Lesson Video: Zeros of Polynomial Functions Mathematics

In this video, we will learn how to find the set of zeros of a quadratic, cubic, or higher-degree polynomial function.

15:45

Video Transcript

In this video, we’ll learn how to find the set of zeros of a quadratic, cubic, or higher-degree polynomial function. We begin by recalling what we actually mean by the zeros of a function. Finding the zeros of a function is the same as finding the roots. And these are the values of 𝑥 that make the polynomial equal to zero.

If we think about this graphically, we know that when the function 𝑦 equals 𝑓 of 𝑥 is set equal to zero, we’re interested in the places where our graph intersects the 𝑥-axis. And so the roots or zeros of the function are the 𝑥-intercepts of the graph. But of course, we don’t always want to draw the graphs of our functions. So we’re going to ask ourselves what other techniques we have to solve a polynomial 𝑓 of 𝑥 is equal to zero. Of course, for factorable functions, we know we can factor the expression and find the roots that way. So let’s see what that might look like.

Find, by factoring, the zeros of the function 𝑓 of 𝑥 equals 𝑥 squared plus two 𝑥 minus 35.

Remember the zeros of the function are the same as the roots. And we find the roots of a function by solving the equation 𝑓 of 𝑥 is equal to zero. And so we find the zeros of our function by setting 𝑥 squared plus two 𝑥 minus 35 equal to zero and solving for 𝑥. Now the question actually tells us the method to use. We’re going to factor the expression 𝑥 squared plus two 𝑥 minus 35.

Firstly, we see that this expression is a quadratic, and so it’s going to factor into a pair of parentheses, as shown. These will consist of a pair of binomials. And the first term in each binomial must be 𝑥 since 𝑥 multiplied by 𝑥 gives us 𝑥 squared. And then to find the numerical part, we want to find two numbers who have a product; that means they multiply to make negative 35 and they have a sum of two. We’ll begin by finding two numbers that have a product of 35. And we’ll consider their signs in a moment.

Of course, we have one times 35. Well, we have five times seven. Since we’re trying to make negative 35 and we know a negative multiplied by a positive gives a negative, we know that one number in our factor pair will have to be negative. So how will we ensure that these numbers add to make two? Well, if we use negative five and positive seven, we know negative five times positive seven is negative 35 but negative five plus seven is two. And so we’re going to put negative five and seven into our parentheses. And so the expression 𝑥 squared plus two 𝑥 minus 35 is equal to 𝑥 minus five times 𝑥 plus seven.

So how does this help us solve the equation 𝑥 squared plus two 𝑥 minus 35 is equal to zero? Well, we know that 𝑥 minus five and 𝑥 plus seven are multiplying one another. And when they do, the answer is zero. So what can we say about one or other of these binomials? We know that for the product of two numbers to be equal to zero, either one or other of the numbers must itself be equal to zero. So in this case, that must mean that either 𝑥 minus five is equal to zero or 𝑥 plus seven is equal to zero. Then we solve as we would any other linear equation. In this case, we add five to both sides. And we get 𝑥 equals five. And for our second equation, we subtract seven. So 𝑥 equals negative seven.

Remember, the zeros of the function are the values of 𝑥 that make the polynomial equal to zero. So the zeros of our function are simply the numbers negative seven and five. Now, of course, we know that the zeros are the values of 𝑥 that make the function equal to zero. So we can check our solutions by substituting them into the expression 𝑥 squared plus two 𝑥 minus 35. In either case, we do get zero as required. So we know that we’ve done our calculations correctly.

In our next example, we’ll look at how to find a set of zeros from a cubic function.

Find the set of zeroes of the function 𝑓 of 𝑥 equals 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81.

We might begin by recalling that the zeros of a polynomial function are the values of 𝑥 that make it equal to zero. And so we need to set the expression 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81 equal to zero and solve for 𝑥.

Now, if we were to distribute our parentheses, we’d probably spot that we have a cubic. That’s a polynomial whose order is three. And so we might be thinking that we need to distribute the parentheses and go from there. However, if we look really carefully, we see that the two terms share a common factor. They share a factor of 𝑥 squared minus 81. And so to solve this equation, we’re going to factor by removing that common factor of 𝑥 squared minus 81.

If we divide the first term by 𝑥 squared minus 81, that leaves us simply with 𝑥. And then if we divide the second term by 𝑥 squared minus 81, we get negative two. And so we can see that the expression 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81 is equal to 𝑥 squared minus 81 times 𝑥 minus two. And so if we can fully factor an expression when finding the zeros, it’s really helpful because we now need to ask ourselves, which values of 𝑥 make this expression equal to zero?

And of course, since we’re multiplying 𝑥 squared minus 81 by 𝑥 minus two, and that gives us zero, we can say that either 𝑥 squared minus 81 must be equal to zero or 𝑥 minus two must be equal to zero. We’ll now solve each of these equations in turn. We’ll solve our first equation by adding 81 to both sides. And that tells us that 𝑥 squared is equal to 81. We’re now going to take the square root of both sides. But we do need to be a little bit careful. We’ll need to take both the positive and negative square root of 81.

And so the solutions to the equation 𝑥 squared minus 81 equals zero are 𝑥 equals nine and 𝑥 equals negative nine. The other equation is a little bit more straightforward. We’re just going to add two to both sides. And so we find another solution to the equation 𝑓 of 𝑥 equals zero and, therefore, a zero of our function to be two. And we found the solutions to the equation 𝑓 of 𝑥 equals zero, but the question asks us to find the set of zeros of the function. And so we use these squiggly brackets to represent the set containing the elements negative nine, two, and nine.

And that’s the answer to our question. The set of zeros of the function 𝑓 of 𝑥 equals 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81 contains the elements negative nine, two, and nine. Note, of course, that we could go back to our original function to check whether these answers are correct. We would need to substitute each one in turn and double-check that we do indeed get an answer of zero.

In our next example, we’ll look at how we can find the roots of a cubic equation by factoring using the factor theorem.

Given that 𝑓 of 𝑥 is equal to 𝑥 cubed plus three 𝑥 squared minus 13𝑥 minus 15 and 𝑓 of negative one equals zero, find the other roots of 𝑓 of 𝑥.

The roots of 𝑓 of 𝑥 or the zeros of the function of the values of 𝑥 that make the function equal to zero. And so we need to find the values of 𝑥 such that the equation 𝑥 cubed plus three 𝑥 squared minus 13𝑥 minus 15 is equal to zero. And one technique that we have is to fully factor that cubic expression. But we can’t do so easily without a little bit of extra information. And that information is given to us in the question. We’re told 𝑓 of negative one is equal to zero. This bit of information means we can use the factor theorem to help us factor our cubic.

The factor theorem says that if 𝑓 of 𝑎 is equal to zero, then 𝑥 minus 𝑎 is a factor of 𝑓 of 𝑥. Comparing this general form with our question, and we see we need to let 𝑎 be equal to negative one. And we can therefore say that 𝑥 minus negative one, which is of course 𝑥 plus one, must be a factor of 𝑓 of 𝑥. So how does this help us? Well, this means that we can divide our cubic 𝑓 of 𝑥 by 𝑥 plus one. Exactly, we won’t have a remainder. And so when we do this division, we should be left with a quadratic, which we can then factor.

There are a number of techniques that we can use. I’m going to use polynomial long division using this bus stop method. I begin by asking myself, what is 𝑥 cubed divided by 𝑥? 𝑥 cubed divided by 𝑥 is 𝑥 squared. Now I multiply 𝑥 squared by each term in my divisor. 𝑥 squared times 𝑥 is 𝑥 cubed. And 𝑥 squared times one is 𝑥 squared. Next, we subtract the two terms we just found from the two terms above them. 𝑥 cubed minus 𝑥 cubed is zero. And then three 𝑥 squared minus 𝑥 squared is two 𝑥 squared.

We bring down the next term. And then we ask ourselves, what’s two 𝑥 squared divided by 𝑥? Two 𝑥 squared divided by 𝑥 is two 𝑥. So we put two 𝑥 above three 𝑥 squared. Next, we multiply two 𝑥 squared by each term in our divisor. Two 𝑥 squared times 𝑥 is two 𝑥 squared. And two 𝑥 times one is two 𝑥. Once again, we subtract. Now two 𝑥 squared minus two 𝑥 squared is zero. So we work out negative 13𝑥 minus two 𝑥, which is negative 15𝑥.

Let’s bring down the last term. We now divide negative 15𝑥 by 𝑥, which is negative 15. And we then multiply negative 15 by both terms in our divisor. That’s negative 15𝑥 minus 15. We subtract once more, but negative 15𝑥 minus 15 minus negative 15𝑥 minus 15 is zero. And that tells us, as we expected, that there’s no remainder when we divide our cubic function by 𝑥 plus one. And it also means we can now rewrite our cubic function as 𝑥 plus one times 𝑥 squared plus two 𝑥 minus 15.

And we’re now able to solve this equation by factoring 𝑥 squared plus two 𝑥 minus 15. We know that there’s going to be an 𝑥 in each of our binomials. And then we want two numbers that multiply to make negative 15 and add to make two. Those numbers are five and negative three. So 𝑥 squared plus two 𝑥 minus 15 is equal to 𝑥 plus five times 𝑥 minus three.

Our last job is to solve this equation. Now for the product of three numbers to be equal to zero, any one of those numbers must itself be equal to zero. So either 𝑥 plus one is equal to zero, 𝑥 plus five is equal to zero, or 𝑥 minus three is equal to zero. If we subtract one from both sides of our first equation, we get 𝑥 equals negative one. Solving our second equation, and we get 𝑥 equals negative five. And solving our last equation, we get 𝑥 equals three. Now we already knew that 𝑥 equals negative one was a root to the equation. And so the other roots of 𝑓 of 𝑥 are 𝑥 equals negative five and 𝑥 equals three.

In our final example, we’ll look at how to find a set of zeros of a quartic function or a polynomial function whose order is four.

Find the set of zeros of the function 𝑓 of 𝑥 equals 𝑥 to the fourth power minus 17𝑥 squared plus 16.

Remember, the zeros of a polynomial function are the values of 𝑥 that make that function equal to zero. And so to find the zeros of our function, we need to solve the equation 𝑥 to the fourth power minus 17𝑥 squared plus 16 equals zero. And one technique we have to find the set of zeros is to factor the expression.

So how do we factor 𝑥 to the fourth power minus 17𝑥 squared plus 16? Well, it’s a little bit tricky to spot, but this does look a little bit like a quadratic function. We’re going to perform a substitution. And we’re going to let 𝑦 be equal to 𝑥 squared. And then if we consider 𝑥 to the fourth power as being equal to 𝑥 squared squared, 𝑥 to the fourth power can be written then as 𝑦 squared. Similarly, negative 17𝑥 squared can be written as negative 17𝑦. So our equation becomes 𝑦 squared minus 17𝑦 plus 16 equals zero.

We now have a rather nice-looking quadratic that we can solve by factoring. We’re going to have two binomials at the front of which we must have a 𝑦. We then need to find two numbers that have a product, they multiply to make 16, and a sum of or they add to make negative 17. Those numbers are negative one and negative 16. Remember, a negative multiplied by a negative is a positive. So our equation is 𝑦 minus one times 𝑦 minus 16 equals zero. And for the product of these binomials to be equal to zero, we can say that either 𝑦 minus one must be equal to zero or 𝑦 minus 16 must be equal to zero. And if we solve as normal, we see that 𝑦 must be equal to one or 16.

But of course, we were trying to find the zeros of our function. And we said those are the values of 𝑥 that make 𝑓 of 𝑥 equal to zero. So we’re going to now replace 𝑦 with our original substitution, 𝑥 squared. And so 𝑥 squared is equal to one or 𝑥 squared is equal to 16. We’ll solve by taking the square root of both sides of each equation remembering, of course, to take both the positive and negative square root of one and 16. And so we get 𝑥 equals positive or negative one and 𝑥 equals positive or negative four. Now, we want to write this using set notation. And so we use these squiggly brackets. The set of zeros of the function 𝑓 of 𝑥 is the set containing the elements negative four, negative one, one, and four.

We’ll now summarize the key points from this lesson. Firstly, we saw that the zeros of a function are the values of 𝑥 that make the function equal to zero. We also call this finding the roots of an equation. And we also saw that if a function is factorable, we’re going to factor that polynomial to help us solve 𝑓 of 𝑥 equals zero.

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