Welcome to a lesson on the complex factorization theorem. In this lesson we'll define the theorem, and then use the theorem to find all the zeros of a polynomial function. A polynomial function of degree n with complex coefficients in this form here has exactly n complex zeros as long as we count multiplicity.

So notice how here we're not just talking about real zeros, we're talking about complex zeros. So if we count complex zeros, the number of complex zeros will always match the degree of the polynomial as long as we count multiplicity.

So if these rs are the complex zeros with multiplicity given by these ms, we can write a polynomial function as a product of linear factors as we see here where again the rs would be the zeros and the ms would be the multiplicities.

It's important to keep in mind that now we're talking about complex zeros or complex numbers where complex numbers are made up of the set of real numbers and the imaginary numbers in the form of a +/- bi.

Let's take a look at our first example. Here we want to write the given polynomial function which is a degree three polynomial function as a product of linear factors or factors of degree one, and then find the zeros and state the multiplicity of each.

We know from our previous lesson on the rational roots theorem that if this polynomial function does have real rational roots it must be a ratio of the factors of the constant term, in this case -80, two of the factors of the leading coefficient, which in this case would be 1.

To save some time, i've already listed the factors of -80 and the factors of 1, but instead of using trial and error to figure out which ratio would be a real zero to the given function, let's go ahead and graph the function and see if it can find a real rational zero as an x intercept.

So here's the graph of the given function. Notice how there is an x intercept here at 5. This would be the (.5, 0) and, therefore, 5 is a zero of the given polynomial function.

And if 5 is a zero, the x 5 must be a factor of the given function. So again, if 5 is a zero, and x 5 is a factor, we can write the given function as p of x would have to be equal to the factor of x 5 x a degree two quadratic factor.

In order to find this degree two quadratic factor, we can take the given polynomial function and divide by x 5. To do this, let's use synthetic division. So if we're dividing by x 5, we'll use the constant 5 when using synthetic division, and now we'll list the coefficients of the polynomial function which will be 1, -5, 16, -80.

Now, to begin, we'll bring down this 1 and start multiplying by 5. 5 x 1 is 5, that goes here. Then we add, that would be zero.

5 x 0 is 0, add, that's 16. And 5 x 16 is 80, and we add, and this last column is zero which is good news. This would be our remainder, and since the remainder is zero, x 5 divide evenly into the polynomial function which is what we expected.

This would be the degree two factor. This represents 1 x squared + 0x + 16, or x squared + 16. But again, our goal here is to write this as a product of linear factors, or factors of degree 1.

And since x squared + 16 doesn't factor, we'll actually have to find the zeros of x squared + 16, and use those zeros to write this as two linear factors.

Let's go ahead and do that part on the next slide. Again, our goal is to write p of x as a product of linear factors. So to find the linear factors of x squared + 16, we're actually going to find the values of x that make this equal to zero or find the zeros of this and then use those zeros to find the remaining two factors.

So we actually want to solve x squared + 16 = 0 to find the remaining two zeros of p of x, and, therefore, find the remaining two linear factors. So if we subtract 16 on both sides and then take the square root of both sides, don't forget we'll have a +/- here.

We'll have x = +/- the square root of -16 which is -1 x 16 which would give us x = +/- 4i. So the two remaining zeros would be x = 4i and x = -4i.

Therefore, the two remaining linear factors would be x 4i and x -4i or x + 4i. Let's go ahead and summarize this. We just found the polynomial function can be written as a product of linear factors as x 5 x x 4i x x + 4i, and the zeros which will have multiplicity one are x = 5, x = 4i, and x = -4i.

Let's take a look at a second example. Same question, but now we have a degree four polynomial function. So again, we'll start by determining the real rational zeros of the given function, which we could use the rational roots theorem by finding the ratio of the factors of the constant term to the factors of the leading coefficient, and use these ratios and trial and error to find a real rational zero of this function, but again, let's go ahead and graph the function instead.

Here's the graph of our function. Notice how there is an x intercept here at (2, 0). But notice how the graph behaves at this x intercept.

It doesn't cross through the x axis like it does when the multiplicity is odd. It touches it and then turns in the other direction which means the zero of x = 2 has an even multiplicity.

And if x = 2 is a zero, x 2 would have to be a factor. And because we have a degree four polynomial function, x = 2 either has multiplicity two or multiplicity four.

Let's begin by assuming it has multiplicity two. If we assume it has multiplicity two, that means we could write the function as p of x = two factor of x 2.

So if we have x 2 squared x what would be a degree two factor. So to find this degree two factor here, we'll actually divide the given polynomial by x 2 twice, and see what factor remains.

So we'll set this up as synthetic division. So we'll use the (0, 2), and then we'll list the coefficients of the polynomial function which are 1, -8, 33, -68 and 52.

So now to begin we'll bring the 1 down, start multiplying by 2. 2 x 1 is 2, add, that's -6. Multiply that's -12, add, that's 21, multiply, that's 42, add, that's -26, multiply, that's 52, and we have a remainder of zero which is good news.

We just divided by x 2 once, and now let's do it again. Bring down the 1, 2 x 1 is 2, add, -4, multiply, -8, add 13, multiply, and add, that's zero.

Again, that's good news again because we have another remainder of zero, so x 2 divided twice into the given polynomial function, this represents a degree two factor that remains which would be 1x squared 4x + 13.

And this doesn't factor, so what we're gonna do is work backwards again, find the zeros of this, and then use those zeros to find the remaining linear factors of the given polynomial function.

So to find the zeros of x squared 4x + 13, we're going to solve this equation by using the quadratic formula given here below where a is equal to 1, b is equal to -4, and c is equal to 13.

So we would have x = -(-4), that's 4 +/- the square root of b squared, -4 squared is 16 4 x a x c, 4 x a is 4 x c, that would be 52. Divide all this by 2 x a or 2 x 1 which is 2.

So we have x = 4 +/- the square root of-- this would be -36 divided by 2, x = 4 +/- the square root of -36 would be 6i divided by 2. So we have x = 4 divided by 2 +/- 6i divided by 2, or x = 2 +/- 3i.

So we can use these two complex zeros to find the remaining two linear factors of our polynomial function. Let's go ahead and finish this on the next slide. We would have p of x = x 2 squared x x the 0 of 2 + 3i and then x the 0 of 2 3i.

So now we have the polynomial function as a product of linear factors, and the zeros are x = 2 that has multiplicity 2, and we have x = 2 + 3i and x = 2 3i, both of which have multiplicity one.

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