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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.

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  • blobby green style avatar for user jeremy radcliff
    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.
    (40 votes)
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    • purple pi pink style avatar for user Sachin
      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.
      (48 votes)
  • leafers ultimate style avatar for user pieboy32
    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?
    (21 votes)
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    • leaf blue style avatar for user ChrisF915
      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.
      (34 votes)
  • blobby green style avatar for user Mike W
    Hi Sal,
    If the polynomial can have complex coefficients then would the complex roots not necessarily come in conjugate pairs?
    Thanks!
    (18 votes)
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  • duskpin ultimate style avatar for user Natalie
    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!
    (7 votes)
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  • duskpin ultimate style avatar for user Rowan Belt
    What about y=0? This has infinite solutions, and it is a zero-degree polynomial.
    (8 votes)
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  • starky ultimate style avatar for user Rafal
    How do I know whenever (x^4)+1 has 4 real solutions or it contains some imaginary one?
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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)
      So your four roots are:
      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.
      (10 votes)
  • leaf orange style avatar for user Marquez
    What exactly is a non-real complex root?
    (4 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (4 votes)
  • blobby green style avatar for user mirandakerr1382
    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?
    (3 votes)
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    • starky ultimate style avatar for user KLaudano
      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)
      (5 votes)
  • hopper happy style avatar for user LetsLearnSomeThings
    Wait, what about x^2?
    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?
    (3 votes)
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    • piceratops ultimate style avatar for user Magistra H
      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.
      (4 votes)
  • piceratops seed style avatar for user andrewmako21
    Doesn't P equal the factors of the constant a.k.a the last number which is the whole number in the polynomial? Also, does Q equal the first number like if you have x to the fourth power, Q, in this case is 1?
    (4 votes)
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Video transcript

Voiceover:The fundamental theorem of algebra. Fundamental, I'll write it out, theorem of algebra tells us that if we have an nth-degree polynomial, so let's write it out. So let's say I have the function p of x and it's defined by an nth-degree polynomial. So let's say it's a x to the n plus b x to the n minus one and you just go all the way to some constant term at the end. So this is an nth-degree polynomial. The fundamental theorem of algebra tells us that this nth-degree polynomial is going to have n exactly n roots, or another way to think about it, there are going to be exactly n values for x, which will make this polynomial, make this expression on the right, be equal to zero. So at first you might say, "Okay, that makes sense." You've seen second-degree polynomials, whose graphs might look something like this. So, that's the y axis and that's the x axis. We know a second-degree polynomial would define a parabola, so it might look something like this and you could buy that. Okay, this is a second-degree, that's second degree, and you see that this function equals zero at exactly two places. It has exactly two roots. It has two roots, so that seems consistent with the fundamental theorem of algebra, and you could also imagine a third-degree polynomial looking like this. So that's my y axis. This is my x axis. You could imagine a third-degree polynomial looking something like this. Bam, bam, bam, and it just keeps going. And here you see it's a third-degree polynomial, and you'll see it has one, two, three roots. And I could have a fourth-degree polynomial. Maybe it looks something like this, where it goes something like this, and you say, "Okay, that makes sense." It will have one, two, three, four roots. But then you might start to remember things that don't always behave in this way. For example, many, many, many times we've seen parabolas, we've seen second-degree polynomials that look more like this, where they don't seem to intersect the x axis. So this seems to conflict with the fundamental theorem of algebra. The fundamental theorem of algebra says if we have a second-degree polynomial then we should have exactly two roots. Now, this is the key. The fundamental theorem of algebra, it extends our number system. We're not just talking about real roots, we're talking about complex roots, and in particular, the fundamental theorem of algebra allows even these coefficients to be complex. And so when we're looking at these first examples, these were all real roots, and real numbers are a subset of complex numbers. So here you had two real roots. Here you had three real roots. In this orange function, you had four real roots. In this yellow function, this yellow parabola right over here, the second-degree polynomial, we have no real roots. That's why you don't see it intersecting the x axis, but we will have two complex roots. So this one right over here will have two complex roots. And the complex roots, the non-real complex roots, because really real numbers are a subset of complex numbers, these always come in pairs, and we'll see that in future videos. So for example, if you have a third-degree polynomial, It might look something like this. A third-degree polynomial might look something like this, where it has one real root, but then the fundamental theorem of algebra tell us it necessarily has two other roots because it is a third degree, so we know that the other two roots must be non-real, complex roots. Now, could you have a situation where you have a third-degree polynomial with three complex roots? So, can you have three non-real complex roots? Is this possible for a third-degree polynomial? Well, the answer is no because complex roots, as we'll see in the next few videos, always come in pairs. They're coming in pairs where they are conjugates of each other. So, you could have a fourth-degree polynomial that has no real roots, for example. Something might look something like this. In this case, you would have two pairs of complex roots or you would have four non-real complex roots and you could group them into two pairs where in each pair you have conjugates, and we'll see that in the next video.