Introduction to The Fundamental Theorem of Algebra
By the Fundamental Theorem of Algebra, every polynomial of degree n has exactly n zeros (some might be counted more than once). Because x = c is a zero implies x − c is a factor, every polynomial can be completely factored in the form a ( x − c _{n} )( x − c _{n} − 1)...( x − c _{1} ), where a is a real number and c _{i} is real or complex. Factors in the form x − c are called linear factors . Factors such as 2 x + 1 can be written in the form x − c by factoring .
To completely factor a polynomial, we usually need to first find its zeros. At times, we will use the Rational Zero Theorem, polynomial division, and the quadratic formula.
Examples
Find all zeros, real and complex.

h(x) = x ^{4} − 16
x ^{4} − 16 = ( x ^{2} − 4)( x ^{2} + 4) = ( x − 2)( x + 2)( x ^{2} + 4)
The real zeros are 2 and −2. We will find the complex zeros by solving x ^{2} + 4 = 0.
x ^{2} + 4 = 0
x ^{2} = −4
The complex zeros are ±2 i.

P(x) = x ^{4} + 6 x ^{3} + 9 x ^{2} − 6 x − 10
The possible rational zeros are ±1, ±2, ±5, and ±10. P (1) = 0.
P(x) = ( x − 1)( x ^{3} + 7 x ^{2} + 16 x + 10)
Because x ^{3} + 7 x ^{2} + 16 x + 10 has no sign changes, there are no positive zeros; x = −1 is a zero for x ^{3} + 7 x ^{2} + 16 x + 10.
P(x) = ( x − 1)( x + 1)( x ^{2} + 6 x + 10)
Solve x ^{2} + 6 x + 10 = 0 to find the complex zeros.
The zeros are ±1, −3 ±i.
Zeros in Complex Numbers and Complex Conjugates of Polynomial Functions
If we know a complex number is a zero for a polynomial, we automatically know another zero—the complex conjugate is also a zero. This gives us a quadratic factor for the polynomial. Once we have this computed, we can use long division to find the quotient, which will be another factor of the polynomial. Each time we factor a polynomial, we are closer to finding its zeros.
Examples
Find all zeros, real and complex.
 f(x) = 3 x ^{4} + x ^{3} + 17 x ^{2} + 4 x + 20 and x = 2i is a zero.

Because x = 2 i is a zero, its conjugate, −2 i , is another zero. This tells us that two factors are x − 2 i and x + 2 i .
( x − 2 i )( x + 2 i ) = x ^{2} + 2ix − 2ix − 4 i ^{2} = x ^{2} − 4(−1) = x ^{2} + 4
We will divide f(x) by x ^{2} + 4 = x ^{2} + 0 x + 4.
f(x) = ( x ^{2} + 4)(3 x ^{2} + x + 5)
Solving 3 x ^{2} + x + 5 = 0, we get the solutions
The zeros are ±2 i , .
 h(x) = 2 x ^{3} − 7 x ^{2} + 170 x − 246, x = 1 + 9 i is a zero.

Because x = 1 + 9 i is a zero, we know that x = 1 − 9 i is also a zero. We also know that x − (1 + 9 i ) = x − 1 − 9 i and x − (1 − 9 i ) = x − 1 + 9 i are factors. We will multiply these two factors.
( x −1 −9 i )( x − 1 + 9 i ) = x ^{2} − x + 9 ix − x + 1 − 9 i − 9 ix + 9 i − 81 i ^{2}
= x ^{2} − 2 x + 1 − 81(−1) = x ^{2} − 2 x + 82
h(x) = (2 x − 3)( x ^{2} − 2 x + 82)
The zeros are 1 ±9 i and (from 2 x − 3 = 0).

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