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## The fundamental theorem of algebra

Current time:0:00Total duration:6:35

# Quadratics & the Fundamental Theorem of Algebra

CCSS Math: HSN.CN.C.9

## Video transcript

Voiceover:Let's say that
we have the function F of X being defined by the
second degree polynomial, 5 X squared plus 6 X plus 5. The fundamental theorem of algebra tells us that because this
is a second degree polynomial we are going to have exactly 2 roots. Or another way of thinking about it, there's exactly 2 values for X that will make F of X equal 0. So I encourage you to pause this video and try to figure out what
those 2 values of X are. So when I, if we're
looking for the X values, that would make this
expression equal to 0. That's essentially trying
to solve this equation. At 5 X squared plus 6
X plus 5 is equal to 0. And there's no obvious way I can think of factoring this right
over here so I'll just resort to the quadratic formula. So the quadratic formula tells us that X, where X is a solution to this equation, is going to be equal to negative B. I'll try to color code it. Negative B, this is right,
this is B right over here. Negative B, so negative 6, plus or minus, plus or minus the square
root of B squared, of B squared, minus 4 times A, times A, times C, times A times C. All of that, all of that over
all of that over 2 times A. All of that over 2 times A. So what does this simplify to? This is going to be equal
to so I'll go back to, this is going to be equal to negative 6. I'll try to keep it color coded still. Negative 6 plus or minus,
plus or minus the square root. Now, what is this going to be equal to? This is 36 minus 100. So negative 64 all of that over 2 times 5. All of that over 10. Now this is interesting. We are trying to take the square
root of a negative number. Or another way of thinking about it, B squared minus 4 A C is less than 0. B squared minus 4 A C which is often, often times called the
discriminant of this quadratic. This is less than 0. So this is less than 0. This part of the quadratic
formula we're going to try to take the square
root of a negative number. This is going to be a negative number. So we're going to result
in an imaginary number. So what that gives us
are not 2 real roots, but 2 non real complex roots. So this is going to be equal to negative 6 plus or minus 8 i, the square root of negative 64 is 8 i. If we extend the principle
square root function to imaginary numbers or complex numbers. All of that over 10. Or we could say X is equal to, let's see if we find a
greatest common divisor. If we try to reduce this,
so we're going to have let's see they're all divisible by 2 so it's negative 3 over 5,
that's the same thing as negative 6 over 10, plus or
minus 4 over 5, 4 over 5 i. Or you could say that the 2 roots, which are non real complex roots, so it's X is equal to
negative 3/5 plus 4/5 i, that's one of the roots. And then the other root is X is equal to negative 3/5 minus 4/5 i. Notice, we satisfied the
fundamental theorem of algebra. We have 2 roots, they're non real, but the fundamental theorem of algebra says, "hey look we're just going to have "at least if we're Nth degree polynomial, "we're going to have n complex roots." They could be real or
they could be non real and we see that right over there. And we also see that they are conjugates. And the quadratic formula kinda sets up a situation where we are, especially if this ends
up being less than 0 and so this when you take the square root you get an imaginary number you see where those conjugates appear. Now let's verify graphically
that this is indeed the case. That this does not have any real roots. So let's get a calculator out. So let me go into graph mode. So Y is equal to. Let me clear what I must have
been doing out here before. So Y1 is equal to 5 times X squared, plus 6 X plus 5 and then let me set a reasonable range here. So I don't know. Actually, I know very little
about this function right over. So I'm going to set minimum
range at a negative 10, X max I don't know positive 10. X scale is 1, let's see Y max let's see, this thing actually does
get pretty big pretty fast. So let's say Y max is and the Y scale, I don't know I'll say 100. 100, the scale I'll make it 10. Each of those notches are going to be 10. Y minimum, we want to see the X axis to make sure it doesn't cross
it so let's go at negative, I don't know, negative 20. And now let's graph this. I'm hoping that I've actually
captured it and I have. So there you have it, you see that this thing does not intersect the X-axis. And in fact, we could,
we could zoom in on it. Let's see it's a little bit
hard to, it's a little actually, let me just change the range a little bit. So let me just change the range. So let's make our X min, let's make it 5, let's make our X max, whoops
not 50, let's make that 5. And let's make, negative 5 and positive 5. And let's see, let's make our Y max, let's make our Y max equal
to 20 and then our Y scale, I don't know we could make it 2. Okay I think this will
get us much closer in, there we see, it does
not intersect the X axis. This does not have any real roots but it has 2 non real complex roots.