If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Zeros of polynomials (with factoring): grouping

When a polynomial is given in factored form, we can quickly find its zeros. When its given in expanded form, we can factor it, and then find the zeros! Here is an example of a 3rd degree polynomial we can factor using the method of grouping.

Want to join the conversation?

  • aqualine tree style avatar for user Eunice Huang
    How do we know which do numbers are suppose to be grouped together?
    (13 votes)
    Default Khan Academy avatar avatar for user
  • purple pi teal style avatar for user lord ghirahim
    what is the derivative of x
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf blue style avatar for user Ryan Rubio
    What would the case be for (x^2-4) ?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user loumast17
      You should eventually learn about the difference of squares, which basically says if you have something like (x^2 - a) you could factor it into (x - sqrt(a)(x - sqrt(a)) so this is really useful if a is a perfect square. It is here in this case.

      So (x^2 - 4) = (x-2)(x+2)

      Does that help? If you want to know why the rule works just look at what happens if you FOIL (x-a)(x+a)

      EDIT

      Also works if a number is in front of the x so (ax+b)(ax-b) = (ax)^2 - b^2
      (5 votes)
  • orange juice squid orange style avatar for user rithikrajesh
    Do you yeet grouping out the math window?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • scuttlebug yellow style avatar for user JemboJet
    I only factored to (x+1)(x^2-9) and then did x+1=0, x=-1; and x^2+9=0, x= +/- 3 (i.e. x=-3 or 3). This led to the same answers even though it wasn't completely factored. Is that bad?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user skodali100
    is (x-3)(x+3) the same thing as (x+3)(x-3)?
    (0 votes)
    Default Khan Academy avatar avatar for user
  • starky tree style avatar for user John
    does anybody know why we need to factor x^2-9 completely, cant u just call an x = 3 and 3^2 - 9 = 0? I supposed u need to find the roots but what really are these roots and why does factoring these to something like (x+b) even become a root?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      You generally start with a quadratic equation f(x)=x^2-9. This forms a parabola. Without a middle term, we know that the y intercept is the vertex, so setting x=0 we know that the vertex (and y intercept) is (0,-9). If we want to find the x intercepts (zeros, roots, or solutions) we set y=0, so we end up with x^2-9=0. Since this is a difference of perfect squares, we get (x+3)(x-3)=0. Using the 0 product rule, we know either x+3=0 or x=-3 or x-3=0 or x=3. We also note that (-3)^2-9=0 as well as 3^2-9=0.
      (3 votes)
  • male robot hal style avatar for user Vinay Sharma
    Does grouping mean fragmenting an expression(numerical or algebraic) into the products of two factors? OR something else?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      No, you're referring to factoring. Grouping is a trick that helps with factoring, it is not factoring itself.

      Say I have x³+x²-2x-2. No two of these terms have a common factor. However, x³ and x² do have a common factor (of x²) as do -2x and -2 (of -2). Grouping refers to factoring only these sub-expressions, like this
      x²(x+1)-2(x+1).

      We can then treat this expression as having only two terms with a common factor of x+1, and factor out the x+1 as
      (x+1)(x²-2).
      (3 votes)
  • duskpin ultimate style avatar for user Pierre Dob
    I read that an irreducible polynomial of the form (x^n) + (x^(n-1))+(x^(n-2))...+x^2+x+1(notice all the coefficients are one) where n can be written as 2^k where k is an integer "can be expressed using a series of quadratic equations. Not by representing it as a product of of simpler factors, because that's not possible[as it is irreducible] but by using equations whose coefficients solve other equations." I cannot make sense of this statement, even though I have tried several times(maybe the quadratics have roots that correspond to this equation, but this seems wrong, as a quartic, for example may have a fourth root, which is impossible to get in a quadratic equation). This statement is absolutely key to what I am reading, so I would be extremely grateful for any help. I'm sorry if I made this question difficult to read, I am trying to avoid run-on sentences but it's hard! Thanks in advance!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      "Can be expressed using a series of quadratic equations" is very ambiguous, since it doesn't say what operations are allowed between the quadratics. I can only guess whether they mean a composition of quadratics, or a polynomial in a quadratic, or something else.

      However, it's worth noting that there are very few polynomials like this. The polynomial which is just a sum of powers like this is irreducible only if the degree is one less than a prime. Since you insist that the degree looks like 2^k, this means we're looking for primes of the form (2^k)+1, or Fermat primes. And there are only five known Fermat primes: 3, 5, 17, 257, and 65537. It is suspected that there are no more, but this is still an open problem in math.

      It would make some sense if your text meant that we are to compose a sequence of quadratic functions, since composing a bunch of quadratics must yield a polynomial whose degree is a power of 2. But I'm not aware of anything useful that could be gained by writing the polynomial like this, and it seems like it is possible for polynomials whose degree is any power of 2, not just the irreducible ones. Also, it seems like the coefficients are just rational numbers, not anything more complex.

      My only other guess is that it means you can write the polynomial as a product of quadratics whose roots are complex conjugate pairs. These quadratics would indeed have 'coefficients that solve other equations', but this is possible for any polynomial of even degree. It's not special for polynomials of your form, or with degree of a power of 2.

      I will say to take heart; it's not you. Unless there is a lot of clarifying context that you left out, this statement is almost meaningless as it is written, and it's all the worse if the statement is indeed 'absolutely key'. I would reject the text you're reading and find a better one.
      (3 votes)
  • purple pi teal style avatar for user vv1101
    At is it right if I just make the term (x^2 - 9) and equal it to zero as follows:
    x^2 - 9 = 0
    x^2 = 9
    take square roots
    x = +3 or -3
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] So we're told that p of x is equal to this expression here, and it says plot all the zeros or x-intercepts, of the polynomial in the interactive graph, and the reason why it says interactive graph is this is a screenshot from this type of exercise on Khan Academy, and on Khan Academy, you'd be able to click on points here and it'd put little dots and you can either delete them and put 'em someplace else, so it'd be an interactive graph. But this is just a screenshot, so I'm just going to draw on top of this. But the main goal is, what are the zeros of this polynomial and then you just to plot it on this graph, so pause this video and have a go at it. All right, now to figure out the zeros of a polynomial, you would essentially have to figure out the x values that would make the polynomial equal to zero. Or another way to think about it is the x values that would make this equation true. X to the third plus x squared minus nine x minus nine is equal to zero. Now, the best way to do that is to try to factor this expression. Now, this is a third degree polynomial, which isn't always so easy to factor, so let's see how we might approach it. The first thing I look for is are there any common factors to all of these terms, and it doesn't look like there is. The next thing I could look for, I could think about whether factoring by grouping could work here, and when I think about factoring by grouping, I would look at the first two terms and I would look at the last two terms, and I would say, is there anything I could factor out of these first two terms that would, or what's the most that I could factor out of these first two terms, and then what's the most that I could factor out of these last two terms, and then it would leave something similar once I've done that factoring. Now, what I mean is, for these first two, we have a common factor of x squared, so let's factor out an x squared and these first two terms become x squared times x plus one, and then for these second two terms, I can factor out a negative nine, so I could rewrite it as negative nine times x plus one. Now, that all worked out quite nicely, because now we see, if we view, if we view this as our now our first term and this as our second term, we can see that x plus one is a factor of both of them. And so we can factor that out. We can factor out the x plus one, and I'll do that in this light blue color, actually let me do it with slightly darker blue color. And so if you factor out the x plus one, you're left with x plus one times x squared, x squared, minus nine. Minus nine. And that is going to be equal to zero. Now, we are not done factoring yet, because now we have a difference of squares. X squared minus nine, this is going to be equal to, and let me just write it all out, so I have this x plus one here, so I have x plus one, and then the x squared minus nine, I can write as x plus three times x minus three. If any of what I'm doing feels unfamiliar to you, if that first factoring feels unfamiliar, I encourage you to review factoring by grouping, and if what I just did looks unfamiliar, I encourage you to look at factoring differences of squares. But anyway, all of that would be equal to zero. Now, if I have the product of several things equaling zero, any, if any one of those things is equal to zero, that would make the whole expression equal to zero. So we have a situation where one solution would be the solution that makes x plus one equal to zero, and once again, I'm gonna do that in darker color, x plus one equal zero, and that of course is x is equal to negative one. Another solution is what would make x plus three equal to zero? And that of course is x is equal to negative three, subtract three from both sides, and then another solution is going to be whatever x value makes x minus three equal to zero, add three to both sides, you get x is equal to three. So there you have it. We have our three zeros. Our polynomial evaluated at any of these x values will be equal to zero. So we can plot it here on this interactive graph, I'm just gonna draw on it. So we have x equals negative one, which is right over there, x equals negative three, which is right over there, and x equals three, which is right over there. And the reason why you might want to do this type of thing, this exercise just asks us to do this, and we're done, but the reason why this is useful is this can help inform what the graph looks like. This tells us where our graph intersects the x-axis. So our graph might do something like this, or it might do something like this, and we would have to look at other information to think about what that might be. But I'll leave you there.