The Factor Theorem

For a polynomial f(x) and a constant c,

a. If f(c) = 0, then x – c is a factor of f(x).

b. If x – c is a factor of f(x), then f(c) = 0.

The Factor Theorem tells us that if we find a value of c such that f(c) = 0, then x – c is a factor of f(x). And, if x – c is a factor of f(x), then f(c) = 0.

Fundamental Theorem of Algebra

Every polynomial of degree 1 or more has at least one complex zero.

Number of Zeros Theorem

A polynomial of degree n has at most n distinct zeros.

Conjugate Zeros Theorem

If f(x) is a polynomial having only real coefficients and if a + bi is a zero of f(x), then a – bi is also a zero of f(x).

Example. Use the Factor Theorem to decide whether x + 1 is a factor of

f(x) = 2x + x + 2.

Solution. According to the Factor Theorem, x + 1, which we can rewrite as (x – (-1)), is a factor of f(x) if f(-1) = 0. We can find out if f(-1) = 0 by using the Remainder Theorem and synthetic division:

The remainder is -1. So, f(-1) = -1. Since f(-1) is not equal to 0, x + 1 is not a factor of f(x).

Example. For each polynomial, one zero is given. Find the other zeros:

a. f(x) = ; 1 is a zero

b. f(x) = ; i is a zero

Solution.

a. Since 1 is a zero of f(x), we will begin by dividing f(x) by x – 1:

The quotient is x + 5x + 5. Using the Division Algorithm, we factor f(x) as:

f(x) = (x – 1)(x + 5x + 5)

We can find the remaining zeros by setting the quotient equal to 0 and solving the equation:

x + 5x + 5 = 0
x =	Use the quadratic formula: x = with a = 1, b = 5, and c = 5
x =	Simplify
x =

The zeros of f(x) are 1, , and

b. We will begin by dividing f(x) by the given zero, i.

The Complex Conjugate Theorem tells us that since i is a zero of f(x), then its conjugate, -i, is also a zero of f(x). Therefore, we will divide the quotient obtained above by –i:

The quotient is x + 10x + 26. We can find the remaining zeros by setting this quotient equal to 0 and solving:

x + 10x + 26 = 0
x =	Use the quadratic formula with a = 1, b = 10, and c = 26.
x = x = x = x = -5 ± i	Simplify

The four zeros of f(x) are i, -i, -5+i, and –5-i.

Example. Find a polynomial f(x) of degree 3 with only real coefficients that has zeros 2, -3, and 5; such that f(3) = 6.

Solution. From the Factor Theorem, we know that if 2, -3, and 5 are zeros of f(x), then

x – 2, x + 3, and x – 5 are factors of f(x). So, we can write f(x) in the following way:

f(x) = a(x – 2)(x + 3)(x – 5) where a is a constant.

To find a, we use the fact that f(3) = 6, and substitute 3 for x in f(x):

f(x) = a(x – 2)(x + 3)(x – 5)
f(3) = a(3 – 2)(3 + 3)(3 – 5)	Replace x with 3
6 = a(1)(6)(-2) 6 = -12a = a	Replace f(3) with 6; simplify

So, f(x) = (x – 2)(x + 3)(x – 5).

Example. Factor f(x) = into linear factors, given that -5 is a zero of f(x).

Solution. Let f(x) be the dividend, and use synthetic division with divider -5:

The quotient, q(x) is and the remainder, r(x), is 0. Using the division algorithm, we get that f(x) = (x + 5)( ). We can factor the quotient into two linear factors to obtain: f(x) = (x + 5)(6x – 1)(x – 2)

Example. For the polynomial f(x) = , find all zeros and their multiplicities.

Solution. According to the Factor Theorem, since x + 1 is a factor of f(x), then -1 is a zero of f(x). Since the factor x + 1 is raised to the 2nd power, it is a factor two times and we say that -1 is a zero of multiplicity 2. Similarly, since x – 1 is a factor, 1 is a zero of f(x). Since x – 1 is a factor 3 times, 1 is a zero of multiplicity 3. Since the remaining factor of f(x), x - 10, is quadratic, we will set it equal to zero to find the remaining zeros:

x - 10 = 0
x = 10	Add 10 to both sides
x = ±	Use the Square Root Property

and - are both zeros of multiplicity 1.

Example. Find a polynomial of lowest degree with only real coefficients having zeros -1 and 6 – 3i.

Solution. We know from the Conjugate Zeros Theorem that since the complex number

6 – 3i is a zero of f(x), then its conjugate, 6 + 3i, is also a zero. By the Factor Theorem, f(x) must have three factors: (x – (-1)), (x – (6-3i)), and (x – (6+3i)). We can express f(x) in the following way:

f(x) = (x – (-1))(x – (6-3i))(x – (6+3i))
f(x) = (x + 1)(x – 6 + 3i)(x – 6 - 3i)	Use Distributive Property
f(x) = (x + 1)(x - 6x – 3ix – 6x + 36 + 18i + 3ix – 18i – 9i)	Multiply the 2nd and 3rd factors
f(x) = (x + 1)(x - 12x + 45)	Simplify
f(x) =	Multiply
f(x) =