Main content

## Algebra (all content)

### Course: Algebra (all content) > Unit 10

Lesson 32: Finding zeros of polynomials# Finding zeros of polynomials (1 of 2)

CCSS.Math: , , ,

Sal finds all the zeros (which is the same as the roots) of p(x)=x⁵+9x³-2x³-18x=0..

## Want to join the conversation?

- What did Sal mean by imaginary zeros?(25 votes)
- Some quadratic factors have no real zeroes, because when solving for the roots, there might be a negative number under the radical. The only way to take the square root of negative numbers is with imaginary numbers, or complex numbers, which results in imaginary roots, or zeroes. They always come in conjugate pairs, since taking the square root has that + or - along with it.(46 votes)

- So why isn't x^2= -9 an answer? Like why can't the roots be imaginary numbers? My teacher said whatever degree the first x is raised is how many roots there are, so why isn't the answer this:

X= { 0, +- (square root of) 2, +- 3i?(13 votes)- The imaginary roots aren't part of the answer in this video because Sal said he only wanted to find the real roots.

You're totally right about if the problem was finding all of the roots, 3i and -3i would be included in the answer.(42 votes)

- Since it is a 5th degree polynomial, wouldn't it have 5 roots?(12 votes)
- It does it has 3 real roots and 2 imaginary roots.

Sal didnt really solve for x^2=-9 if you do, you get +or - 3i that is 2 complex roots and 3 real roots.(4 votes)

- I'm lost where he changes the (x^2- 2) to a square number was it necessary and I also how he changed it. In total, I'm lost with that whole ending.(7 votes)
- I don't understand anything about what he is doing. How did Sal get x(x^4+9x^2-2x^2-18)=0? And how did he proceed to get the other answers?(5 votes)
- for x(x^4+9x^2-2x^2-18)=0, he factored an x out

after this, you can see, 9x^2 and -2x^2 are both 2nd degree terms

further, he factored the new terms to simplify the polynomial

after this he factorized (x^2-2) [@6:08] and then equated it all with zero because that's when we get factors of a polynomial. if a*b*c*d=0 then atleast one or more terms would need to be 0 (ain't that right?). so he continued to equate each of them with zero and the ones that gave out a real value of x were valid zeroes(2 votes)

- This is not a question. It is a statement. Put this in 2x speed and tell me whether you find it amusing or not.(4 votes)
- At0:09, how could Zeroes and Roots be the same things?(3 votes)
- Yes, as kubleeka said, they are synonyms They are also called solutions, answers,or x-intercepts.(2 votes)

- Why are imaginary square roots equal to zero?(1 vote)
- Same reply as provided on your other question. It is not saying that the roots = 0. A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0 (or make Y=0). It is an X-intercept. The root is the X-value, and zero is the Y-value. It is not saying that imaginary roots = 0.(6 votes)

- How do you graph polynomials?(2 votes)
- There are many different types of polynomials, so there are many different types of graphs.

But the most simple polynomial x^2+x, looks like a U centered at the origin.

Hope this helps! :)(3 votes)

- How do you find the zeroes of f(x)=x^5?

From the graph it seems that f(x) is equal to 0 at a lot of points.(2 votes)- The graph has one zero at x=0, specifically at the point (0, 0).

The other points that appear to be x=0 are actually small fractions or decimal values. You would need a graph with a very small scale to show that these are not actually zero.(2 votes)

## Video transcript

- [Voiceover] So, we have a
fifth-degree polynomial here, p of x, and we're asked
to do several things. First, find the real roots. And let's sort of remind
ourselves what roots are. So root is the same thing as a zero, and they're the x-values
that make the polynomial equal to zero. So the real roots are the x-values where p of x is equal to zero. So, the x-values that satisfy this are going to be the roots, or the zeros, and we want the real ones. As you'll learn in the future,
there's also going to be imaginary roots, or
zeros, or there might be. Then we want to think
about how many times, how many times we intercept the x-axis. Well as we'll see, however
many real roots we have that's how many times we
are going to intercept... However many unique real roots we have, that's however many times we're going to intercept the x-axis. How do I know that? Well, let's just think about an arbitrary polynomial here. So those are my axes. This is the x-axis, that's my y-axis. And let me just graph an
arbitrary polynomial here. So, let's say it looks like that. Well, what's going on right over here. At this x-value, we see, based
on the graph of the function, that p of x is going to be equal to zero. So that's going to be a root. This is also going to be a root, because at this x-value, the
function is equal to zero. At this x-value the
function's equal to zero. At this x-value the
function is equal zero. If we're on the x-axis
then the y-value is zero. So the function is going
to be equal to zero. This is a graph of y is equal, y is equal to p of x. Not necessarily this p of x, but I'm just drawing
some arbitrary p of x. So there's some x-value
that makes the function equal to zero. Well, that's going to be a point at which we are intercepting the x-axis. So we want to know how many times we are intercepting the x-axis. As we'll see, it's
gonna be the same number of real roots, or the same
number of real zeros we have. And then they want us to
figure out the smallest of those x-intercepts,
and we'll figure it out for this particular polynomial. So, let me give myself
a little bit more space. So, let's get to it. So we really want to solve
p of x is equal to zero. So we really want to set,
that right over there, equal to zero, and solve this. So we want to solve this equation. The x-values that make this equal to zero, if I input them into the function I'm gonna get the function equaling zero. All right. So the first thing that
might jump out at you is that all of these
terms are divisible by x. So I like to factor that
out from the get-go. So, we can rewrite this as x times x to the fourth power plus nine x-squared minus two x-squared minus 18 is equal to zero. Now there's something else that might have jumped out at you. It actually just jumped out of me as I was writing this down is that we have two third-degree terms. After we've factored out an x, we have two second-degree terms. Now, it might be tempting to
just add these two together, and actually that it would be
a completely legitimate way of trying to factor this so
that we can solve this equation. But instead of doing it that way, we might take this as a clue that maybe we can factor by grouping. Remember, factor by grouping, you split up that middle degree term
and see if you can reverse the distributive property twice. So, let's see if we can do that. Can we group together
these first two terms and factor something interesting out? And group together these second two terms and factor something interesting out? And then maybe we can factor
something out after that. What am I talking about? Well, this is going to be
the same thing as x times... Well, this one, actually let
me write a big parenthesis, here, this one right over
here is the same thing as... I can factor out an x-squared. So, it's gonna be x-squared plus... Sorry. It's gonna be x-squared, if
I factor out an x-squared, I'm gonna get an x-squared plus nine. And then over here, if I factor out a, let's see, negative two. I don't want to... If I factor out a, yep, negative two, I'm gonna get, so minus two times... I'm gonna get an x-squared
plus nine, again. Now this is interesting,
because this is telling us maybe we can factor out
an x-squared plus nine. So, let me factor out
an x-squared plus nine from each of these terms,
and I'm going to get, I am going to get x... I'll leave these big green
parentheses here for now, If we factor out an x-squared plus nine, it's going to be x-squared plus nine times x-squared, x-squared minus two. X-squared minus two, and I gave myself a
little bit too much space. So, let me delete that. So let me delete that right over there and then close the parentheses. Then close the parentheses. Actually, I can even get rid
of those green parentheses now, if I want to, optimally, make
this a little bit simpler. So far we've been able to factor it as x times x-squared plus nine
times x-squared minus two. And the whole point
that I'm factoring this is if I can find the product of a bunch of expressions equaling zero, then I can say, "Well, the
product of those expressions "are going to be zero if one
or more of those expressions "are equal to zero",
and I can solve for x. Well, let's see. This one's completely factored. This one is completely
factored if we're thinking about real roots. This one, you can view it
as a difference of squares if you view two as a
square root of two-squared. So, we can rewrite this as, and of course all of
this is equal to zero. Let me just write equals. So we could write this as equal to x times times x-squared plus nine times... Let's see, I can factor this business into x plus the square root of two times x minus the square root of two. I'm just recognizing this
as a difference of squares. And, once again, we just
want to solve this whole, all of this business, equaling zero. All of this equaling zero. So how can this equal to zero? Well any one of these expressions, if I take the product, and if
any one of them equals zero then I'm gonna get zero. So, x could be equal to zero. X could be equal to zero, and that actually gives us a root. When x is equal to zero, this
polynomial is equal to zero, and that's pretty easy to verify. Let's see, can x-squared
plus nine equal zero? X-squared plus nine equal zero. Well, if you subtract
nine from both sides, you get x-squared is
equal to negative nine. And that's why I said, there's
no real solution to this. So, no real, let me write that, no real solution. There are some imaginary
solutions, but no real solutions. Now, can x plus the square
root of two equal zero? X plus the square root of two equal zero. Sure, if we subtract square
root of two from both sides, you get x is equal to the
negative square root of two. And can x minus the square
root of two equal zero? Sure, you add square root
of two to both sides, you get x is equal to
the square root of two. So, there we have it. We have figured out our zeros. X could be equal to zero. P of zero is zero. P of negative square root of two is zero, and p of square root of
two is equal to zero. So, those are our zeros. Their zeros are at zero,
negative squares of two, and positive squares of two. And so those are going
to be the three times that we intercept the x-axis. And what is the smallest
of those intercepts? Well, the smallest number here is negative square root, negative square root of two. And you could tackle it the other way. You could take this part right over... Which part? Yeah, this part right over here and you could add those two middle terms, and then factor in a non-grouping way, and I encourage you to do that. But just to see that this makes sense that zeros really are the x-intercepts. I went to Wolfram|Alpha and
I graphed this polynomial and this is what I got. So, this is what I got, right over here. If you see a fifth-degree polynomial, say, it'll have as many
as five real zeros. But, if it has some imaginary zeros, it won't have five real zeros. Instead, this one has three. And that's because the imaginary zeros, which we'll talk more about in the future, they come in these conjugate pairs. So, if you don't have five real roots, the next possibility is
that you're going to have three real roots. And, if you don't have three real roots, the next possibility is you're
gonna have one real root. So, that's an interesting
thing to think about. And so, here you see,
your three real roots. You see your three real roots which correspond to the x-values at which the function is equal to zero, which is where we have our x-intercepts.