Find all complex zeros of each polynomial function. Give exact va...

Question

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^6-9x^4-16x^2+144

Accepted Answer

Hello. Today we're going to be solving for the complex zeros of the following polynomial function. So what we are given is X to the sixth minus 25 X to the fourth minus 81 X squared plus 2025. Now, in order to solve for the complex zeros, we're going to need to set this polynomial function equal to zero, then solve for X. So we're gonna have X to the sixth minus 25 X to the fourth minus 81 X squared plus 2025 equal to zero. Now, we have to think to ourselves, what is the easiest way or what is the easiest approach in order to factor this polynomial function will notice that the function has a total of four terms. Since there are four terms, we can start off by factoring by grouping. So we're gonna group up the first two terms together and we're going to group up the last two terms together. Then what we want to do is determine a common factor that we can factor out from both of the groups. So if we take a look at the first group, we have X to the sixth minus 25 X to the fourth. Both of the terms have at least four multiples of X. So we can go ahead and factor out X to the fourth from the first group. And in the second group, both of the terms have coefficients of multiples of 81. So we can go ahead and factor out an 81 from the second group. However, we want this leading term to be positive after the factory. So let's go ahead and factor out negative 81 from the second group. Factoring out these values allows us to rewrite the function as X to the fourth times the quantity X squared minus minus 81 times the quantity X squared minus 25. Now notice that we have two groups of X squared minus 25. We can combine this into a single group by adding the leading coefficients together and adding the leading coefficients allows us to rewrite the function as X to the fourth minus 81 times the quantity X squared minus 25. Now, we can still further factor these groups in the first group X to the fourth can be rewritten as X squared raised to the power of two and 81 is a perfect square because can be rewritten as nine squared. So we can factor the first group by using the factoring of the difference of squares. And the same thing can be said with the second group X squared is the same thing as X raced to the power of two and 25 is a perfect square because it could be written as five squared. So if we factor both of these groups by using the difference of squares, we can rewrite the polynomial function as X squared minus nine times X squared plus nine times X minus five times X plus five. And that is going to equal to zero. Now, we're not done yet because there is one more quantity that we can factor further. And that is going to be the first quantity which is X squared minus nine. Again, X squared can be rewritten as X rates the power of two and nine is a perfect square because it could be rewritten as three squared. So we can factor X squared minus nine by using factoring of the difference of squares again. And that allows us to rewrite the polynomial as X plus three times X minus three times X squared plus nine times X minus five times X plus five equal to zero. So we have quite a bit of groups or quite a bit of factors of X to solve for. But using our zero property, we can go ahead and set each, each individual factor of X equal to zero in order to solve for the complex zeros. So doing this allows us to rewrite this equation as X plus three equal to zero. X minus three equal to zero, X minus five, equal to zero, X plus five, equal to zero and X squared plus nine equal to zero. Now, what we need to do is to solve for each of the equations in order to get the complex zeros for X. So in the first equation, we're just gonna subtract three to both sides. And that is going to give us X is equal to negative three. In the second equation, we're going to add three to both sides. And that is going to give us X as equal to positive three. In the third equation, we add five to both sides of the equation. And that is going to give us X is equal to positive five. In the fourth equation, we subtract five. And that is going to give us X is equal to negative five. And in the final equation, we have a bit of work to do. The first thing that we can do in order to solve for X is to subtract nine to both sides of the equation that is going to give us X squared is equal to negative nine. And in order to solve for X, we need to eliminate the exponent of two. And in order to do that, we can take the square root of both sides of the equation that is going to leave us with X is equal to the square root of negative nine. But recall that every time you have a negative sign or a negative number inside of a square root that is going to produce an imaginary value. So we can rewrite the square root of negative nine as the square root of nine, I. Now the square root of nine is going to simplify to give us three. But keep in mind that every time you take a square root of a number, it produces a positive and negative value. So that is going to give us X is equal to plus or minus three I. And what we have here is two more complex zeros. We get X is equal to positive positive three I and we get X is equal to negative three I. So what we have in total is six complex six complex zeros for X, we have X is equal to negative three, X is equal to positive three, X is equal to positive five, X is equal to negative five, X is equal to positive three I and X is equal to negative three I and this is going to be the total amount of zeros for X. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this process. And I'll go ahead and see you all in the next video.

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Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^6-9x^4-16x^2+144

Zeros of Polynomial Functions practice set