

21. Example: • Is the factor a root of:

22. You try: • Is the factor a root of:

23. You try: • Is the factor a root of:

24. You try: • Is the factor a root of:

25. For Tonight • Worksheet