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## Precalculus

### Unit 3: Lesson 9

The fundamental theorem of algebra

# The Fundamental theorem of Algebra

Sal introduces the Fundamental Theorem of Algebra, which, as is reflected in its name, is a very important theorem about polynomials. Created by Sal Khan.

## Want to join the conversation?

• What about something like f(x) = x^2? The only real root seems to be the vertex at x = 0 and the Theorem of Algebra states that there should be a second root, but it also states that complex roots come in pairs. I don't understand how both of those constraints can exist at the same time. •  The fundamental theorem of algebra states that you will have n roots for an nth degree polynomial, including multiplicity. So, your roots for f(x) = x^2 are actually 0 (multiplicity 2). The total number of roots is still 2, because you have to count 0 twice.
• What about a second-degree polynomial that is the square of a binomial, such as x^2+2x+1 ? It is a second degree polynomial, yet it has only one root, which is -1. Doesn't this break the fundamental theorem of algebra? •  No it does not. The 2. degree polynomial x² + 2x + 1 is composed of (x + 1) * (x + 1), which can also be written as (x + 1)². Even though both factors have their x-intersection at the same x-value (x = -1), there are still two of them. This is called a double root.

You will get those whenever you multiply a binomial with itself (whenever you square it).

For example: (please check for yourself)
(x + 3)² = (x + 3) * (x + 3) = x² + 6x + 9; has it's double-root at x = -3.
(x - 4)² = (x - 4) * (x - 4) = x² - 8x + 16; has it's double-root at x = 4.
• Hi Sal,
If the polynomial can have complex coefficients then would the complex roots not necessarily come in conjugate pairs?
Thanks! • What about the function f(x) = x^2? This, to me, only has one real root, which is zero. Does it count twice? I remember learning about this at the beginning of the year, but in light of recent knowledge it has quite escaped my mind. Thanks! • What about y=0? This has infinite solutions, and it is a zero-degree polynomial. • How do I know whenever (x^4)+1 has 4 real solutions or it contains some imaginary one? • There is a way to directly calculate whether a fourth degree polynomial with real coefficients has 0, 2, or 4 nonreal roots, but the method is extremely difficult and far more trouble than it is worth for this level of study (it involves using the quartic discriminant). So, instead, you need to try to find one or more of the roots and make determinations from there.

Here are a few helpful notes:

Remember that "complex number" includes both real numbers and nonreal numbers. Some people use "complex number" incorrectly to refer only to nonreal numbers.

A fourth degree polynomial with real coefficients has its real or non-real roots occur in sets of two. Thus, if you know it has one nonreal root, then it must have a total of two or four nonreal roots. Likewise, if you know it has one real root, then it must have a total or two or four real roots.

Remember that, if a + bi is a root, then so is a - bi provided that b is not equal to 0.

So, the approach to take is to try to find a real root. If you find one, then you know you must have at either two or four. Similarly, if you can find a nonreal root, then you know it must have a mate.

However, this particular problem is easy. Just look at the equation for finding the roots:
x⁴ + 1 = 0
Thus, x⁴ = −1
Ask yourself what real number raised to the fourth power is negative? There are not any, so you know all of the roots must be nonreal / imaginary.
Here is how to find its acual roots:
x⁴ + 1 = 0
x⁴ = −1
√x⁴ = √−1
± x² = i
x² = ± i
Split into two equations
x² = i and x² = − i
x = ±√i and x = ±√(−i)
NOTE: √i = −i√i NOT i√i (there are different rules involved when i is square rooted)
x = √i, x= −√i, x = i√i and x = −i√i
This can be simplified further, but I don't think you've had that level of math yet.
• • Real numbers are technically speaking a subset of the complex numbers (in which the imaginary component is 0). Thus to speak of complex number and to be clear that you are non including real numbers you should say, "nonreal complex number".

Thus a nonreal complex root is a root is complex but is not real.

Unfortunately, there are some who use the term "complex" ambiguously, sometimes meaning it as "complex including reals" and sometimes using it as "complex not including reals". As the term is used in the Fundamental Theorem of Algebra, complex root DOES include real roots.
• Hi! I have four questions.
1. Is this applicable to equations where the coefficient of x is zero?
2. How is FTA proven?
3. Can it be extended to functions like log (x)?
4. If the degree of the equation is odd, does that mean that it has a double root considering the conjugate also being a root? • 1. The coefficient of x can be 0 provided that the degree of the polynomial is greater than 0.

2. There are a number of different proofs for the Fundamental Theorem of Algebra, all of which rely on some math beyond algebra. (The Wikipedia page on the Fundamental Theorem of Algebra provides many proofs if you would like specific examples.)

3. The Fundamental Theorem of Algebra only applies to polynomials.

4. An odd-degree polynomial does not necessarily imply that one or more of its roots have a multiplicity greater than 1. (e.g. (x+1)*(x+2)*(x+3) has a degree of three, but each of its roots has a multiplicity of 1)
I only see one way that x^2 can equal 0. (x=0)
Is there any complex numbers that when squared, equals 0? Or is there actually only one root? • Short answer: 0 is the only square root of 0.
Longer answer, let's set the square of a complex number equal to 0 and see what it can be:
Let a + bi be a complex number (so a and b are real numbers), and let (a + bi)^2 = 0.
Square the complex number: a^2 + 2abi - b^2 = 0
Rewrite in standard complex form: (a^2 - b^2) + (2ab)i = 0 + 0i
Since complex numbers are equal only if their components are equal, 2ab = 0
By the 0 product property (remember a and b have to be real), 2ab = 0 means a = 0 or b = 0.
If b = 0, then (a + bi)^2 = 0 means a^2 = 0, so a = 0 and a + bi = 0.
If a = 0, then (a + bi)^2 = 0 means (bi)^2 = 0, which means -b^2 = 0. So b = 0 , and again a + bi = 0. 