Proved by Carl Friedrich Gauss
If f (x) is a polynomial of a degree “n”, where n is greater than 0,
“ f ” has at least one zero in the complex number system.
I. Linear Factorization Theorem:
If f(x) is a polynomial of degree “n” where “n is greater than zero”, “f” has precisely “n” linear factors.
f (x) = an (x – c1)( x – c2)( x – c3) … ( x – cn) where
c1, c2, c3, … cn are complex numbers.
Factors of a Polynomial:
Example 1: f (x) = t4 - 5t3 + 15t2 - 45t + 54
4th degree n = 4, n is greater than zero is true
Step 1: find all the possible zeros of the function:
(factors of 54)/(factors of 1) = ±1, ± 2, ± 3, ±6, ± 9, ± 18, ± 27, ± 54
Step 2: Use synthetic division to find out which ones are factors
f (x) = (x – 2)(x – 3)(x2 + 9)
0 = (x – 2)(x – 3)(x2 + 9)
0 = x – 2
x = 2
0 = x – 3
x = 3
0 = x2 + 9
- 9 = x2
±√(-9) = x
± 3i = x
So to write this function in linear form:
f(x) = (x – 2)(x – 3)(x + 3i)(x – 3i)
So the following zeros of “f” are:
x = {2, 3, 3i, and -3i}
II. Complex Zeros Occur in Conjugate Pairs
In the last example, “± 3i”, the pair is 3i and -3i so we can conclude:
Let f(x) be a polynomial function that has real coefficients.
If a + bi, where b ≠ 0, is a zero of the function,
the conjugate a – bi is also a zero of the function.
Example: Your zeros are {2, 4 + i, and 4 – i}, find the polynomial function:
f(x) = (x – 2)(x – (4 + i))(x – (4 – i))
= (x – 2)((x – 4) – i)((x – 4) + i)
= (x – 2)( (x – 4)2 - (i)2)
= (x – 2)(x2 - 8x + 16 – (-1))
= (x – 2)(x2 - 8x + 17)
= x3 – 8x2 + 17x – 2x2 + 16x – 34
= x3 – 10x2 + 33x – 34
III. If you are given one zero, can you find the rest?
g(x) = 4x3 + 23x2 + 34x – 10 given zero: -3 + i
Recall both conjugates:
– 3 + i and – 3 – i so
(x + 3 – i) (x + 3 + i)
= ((x + 3) – i)((x + 3) + i)
= ((x + 3)2 - (i)2)
= x2 + 6x + 9 – (-1)
= x2 + 6x + 10
So using long division:
Every polynomial of degree n is greater than zero with real coefficients can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros.
A quadratic factor with no real zeros is said to be irreducible over the real numbers.
x2 + 1 = (x + i)(x – i) is irreducible over the real numbers
x2 - 2 = (x + √2)(x - √2) is irreducible over the rational numbers but reducible over the real numbers.
Example: The complex number 4i is a zero of f(x) = x4 + 13x2 - 48.
Find the remaining zeros of f(x), and write it in its linear factorization.
Since 4i is a zero, then -4i is also a zero so
(x-4i)(x+4i) = x2 - (4i)2
= x2 - 16i2
= x2 + 16
therefore a factor of f(x) is x2 + 16
f(x) = x4 + 13x2 - 48
= (x2 + 16)(x2 - 3)
= (x + 4i)(x – 4i)(x + √3)(x - √3)
So the zeros are:
x = {-4i, 4i, -√3, √3}
Homework: Pg. 170/ 57, 65, 69
*Pg. 187/ 1 – 5odd, 9 - 19odd, 33, 37, 41, 45, 53, 55, 61, 65
