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

Lesson 33: Zeros of polynomials and their graphs# Zeros of polynomials & their graphs

Sal uses the zeros of y=x^3+3x^2+x+3 to determine its corresponding graph. Created by Sal Khan.

## Want to join the conversation?

- Why do complex solutions come in pairs?(19 votes)
- A polynomial of degree n has n solutions. So let's look at this in two ways, when n is even and when n is odd.

1. n=2k for some integer k. This means that the number of roots of the polynomial is even. Since the graph of the polynomial necessarily intersects the x axis an even number of times. If the graph intercepts the axis but doesn't change sign this counts as two roots, eg: x^2+2x+1 intersects the x axis at x=-1, this counts as two intersections because x^2+2x+1=(x+1)*(x+1), which means that x=-1 satisfies the equation twice. This means that for an even n we have an even number of real solutions.

2. n=2k+1 for some integer k. This means that the graph of the polynomial intersects the x axis an odd amount of times, but at least once. So there is at least one real solution to the polynomial, meaning that we have an even amount of unspecified solutions, which brings us back to the n=2k train of thought. This means that for an odd n we have an odd number of solutions.

Putting together 1. and 2. we get that there always has to be an even number of complex solutions. I hope this answers your question :)(24 votes)

- Why did Sal neglect to evaluate the situation where x=0? With much less computation, that would have simplified the equation to y=3 and the y-intercept of (0,3) is only found on graph A.(14 votes)
- I think because he wanted to mention the roots, etc. Without that information, we the viewers would be less capable of solving similar problems in the future. Sal is always good about trying to impart some insight, rather than just showing the 'easy' way to solve a particular problem.(13 votes)

- So what are the two complex solutions for the
`x^2 = -1`

? I understand that the square root of -1 is i. But why is it also -i ? Is it because`-i * -i = -1`

, and also`i*i = -1`

?

Basically the same reason that`1*1 = 1`

and`-1*-1 = 1`

?(7 votes)- Yes, you are correct. Remember: when taking the square root of both sides, if a side is not yet squared, you have to consider +- the root.

For example:

x^2 = 256

x = +-√256

x = 16 or x = -16(3 votes)

- At3:00he says that
`C`

only has two real roots so it's excluded. But how can we say that for sure? Couldn't the graph curve back down off to the left or the right side somewhere that we don't see?(6 votes)- As always, there are some unspoken assumptions. In this type of problem, the implication is that all of the interesting features of the graph are visible. In practice, if you saw a curve of this nature and did not know the formula, you could say that it is at least locally approximated by a curve with only two zeroes.(7 votes)

- What would happen if an imaginary number is a zero too? Then how would we graph it?(4 votes)
- Imaginary numbers and complex numbers with imaginary parts can be graphed on what we call an "Argand diagram" which is just a coordinate plane but with the y-axis being imaginary and the x-axis being real.(3 votes)

- What is the difference between complex and imaginary?(3 votes)
- Imaginary numbers are just a multiple of i;( For example: 3i). Complex numbers are imaginary numbers and another number; (For example: 3i-8.)(4 votes)

- It seems to me that the complex numbers Sal found at the end caused the curved line from points (-1, 5) to (0, 3), but how do we know where the effect of the complex numbers will take place on our graph?(4 votes)
- If you think of it as (x^2+1)*(x+3) you have the product of a parabola (x^2 translated up 1 ) and a straight line x, slope 1, translated up 3).

Or, you can picture it as (this is more helpful) the sum of x^3, the parabola 3x^2, the straight line x, and horizontal line 3.

As for the complex factors that "x^2+1" can be factored into, I can't visualize them at all.(2 votes)

- x^2=-1 is analogous to i, yes? x=sqrt(-1)?(3 votes)
- How do you know how many zeros a polynomial function has?(2 votes)
- The Fundamental Theorem of Algebra tells us that every n-degree polynomial has exactly n complex roots. Keep in mind, that this theorem does not account for multiplicity. In other words, some of the roots may not be unique. For example, the quadratic:

𝑥² + 6𝑥 + 9 = 0

Has only one unique solution: 𝑥 = -3. However the Fundamental Theorem of Algebra says that since the degree of the polynomial was 2, then there must have been 2 solutions. We call 3 a DOUBLE ROOT of the polynomial, because it accounts for both roots of the polynomial. Furthermore, that quadratic could have been factored as:

(𝑥 + 3)(𝑥 + 3) = 0

Here, two binomial factors both yield 𝑥 = -3 so again we can see that -3 must be a double root. We can also say that the root -3 has a multiplicity of 2. A triple root would have a multiplicity of 3 etc. Therefore, the Fundamental Theorem of Algebra can be used to find the number of zeros a polynomial function and if there is an apparent "contradiction" then it is because of one or more of the roots may have a multiplicity greater than 1. Comment if you have questions.(2 votes)

- Can the graph of an odd degreed (like a cubic function) polynomial function have no x-intercepts?(1 vote)
- No. Since the range of an odd degree polynomial function is all real numbers it must equal 0 at some point. Also, we see that according to the Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra, the lowest amount of real roots possible for an odd degree polynomial function is 1 because complex/imaginary roots come in conjugate pairs and a odd degree polynomial function has a corresponding odd number of roots (the same as its degree). This means that after all the complex conjugate pairs are accounted for, there must be
*at least*one root left over and that root must be real. Therefore all odd degree polynomial functions must have at least one real 𝑥-intercept.(3 votes)

## Video transcript

Use the real 0's of the
polynomial function y equal to x to the third plus
3x squared plus x plus 3 to determine which of the
following could be its graph. So there's several ways
of trying to approach it. One, we could just look at what
the 0's of these graphs are or what they appear to be and
then see if this function is actually 0 when x
is equal to that. So for example, in graph
A-- and first of all, as always, I encourage
you to pause this video and try it before I show
you how to solve it. So I'm assuming you've
given a go at it. So let's look at this
first graph here. Its 0, it clearly has a
0 right at this point. And just by trying to
inspect this graph, it looks like this is at
x is equal to negative 3, if I were to estimate. So that looks like the
point negative 3, 0. So let's see, if we substitute
x equals negative 3 here, whether we get y equaling 0. So let's see, negative 3
to the third power plus 3 times negative 3 squared
plus negative 3 plus 3. What does this give us? This gives us negative 27. This gives us positive 27. This gives, of
course, negative 3. This is plus 3. These two cancel out. These two cancel out. This does indeed equal 0. So this was actually
pretty straightforward. Graph A does indeed work. You could try
graph B right here, and you would have to
verify that we have a 0 at, this looks like negative 2. Another one, this looks like at
1, another one that looks at 3. And since we already know that
A is the answer, none of these-- if you were to input x equals
negative 2, x equals 1, or x equals 3 into this function
definition right over here, you should not get 0. And you'll see that
this doesn't work. Same thing for this one. If you tried 4 or
7 for your x's, you should not get 0
over here, because we see that the real function
does not equal 0 at 4 or 7. Another giveaway that this is
not going to be the function is that you are going to
have a total of three roots. Let me write this down. So you're going to have
a total of three roots. Now, , those three roots could
be real or complex roots. And the big key is complex
roots come in pairs. So you might have a situation
with three real roots. And this is an example
with three real roots, although we know
this actually isn't the function right over here. Or if you have one
complex root, you're going to have
another complex root. So if you have
any complex roots, the next possibility is one
real and two complex roots. And this right over
here has two real roots. That's not a possibility. That would somehow imply that
you have only one complex root, which that is not a possibility. Now another way that you could
have thought about this-- and this would have
been the longer way. But let's say you didn't
have the graphs here for you, and someone asked you to
just find the roots-- well, you could have attempted
to factor this. And this one actually
is factorable. y is equal to x to the third
plus 3x squared plus x plus 3. As mentioned in previous videos,
factoring things of a degree higher than 2, there is
something of an art to it. But oftentimes, if
someone expects you to, you might be able to group
things in interesting ways, especially when you see
that several terms have some common factors. So for example, these first
two terms right over here have the common
factor x squared. So if you were to
factor that out, you would get x squared
times x plus 3, which is neat because that looks a lot
like the second two terms. We could write that as
plus 1 times x plus 3. And then you can factor
the x plus 3 out. We could factor
the x plus 3 out, and we would get x plus
3 times x squared plus 1. And now, your 0's
are going to happen, or this whole y--
remember this is equal to y-- y is
going to equal 0 if either one of these
factors is equal to 0. So when does x plus 3 equal 0? Well, subtract 3
from both sides. That happens when x is
equal to negative 3. When does x squared plus
1 equal 0, I should say? Well, when x squared
is equal to negative 1. Well, there's no real
x's, no real valued x's. There's no real
number x's such that x squared is equal to negative 1. x is going to be
an imaginary-- or I guess I'll just say it in
more general terms-- it's going to be complex valued. So once again, you
see you're going to have a pair of
complex roots, and you have one real root at x
is equal to negative 3.