Welcome to a lesson on the zeros, factors, and graphs of polynomial functions. In this lesson, we'll understand the connections among the real rational zeros, the factors, and the graphs of polynomial functions.

To begin, let's assume we have a polynomial function of degree n given in this form here, and c is a real number such that p of c = 0. So if we sub the value of c into the function, the function value is zero.

So we say the real number c is a zero or root of the polynomial function p of x. It's also true and x = c is a solution to the equation p of x = 0.

So if we take this polynomial function and replace p of x with zero forming an equation, c would be a solution to that equation. The quantity x c is a factor of the polynomial function which means it would divide evenly into the polynomial function, and then finally, the (.C, 0) is an x intercept of the graph of the polynomial function p of x.

To better understand all of this, let's take a look at some examples. A degree four polynomial function has zeros or roots of multiplicity one at x = 0 and x = 3, and a zero of multiplicity 2 at x = -1 and has a leading coefficient of -2.

We're first asked to find the x intercepts of the polynomial function. The x intercepts will occur at the real zeros of the polynomial function. And since we're given the zeros are x = 0, 3, and -1, these would be the x intercepts.

So if we give them as ordered pairs, if the x intercept is zero, that would be the origin with coordinates (0, 0). If the zero is 3, the x intercept would be (3, 0), and if the zero is -1, the x intercept would be (-1, 0).

Next, we're asked to write the polynomial function in factored form where the rs would be the zeros of the polynomial function, and the ms would be the multiplicity. So if the multiplicity is 1, we'd only have one factor containing the zero, and if the multiplicity is 2, we'd actually have two factors containing the zero.

So for our polynomial function, we'd have p of x = notice in this from a would be the leading coefficient which we know is -2, so a is -2, and then we have two zeros with multiplicity 1, one of them is 0 is one of them is 3.

So one factor would have to be x 0. We only have one factor of this because the multiplicity is 1. We also have one factor of x 3, but because -1 has a multiplicity of 2, we'd actually have two factors of x -1, or x + 1, so this would be squared.

Notice if we sub any of these x values into this polynomial function, it would make one of these factors equal to zero, and, therefore, the polynomial function would be equal to zero. Let's go ahead and clean this up a little bit.

This is p of x = of course, x 0 is just x. So we can write -2x x x 3 x x + 1 squared. And now we're ask to graph the polynomial function.

And because we're given three real zeros of the polynomial function, our polynomial function will have three x intercepts. But whenever the multiplicity is odd, the graph will cross the x axis, but if it's even, the graph will touch the x axis at that point, and then turn in the other direction.

So if we take a look at the graph of our function, notice how at x = -1 we have an x intercept, but it does not cross the x axis, just touches it and turns direction.

And that's because the multiplicity of the 0x = -1 is 2 or even. But then notice that x = 0 here and that x = 3 here, the graph does cross the x axis at those two points, because the multiplicity is odd.

Let's take a look at another example. Here we're given the graph of a degree three polynomial function, and we're first asked to find the x intercepts. Well, the graph crosses the x axis here at -2, and it also touches the x axis here at +1.

So those would be the x intercepts, so the coordinates would be (02, 0) and (1, 0). And now we're asked to find the real zeros or real roots of the function, and we'll also give them multiplicity.

Well, if there's an x intercept that x = -2, that would be one of the zeros, and because it crosses the x axis here, the multiplicity must be odd, and, therefore the multiplicity would be 1 here.

It couldn't be 3 because there's another x intercept over here. Our other zero would be x = 1, and because it does not cross the x axis, just touches this and then turns, this has an even multiplicity, and because we have a degree three polynomial, that means the multiplicity would be 2.

And now we're asked to find the equation of the polynomial function. Let's set this up over here on the side. We would have p of x. Here we're not given the leading coefficient, so let's go ahead and just leave it as a x if x = -2 is a zero, then x + 2 would have to be a factor.

It has multiplicity one so we only have one factor of x + 2, but then the zero of x = 1 has multiplicity 2, so we'd have to have two factors of x 1.

And now to find the value of a though, we'll have to use one of the points given on the function. Let's go ahead and use this point here where the coordinates would be (2, -2), which means that p of 2 = -2.

So now what we're going to do is substitute 2 for x and set this equal to the function value of -2, and that will allow us to find a.

So we would have a x 2 + 2, that would be 4 x 2 1 squared, that would be 1 squared = -2. So this would just be 4a = -2, divide both sides by 4.

We have a = -1/2, which means our polynomial function would be p of x = a, which is -1/2 x the quantity x + 2 x the quantity x 1 squared. I think we have time for one more example.

Here we're actually given a polynomial equation, and the first question is what polynomial function would you graph to verify your solutions? This equation would be equivalent to the polynomial function where p of x = 0, so we'll substitute p of x for 0 to form the polynomial function.

So our polynomial function we would use to verify our solutions would just be p of x equals 0.25 x the quantity x 1 x the quantity x 3 x the quantity x + 4 x the quantity x + 2 squared.

Next it says, where would you look to verify the solutions? The solutions would be the x intercepts. Next we're asked what are the x intercepts of the function. Well, the x intercepts would be the x values where the function value would be zero, or also it would be the solution to this equation here.

And we can see this would equal 0 when x = 1, 3, -4, and -2, which would be the x intercepts. So we'd have the point (1, 0), we'd have the point (3, 0), we'd have the point (-4, 0), and we'd have the point (-2, 0).

And the last question, what are the zeros of the polynomial function? Well, these would be the same as the x intercepts, except we will also include the multiplicity.

So the zeros are x = 1, x = 3, and x = -4 with multiplicity 1, and then we have the 0x = -2, with multiplicity 2 because notice how we have two factors of x + 2.

Let's go ahead and verify all of this by looking at the graph of our polynomial function. Again, notice how we have 1, 2, 3, 4 x intercepts, three of which cross through the x axis, and one that touches and turns, so the zeros of -4, 1, and 3 have multiplicity 1, and the zero of x = -2 has an even multiplicity, or a multiplicity of two.

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