Warm-Up Simplify the following terms: 81 40 20 i 3 i 16 i 82
TEST Our Ch. 9 Test will be on 5/29/14
Complex Number Operations
Learning Targets Adding Complex Numbers Multiplying Complex Numbers Rules for Adding and Multiplying Conjugates
Addition of Complex Numbers When adding imaginary numbers we combine like terms Ex: 12 + 5i + 7i 12 + 12i
Multiplication of Complex Numbers When multiplying complex numbers we will distribute the factors throughout Ex: 5 2 4i 10 20i
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: 3i 6 + 7i = 18i + 21i 2 = 21 + 18i ***i 2 = 1
You Try a. 14i + 10 2i b. i 5i + 3 17i 3 12i + 10 21i c. 6 5i + 6 + 5i 12 d. 4x 16ix + 51 + 16ix 4x + 51
You Try a. 2i(14i + 10) 28 20i b. 3 + 2i c. 6 5i 2 d. 3 + 7i 9 14i 3 7i 55 24i 11 60i 58
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
Sum of Complex Conjugates The sum of our conjugates will always result in twice the value of our real terms a + bi + a bi 2a
Multiplication of Complex Conjugates Multiplying the conjugates will always result in the sum of our a terms squared and b terms squared a + bi a bi a 2 abi + abi b 2 i 2 a 2 b 2 1 a 2 + b 2
Why is the Conjugate Important The conjugate is important because our non real roots of polynomials always come in pairs x 2 + 4x + 5 = 0 x = 4 ± 42 4 5 2 x = 4 ± 4 2 x = 2 ± i Our pairs of complex numbers will always be conjugates
Conjugate cont. So if we multiply our roots we should get our polynomial in standard form x 2 + 2i x 2 2i x 2 + 4x + 5
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
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
Types of Division There are two methods that we can use to divide polynomials Long Division Synthetic Division (preferred method)
Long Division 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. 21 Now Subtract multiply (which by x + 11 the changes Bring divisor down the sign the x 32 698 Multiply and put x - 3 x 2 + 8x - 5 the of next each answer put number below term below. in or 64 the subtract polynomial) term x 2 3x First divide 58 3 into 6 or x into x 2 32 11x - 5 Now divide 26 3 into 5 This or is the x into 11x 11x - 33 remainder 28 28 3 Remainder added here over divisor So we found the answer to the problem x 2 + 8x 5 x 3 or the problem written another way: 2 x x 8x 5 3
Let's try a problem where we factor the polynomial completely given one of its factors. 4x 3 8x 2 25x 50 factor - 2 4 8-25 -50 You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. Multiply Bring - 8 first Add number 0 these Add down up 50 these below Add up these line up Multiply these these and and No remainder so x + 2 put answer 4 x 0-25 0 put answer + x IS a factor because it above line above line divided in evenly in So next the List answer all Put coefficients variables is the divisor back (numbers times in (one in the x front was quotient: of divided x's) and out the in in next column constant You could process along check so the first this top. number by If a term is one is missing, less power put than in a 0. column original problem). multiplying them out and getting x 2 4x 25 original polynomial : x 2
Comparison Between Synthetic and Long Division Why Synthetic Division Works
Example: Is the factor x 2 a root of: x 3 + 5x 2 2x 24
You try: Is the factor x + 5 a root of: x 3 x 2 37x 35
You try: Is the factor x + 4 a root of: 2x 3 + 3x 2 7x + 14
You try: Is the factor x + 6 a root of: 2x 3 + 6x 2 116x 480
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