6.7 – Using the Fundamental Theorem

of Algebra

Objectives:• Use the fundamental theorem of algebra

to determine the number of zeros of a polynomial function.

• Write a polynomial function given the zeros of the function.

Review:

• Find all the zeros:

1. f(x) = x3 + x2 – 2x – 2

Answer: , , -12

A polynomial to the nth degree will have n zeros.

f(x) = x3 – 6x2 – 15 x + 100 = (x + 4)(x – 5)(x – 5) the zeros are: -4, 5, 5

5 is a repeated solution

2

Find zeros of functions/solutions to polynomial equations

1. Factor the polynomial2. Rational Zero Theorem (p/q), then factor using synthetic division3. Set unsolved factors to zero and solve

f(x) = x4 + 4x3 – 6x2 – 36x + 27

f(x) = x4 – 5x2 – 36

Review:

1. Use the fundamental theorem to determine the number of roots

The Fundamental Theorem of Algebra

If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one root in the set of complex numbers.

That means if there is a variable, there is at least one solution in the complex numbers, a + bi.

1. Use the fundamental theorem to determine the number of roots

We know:· the degree of the equation tells you the

number of solutions· imaginary solutions come in pairs

A new thing to know:• if an imaginary number is a zero, then its

conjugate is also a zero

so, if 4 + 3i is solution, then 4 – 3i is also a solution

Using a Graphing Calculator to Find the Real Zeros

• Under “Y =” type in the equation.• Go to [2nd]; “CALC”; “2: zero”• Left bound: you need to place the cursor to

the left of the intersection and press enter.• Right bound: you need to place the cursor

to the right of the intersection and press enter; and enter again.

Type in the equation y = x3 + 3x2 + 16x + 48.

Example: find all zeros of x3 + 3x2 + 16x + 48 = 0

• (should be 3 solutions total)• CT = ±1 ±2 ±3 ±4 ±6 ±8 ±12 ±16 ±24 ±48• LC ±1• Graph and you’ll see -3 is the only one on the graph you can see so

synthetic divide with -3 • 1 3 16 48• -3 -3 0 -48

1 0 16 0 x2 + 16 = 0

x2 = -16x = ±√-16 = ±4i

The three zeros are -3, 4i, -4i

Now write a polynomial function of least degree that has real coefficients, a leading coefficient

of 1 and 1, -2 + i, -2 – i as zeros.• f(x) = (x – 1)(x – (-2 + i))(x – (-2 – i))• f(x) = (x – 1)(x + 2 – i)(x + 2 + i)• f(x) = (x – 1){(x + 2) – i} {(x + 2) +i}• f(x) = (x – 1){(x + 2)2 – i2} FOIL• f(x) =(x – 1)(x2 + 4x + 4 –(-1)) Take care of

i2

• f(x) = (x – 1)(x2 + 4x + 4 + 1)• f(x) = (x – 1)(x2 + 4x + 5) Multiply• f(x) = x3 + 4x2 + 5x – x2 – 4x – 5• f(x) = x3 + 3x2 + x – 5

Writing Polynomial Functions:

Now write a polynomial function of least degree that has real coefficients, a leading coefficient of 1 and 4, 4, 2 + i as zeros.

• Note: 2+i means 2 – i is also a zero• f(x) = (x – 4)(x – 4)(x – (2 + i))(x – (2 – i))• f(x) = (x – 4)(x – 4)(x – 2 – i)(x – 2 + i)• f(x) = (x2 – 8x + 16)((x – 2) – i)((x – 2) + i)• f(x) = (x2 – 8x +16)((x – 2)2 – i2)• f(x) = (x2 – 8x +16)(x2 – 4x + 4 –(-1))• f(x) = (x2 – 8x +16)(x2 – 4x + 5)• f(x) = x4 – 4x3 + 5x2 – 8x3 + 32x2 – 40x + 16x2 – 64x + 80• f(x) = x4 – 12x3 + 53x2 – 104x + 80

Writing Polynomial Functions:

Assignment:• pgs. 369-370 #15, 16, 21, 38, 39, 41, 42, 47