The Fundamental Theorem of Algebra can be used in order to determine how many real roots a given polynomial has. Check it out! Created by Sal Khan.
Want to join the conversation?
- Can't the number of real roots of a polynomial p(x) that has degree 8 be 7? I mean, there are equal roots, and there can be exist 7 real roots with an equal real roots and 6 distinct real roots, or 5 with 2 equal real roots and 2 distinct real roots and 2 complex roots, or anything else can be, isn't it?(15 votes)
- In terms of the fundamental theorem, equal (repeating) roots are counted individually, even when you graph them they appear to be a single root. You have to consider the factors:
x^3(ax^2 + bx + c)
If x^3 = 0, this is the same thing as x * x * x = 0, or x = 0, x = 0, x = 0. Even though they are equal and repeating, they are still 3 factors of x in the original polynomial, and can be counted 3 times.
I hope this makes sense.(25 votes)
- Why can't you have an odd number of non-real or complex solutions?(9 votes)
- It's demonstrated in the previous video that you get them in second degree polynomials by solving quadratic equations with negative discriminant (the part under the square root in the quadratic formula) and taking the "plus or minus" of the resulting imaginary number. So the quadratic formula (which itself arises from completing the square) sets up the situation where imaginary roots come in conjugate pairs. For higher degree polynomials, I guess you just can factor them into something that I've described and something that obviously has a real root. Any odd-degree polynomial must have a real root because it goes on forever in both directions and inevitably crosses the X-axis at some point. Hope it makes sense! : )(13 votes)
- OK. Why doesn't this work with quadratic functions. I remember that quadratic functions could have one real root which would mean they would have one real root and one non real root. Which is clearly not possible since non real roots come in pairs.(2 votes)
- From the quadratic formula, x = -b/2a +/-(sqrt(bb-4ac))/2a
If 1 root is non-real, then the discriminant is negative, and both roots have an imaginary component; in one root it's added to -b/2a, in the other subtracted. So there must be 2 non-real roots.
If 1 root is real, then the discriminant is either + or 0. If it's +, then there are 2 real roots; in 1 (sqrt(bb-4ac))/2a is added to, in the other subtracted from, -b/2a.
If the discriminant is 0, then there can be only 1 root, -b/2a, +/-0, which must be subtracted from x in both of the binomial factors of the quadratic; so both factors are identical and we get a perfect square. The vertex form of the equation is (x-r)^2 + 0 = 0. The y coordinate of the vertex is 0.(7 votes)
- Would the fundamental theorem of algebra still work if we have situation like p(x)=gx^5+hx^2+j, where the degrees of the terms are not consecutive?
thanks very much(4 votes)
- Of course. Note that we can't really say "degree of the term" because the degree of a univariate polynomial is just the highest exponent the variable is being raised - so we can only use degree to describe a polynomial, not individual terms. Also note that the Fundamental Theorem of Algebra does not accounts for multiplicity meaning that the roots may not be unique.(5 votes)
- Why do the non-real, complex numbers always come in pairs? I know about complex conjugates and what they are but I'm confused why they have to be both or it's not right.(3 votes)
- To end up with a complex root from a polynomial you would have a factor like (x^2 + 2). To solve this you would end take the square root of a negative and, just as you would with the square root of a positive, you would have to consider both the positive and negative root.
This is not true of real roots because real root do not have to come from a square root. You can get real roots from linear factors.
I hope that explanation made sense. :)(5 votes)
- I heard somewhere that a cubic has to have at least one real root. Why is this true?(3 votes)
- The Fundamental Theorem of Algebra says that a polynomial of degree n has exactly n roots. If those roots are not real, they are complex. But complex roots always come in pairs, one of which is the complex conjugate of the other one
Also, if you graph a cubic, you'll see that it has two "bends" in it, and one side goes off to -∞ (negative if the degree three term is positive) and the other goes off to +∞. You can imagine adding different constant terms, positive or negative to the basic x³ , but no term will ever keep it from crossing the x-axis in at least one place.(3 votes)
- How does y = x^2 have two roots?(3 votes)
- It makes more sense if you write it in factored form. in this case it's xx. Or if you'd rather (x-0)(x-0). Finding roots is looking at the factored form of the polynomial, where it is also factored into its complex/ imaginary parts, and finding how to make each binomial be 0. In a degree two polynomial you will ALWAYS be able to break it into two binomials. So it has two roots, both of which are 0, which means it has one ZERO which is 0.
A similar case is something like (x-1)^2, which is x^2 moved to the right 1 unit. breaking it into its binomials gets (x-1)(x-1) so the two roots are both 1 which means a single zero which is 1
does that make sense?(2 votes)
- Shouldn't complex roots not in pairs be possible? It would just mean that the coefficients are non real(2 votes)
- That's correct. But all the polynomials we work with have real coefficients, so given that, we can only have conjugate pairs of complex roots.(3 votes)
- The roots of the equation x^3-3x-2=0 are(2 votes)
- If you wanted to do this by hand, you would need to use the following method:
1) Use the rational root theorem :
Possible rational roots = (±1±2)/(±1) = ±1 and ±2. (To find the possible rational roots, you have to take all the factors of the coefficient of the 0th degree term and divide them by all the factors of the coefficient of the highest degree term.)
2) Now you have to use synthetic division, using trial and error to determine which of the possible roots is an actual root:
I'll save you the math, -1 is a root and 2 is also a root. Now you have to use synthetic division to factorize the polynomial:
x^3 - 3x - 2=0
Evaluating each factor equal to zero, you get that -1 (multiplicity of 2) and 2 are the solutions to this equation.(2 votes)
- at2:08sal says "conjugate pairs". what does it mean?(1 vote)
- For a nonreal number, you can write it in the form of
a + bi
To form its conjugate, just change the sign of the imaginary portion:
a − bi(4 votes)
Voiceover:So we have a polynomial right over here. We have a function p(x) defined by this polynomial. It's clearly a 7th degree polynomial, and what I want to do is think about, what are the possible number of real roots for this polynomial right over here. So what are the possible number of real roots? For example, could you have 9 real roots? And so I encourage you to pause this video and think about, what are all the possible number of real roots? So I'm assuming you've given a go at it, so the Fundamental Theorem of Algebra tells us that we are definitely going to have 7 roots some of which, could be actually real. So we're definitely not going to have 8 or 9 or 10 real roots, at most we're going to have 7 real roots, so possible number of real roots, so possible - let me write this down - possible number of real roots. Well 7 is a possibility. If you graphed this out, it could potentially intersect the x-axis 7 times. Now, would it be possible to have 6 real roots? Is 6 real roots a possibility? Is this a possibility? Well, let's think about what that would imply about the non-real complex roots. If you have 6 real, actually let's do it this way. Let me write it this way. So real roots and then non-real, complex. The reason I'm not just saying complex is because real numbers are a subset of complex numbers, but this is being clear that you're talking about complex numbers that are not real. So you could have 7 real roots, and then you would have no non-real roots, so this is absolutely possible. Now could you have 6 real roots, in which case that would imply that you have 1 non-real root. Well no, you can't have this because the non-real complex roots come in pairs, conjugate pairs, so you're always going to have an even number of non-real complex roots. So you can't just have 1, so let's rule that out. Now what about having 5 real roots? That means that you would have 2 non-real complex, adding up to 7, and that of course is possible because now you have a pair here. So I think you're starting to see a pattern. Essentially you can have an odd number of real roots up to and including 7. So for example,this is possible and I could just keep going. I could have, let's see, 4 and 3. This is not possible because I have an odd number here. You're going to have to have an even number of non-real complex roots. So rule that out, but then if we go to 3 and 4, this is absolutely possible. You have two pairs of non-real complex roots. And then we can go to 2 and 5, once again this is an odd number, these come in pairs, so this is impossible. And then you could go to 1 real and 6 non-real. Completely possible, this is an even number. And then finally, we could consider having 0 real and 7 non-real complex and that's not possible because these are always going to come in pairs, so you're always going to have an even number here. So the possible number of real roots, you could have 7 real roots, 5 real roots, 3 real roots or 1 real root for this 7th degree polynomial.