Download Ex1: Find all the zeros of f(x) = x4 - 3x3 + x

September 25, 2014
Section 2.5 The Fundamental Theorem of Algebra
Objective: Know how to find zeros of a polynomial function.
The Fundamental Theorem of Algebra
If f(x) is a polynomial function of degree n, where n > 0,
then f has at least one zero in the complex number
system.
Existence Theorems - Don't tell you how
to find zeros, just that they exist.
The Linear Factorization Theorem
If f(x) is a polynomial function of degree n, where n > 0,
then f has precisely n linear factors
f(x) = a (x - c ) (x - c )...(x - c ) where c , c ,..., c are
complex zeros.
n
1
2
n
1
2
n
Note: Zeros can repeat.
Ex1: Find all the zeros of f(x) = x4 - 3x3 + x - 3
How many zeros?
How do you find them?
Start by using Descartes's Rule of Signs to find # of pos.
and neg. real zeros.
3 sign variations -> 3 or 1 pos. real zeros
f(-x), 1 sign variation -> 1 neg. real zero
Use Rational Zero Test to find possible zeros and Synthetic
Division to verify.
-1 1 -3 0 1 -3
p/q: + 1, + 3
-1 4 -4 3
1 -4 4 -3 0
September 25, 2014
Next, we look for the positive zero, since no other negative numbers can
be zeros. Testing 1 does not work, so 3 must be a zero.
3
1 -4 4 -3
3 -3 3
1 -1 1 0
Now we solve 1x - 1x + 1 = 0 using the Quadratic Formula.
2
Therefore, all four of the zeros of f are -1, 3, and
.
Remember...
Descartes’ Rule of Signs
For f(x) = a x + a x
n
n
n-1
n-1
+...+ a x + a be a polynomial with real coefficients
1
0
and a =0,
• The # of positive real zeros of f is either equal to the # of sign
variations in f (x) or less than that# by an even integer.
• The # of negative real zeros of f is either equal to the # of sign
variations in f (-x) or less than that number by an even integer.
0
Back
September 25, 2014
Note: In Example 1, the two complex zeros were conjugates.
!! Complex zeros occur in conjugate pairs !!
If f is a polynomial function with real coefficients, then
whenever a+bi is a zero of f, a-bi is also a zero of f.
Ex2: Find a fourth degree polynomial function with real
coefficients, that has 0, 1, and i as zeros.
Linear Factorization Theorem
Any nth degree polynomial can be written as the product of n
linear factors.
f(x)=a(x-c1)(x-c2)(x-c3)...(x-cn)
If you don't want complex
factors use...
Every polynomial of degree n > 0 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.
Note: A quadratic factor with no real zeros is said to be
irreducible or prime over the reals. (This is different from "being
irreducible over the rationals.")
Since i is a zero
f(x)=x(x-1)(x-i)(
September 25, 2014
Ex3: Factor f(x) = x - 12x - 13
4
2
a) as the product of factors that are irreducible over the
rationals.
Ans: f(x)=(x - 13)(x + 1)
b) as the product of factors that are irreducible over the
reals.
2
2
Ans: f(x)=(x-√13)(x+√13)(x +1)
2
c) completely.
Rational
Real
Irrational
Complex
Ans: f(x)=(x-√13)(x+√13)(x+i)(x-i)
Ex4: Find all zeros of f(x) = x4 - 4x3 + 12x2 + 4x - 13,
given that 2 + 3i is a zero.
Since 2 + 3i is a zero, 2 - 3i is also a zero.
That means (x-(2 + 3i))(x-(2 - 3i)) = x2 - 4x + 13,
so we can find the other factors by dividing.
All the zeros of f are
-1, 1, 2 + 3i, 2 - 3i.