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Contents

## Topics

Hi everyone. This week we continued our work with functions, learning how to find solutions of higher order polynomials. We also worked on dividing polynomials, specifically using long division. You can now find all the possible real and complex zeros of a function. Let's try a few problems to practice what we've learned.

## Zeros

We begin by finding the real roots or the real zeros, and sketching the graph of polynomial functions. So let's start by looking at this first function. Please take some time to find the roots of the function g of x equals the quantity x plus 2 times the quantity x Squared plus 1. Remember that the word root and the word zero mean the same thing, values of x that make the function

## Zeros

So, luckily, this is already factored as much as it can be. We need to solve the equations, x squared plus 1 equals zero. And x plus 2 equals zero. The first one gives us i and negative i. And the second one tells us that the only real zero, or root, in this case, is negative 2. Remember, though, there are three solutions to this problem. I, negative i, and negative 2.

## Long Divisionx

For a second review problem, let's take a look at division of polynomials. We've learned two specific new strategies for determining all possible real and complex zeros of a function. Let's next take a look at long division. Divide the following. 4 x squared plus 20 x plus 24 by 2 x plus 6.

## Long Division

We can see that when we divide 4x squared plus 20x plus 24 by 2x plus 6, we obtain 2x plus 4.

## Intermediate Value Theorem

Lastly, this problem will be a bit more challenging than the others we've done today. As we're going to use the intermediate value theorem to approximate the zeros of the following function. Please note that we are calling the independent variable here p instead of x. Remember that we can use any letters we want.

## Intermediate Value Theorem

When you substituted in for p, you should have gotten negative 11, 1 and negative 11 for solutions. Since we found that f of negative 2 is negative, and f of 0 is positive, then we can conclude that the function has a 0 between negative 2 and 0. And likewise, between f of 0 and f of 2.

## Smaller Sections

So, let's take a minute to look at the graph of f of p equals 1 minus 3p squared. We can see there are two different roots here. But for this problem, we're just going to focus on this one. You can choose any intervals you want to approximate the location of x intercepts, depending on how accurately you want to know their values. To approximate this 0 a bit more closely, we need to divide our interval into smaller sections than we did before. So, let's use a distance of 0.5 this time for the interval between -2 and 0. Please take a minute to fill in the missing information for f of x, and then answer the question, as zero is between blank and blank.

## Smaller Sections

Let's look at what we found when we substituted in our values of x for f of x. We obtain the values of negative 11, negative 23 fourths, negative 2, one quarter and 1. And from this information we find that a 0 is located between negative 1 and negative 0.5