Fundamental Theorem of Algebra

If f(x) is a polynomial of degree n,

where n≥1, then the equation

f(x) = 0

has at least one complex root.

Date:2.6 Topic: Complex Zeros and the Fundamental Theorem of Algebra

Properties of Polynomial Equations

• If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots:

x4+3x2+2x-1 has possibly 4 roots

• If a+bi is a root of the equation, then a-bi is also a root.

Text Example

State how many complex and real zeros the function has f(x) = x4 5x3 x2 3x + 6

Graph the function on a graphing calculator to find out how many are real zeros.

Degree is 4, therefore there are 4 complex zeros.

Imaginary roots never cross the x-axis.How many x-intercepts are there? 2

Because there are 2 x-intercepts there are 2 real zeros.

Linear Factorization Theorem

If f (x) anxn an-1x

n-1 … a1x a0

and n≥1, and an≠ 0 , then

f(x) =an (x - c1 )(x - c2 )…(x - cn )

where c1, c2 , … , cn are complex numbers (possibly real roots with multiplicity greater than 1)

f(x)= (x + 2)2 (x - 4)2

f(x)= (x - (-2))2 (x - 4)2 Use the given zeros and multiplicity in the equation.

f(x)= (x2 + 4x + 4) (x2 - 8x + 16) Multiply.

f(x)= x4 - 4x3 - 12x2 + 32x + 64 Multiply.

ExampleWrite a polynomial function with real coefficents whose zeros and multiplicity are -2(multiplicity 2) and 4(mulitiplicity 2)

f(x)= an(x + 2) (x - 2) (x - i) (x + i)

f(x)= an(x - (-2)) (x - 2) (x - i) (x - (-i)) Use the given zeros in the

equation. If i is a zero, so is -i

f(x)= an(x2 - 4) (x2 + 1) Multiply.

f(x)= an(x4 - 3x2 - 4) Multiply.

ExampleFind a fourth-degree polynomial function with real coefficents that has -2, 2, and i as zeros and such that f(3) = -150

50an= -150

an= -3

f(x)= -3x4 + 9x2 + 12

an(34 - 3•32 - 4) = -150

f(x)= -3(x4 - 3x2 - 4) substitute an in the equation.

Find an, use that f(3) = -150

Complete Student CheckpointFind a third-degree polynomial function f(x) with real coefficients that has -3 and i as zeros and such that f(1)=8

f (x) an(x - )(x - )(x - ) ( 3) i i

f (x) an(x 3)(x2 +1)

f (1) an(13)(12 +1)

8 an(13)(12 +1)

8 an(4)(2)

8 8an

1an

f (x) 1(x 3)(x2 +1)

f (x) x3 3x2 x 3

Text Example

Solve: x4 6x2 8x + 24 0.

Solution The graph of f (x) x4 6x2 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation.

x-intercept: 2

The zero remainder indicates that 2 is a root of x4 6x2 8x + 24 0.

2 1 0 6 8 24 2 4 4 241 2 2 12 0

Day 2

Solution Now we can rewrite the given equation in factored form.

(x – 2)(x3 2x2 2x 12) 0 This is the result obtained from the synthetic division.

x – 2 0 or x3 2x2 2x 12 Set each factor equal to zero.

x4 6x2 8x + 24 0 This is the given equation.

Text Example cont.

Solve: x4 6x2 8x + 24 0.

2 1 0 6 8 24 2 4 4 241 2 2 12 0

Solution We can use the same approach to look for rational roots of the polynomial equation x3 2x2 2x 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4 6x2 8x + 24 0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity (x - 2)2. Thus, 2 must also be a root of x3 2x2 2x 12 = 0, confirmed by the following synthetic division.

x-intercept: 2

2 1 2 2 12 2 8 12 1 4 6 0

These are the coefficients of x3 2x2 2x 12 = 0.

The zero remainder indicates that 2 is a root of x3 2x2 2x 12 = 0.

Text Example cont.

Solve: x4 6x2 8x + 24 0.

So, x3 x2- 2x 12 = (x – 2)(x2 4x 6)

Solution Now we can solve the original equation as follows.

(x – 2)(x3 2x2 2x 12) 0 This was obtained from the first synthetic division.

x4 6x2 8x + 24 0 This is the given equation.

(x – 2)(x – 2)(x2 4x 6) 0 This was obtained from the second synthetic division.

x – 2 0 or x – 2 0 or x2 4x 6 Set each factor equal to zero.

x 2 x 2 x = -2+ i2 Solve, use the quadratic

formula to solve

x2 4x 6

Text Example cont.

Solve: x4 6x2 8x + 24 0.

The solution set of the original equation is: {2, -2+ i2}

Complete Student Checkpoint

Factor as the product of factors that are irreducible over a. the rational numbers

b. the real numbers

c. the complex imaginary numbers

x4 4x2 5

hint: factor

x2 5 x2 1

hint: factor the binomial that can be factored into real numbers

x 5 x 5 x2 1

hint: factor the binomial that can be factored into imaginary numbers

x 5 x 5 x i x i

x 1 x 1 x2 4x 13 0

Complete Student CheckpointSolve: x4 6x3 22x2 30x 13 0

p/q = +1, +13 are possible zerosPlug them in to find the zeros (use synthetic division): f(1) = 0 1 is a zero

1 -6 22 -30 13 1 1 -5 17 -13 1 -5 17 -13 0

x4 6x3 22x2 30x 13 0

x 1 x3 5x2 17x 13 0

1 -5 17 -13 1 1 -4 13 1 -4 13 0

Use possible zeros and synthetic division to factor again

Use quadratic formula:

x

( 4) ( 4)2 4(1)(13)

2(1)

x

x

4 36

2

x

4 6i

2 2 3i, 2 3i 1,

Complex Zeros and the Fundamental

Theorem of Algebra