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Solving Polynomial Functions involving Complex Numbers
1. Find a polynomial function given zeros of -4, -2 and 4 + i.
2. Find a polynomial function given zeros of;
3. Given the polynomial function below has a zero of 2 i, find the remaining zeros.
4. Given the polynomial function below has a zero of 3 - 2 i, find the remaining zeros.
5. Find the zeros for the following polynomial function. Both real AND complex. Leave in exact value.
6. Find the zeros for the following polynomial function. Both real AND complex. Leave in exact value.
Polynomial Inequalities in One Variable
Notes 6.6 Fundamental Theorem of Algebra
2.4 – Zeros of Polynomial Functions
5.5 Apply the Remainder and Factor Theorem
Section 5.6 – Complex Zeros; Fundamental Theorem of Algebra Complex Numbers Standard form of a complex number is: a + bi. Every complex polynomial function.
Lesson 2.5 The Fundamental Theorem of Algebra. For f(x) where n > 0, there is at least one zero in the complex number system Complex → real and imaginary.
The Fundamental Theorem of Algebra And Zeros of Polynomials
Objectives Fundamental Theorem of Algebra 6-6
5.7 Apply the Fundamental Theorem of Algebra
Complex Zeros; Fundamental Theorem of Algebra
9.9 The Fundamental Theorem of Algebra
The Rational Root Theorem The Rational Root Theorem gives us a tool to predict the Values of Rational Roots:
6.6 The Fundamental Theorem of Algebra
Linear Factorizations Sec. 2.6b. First, remind me of the definition of a linear factorization… f (x) = a(x – z )(x – z )…(x – z ) An equation in the following.
Solving Quadratic Equations using the Quadratic Formula
2.5 The Fundamental Theorem of Algebra Students will use the fundamental theorem of algebra to determine the number of zeros of a polynomial. Students.
1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 3 Polynomial and Rational Functions.
EQ: How can all of the roots of a polynomial (both real & imaginary) be found?
Solving equations with polynomials – part 2. n² -7n -30 = 0 ( )( )n n 1 · 30 2 · 15 3 · 10 5 · n + 3 = 0 n – 10 = n = -3n = 10 =
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