4 5 The Fundamental Theorem of Algebra 1

4. 5 The Fundamental Theorem of Algebra (1 of 2) • Apply the fundamental theorem of algebra, number of zeros theorem, and conjugate zeros theorem. • Factor polynomials having complex zeros • Solve polynomial equations having complex solutions Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Fundamental Theorem of Algebra (2 of 2) A polynomial f(x) of degree n, with n ≥ 1, has at least one complex zero. The fundamental theorem of algebra guarantees that every polynomial has a complete factorization, if we are allowed to use complex numbers. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Number of Zeros Theorem A polynomial of degree n, with n ≥ 1 has

Example: Classifying zeros (1 of 4) All zeros for the given polynomials are distinct. Use the figures to determine graphically the number of real zeros and the number of nonreal complex (imaginary) zeros. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Classifying zeros (2 of 4) The graph of ƒ(x) crosses the x-axis once, so there is one real zero. Since ƒ is degree 3 and all zeros are distinct, there are two nonreal complex zeros. Solution Real zeros correspond to x-intercepts; complex nonreal (imaginary) zeros do not. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Classifying zeros (3 of 4) The graph of g(x) never crosses the x-axis. Since g is degree 2, there are no real zeros and two nonreal complex zeros. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Classifying zeros (4 of 4) The graph of h(x) is shown. Since h is degree 4, there are two real zeros and the remaining two zeros are nonreal complex. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Constructing a polynomial with prescribed zeros Determine a polynomial f(x) of degree 4 with leading coefficient 2 and zeros − 3, 5, i and −i in (a) complete factored form and (b) expanded form. Solution a. Let an = 2, c 1 = − 3, c 2 = 5, c 3 = i, and c 4 = −i. f(x) = 2(x + 3)(x − 5)(x − i)(x + i) Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Conjugate Zeros Theorem If a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a − bi is also a zero of f(x). Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Finding complex zeros of a polynomial (1 of 3) Solution By the conjugate zeros theorem it follows that i must be a zero of f(x). (x + i) and (x − i) are factors (x + i)(x − i) = x 2 + 1, using long division we can find another quadratic factor of f(x). Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Finding complex zeros of a polynomial (2 of 3) Long division Copyright ©

Example: Finding complex zeros of a polynomial (3 of 3) Use the quadratic formula

Polynomial Equations with Complex Solutions Copyright © 2018, 2014, 2010 Pearson Education, Inc. All

Example: Solving a polynomial equation (1 of 3) Solve x³ = 3 x² − 7 x + 21. Solution Rewrite the equation as f(x) = 0, where f(x) = x³ − 3 x² + 7 x − 21. We can use factoring by grouping or graphing to find one real zero of f(x). Let’s use graphing in this example. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved

Example: Solving a polynomial equation (2 of 3) The graph shows a zero at 3. So, x − 3 is a factor. Now use synthetic division by 3. Copyright © 2018, 2014, 2010 Pearson Education, Inc. All Rights Reserved