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)=2x^5+11x^4+16x^3+15x^2+36x

Accepted Answer

Welcome back. I am so glad you're here. We are asked for the following polynomial function to determine all the complex zeros, right. The answer in exact form our polynomial function is F of X equals three X to the fifth plus four, X to the fourth plus seven X cubed plus 14 X squared plus eight X. Answer choice A is zero negative one with a multiplicity of two, one third plus square rut of 23 3rd. I and one third minus square rut of 23 3rd. I answer choice B is 01 with a multiplicity of 21 3rd plus square root of 23 3rd. I and one third minus square root of 23 3rd. I answer choice C negative one with a multiplicity of two, one third plus square at 23 3rd I and one third minus square 3rd. I and answer choice D zero with a multiplicity of two negative one with a multiplicity of two and one third minus square. It 23 3rd. I, all right. So we've got a polynomial with a degree of five. That's pretty big. And we want to figure out all of the zeros. And what we notice here, these are in descending order correctly from X to the fifth all the way down to just an X. But there is no constant term here. And that means that every term here could have an X factored out of it. So that's what we do. First, we take X factor that out and then we'd be left with three X to the fourth plus four X cubed plus seven X squared plus 14 X plus eight as our function. And so that gives us this X out in front, that is a factor and we set that X equals to 0 to 0 and we have one of our zeros is zero. So thank you X. And now we look at the rest of that inside of the parentheses and we want to figure out what its zeros are. And the way we do that when we've got this fourth degree here is to go to the rational zeros theorem. As we recall from previous lessons, we're going to figure out all the possible rational zeros by taking P which is all of the factors of the constant term. And the constant term here is eight. And we're going to divide those by Q which is all of the factors of the leading coefficient, which in this case is three. So what are all of the possible factors of our constant term? Eight? Well, it's a plus or minus eight which would be multiplied by A plus or minus one, but also a plus or minus four, which would be multiplied by a plus or minus two. How about all of the possible factors of Q of our leading coefficient three? Well, that would be a plus or minus three, which would be multiplied by a plus or minus one. And what we do with these PS and Qs is we're going to use the remainder theorem. We recall from previous lessons that we use the remainder theorem to set up synthetic division to test out all of our possible rational zeros to see if we get a remainder of zero. And that means that it is one of our zeros. So we set this up with the coefficients and the constant term from our polynomial. We'll have a three 47, 14 8 and then we would set this up testing out and we just use our guess and check, we could start with eight thirds, eight divided by three. However, this is multiple choice and we see that we've got negative one and one as the only possible real rational zeros that we need that could possibly be answers. So for answer choice A they start with negative one. So let's test out negative one. Now we do the synthetic division, we bring down this three and will then multiply negative one, multiplied by three is a negative three. Then we add four plus negative three is a positive one multiply negative one multiplied by one is a negative one add seven plus negative one is six, multiply negative one multiplied by six is negative six. Add 14 plus negative six is eight multiply negative one multiplied by eight is negative eight, add eight plus negative eight is zero. Thank you. Remainder theorem. We have a remainder of zero which tells us that negative one is another of our zeros. And now we have this new polynomial with 316 and eight as our coefficient and constant term. And again, we can test out with the remainder theorem. We have the same PS and QS P s of the factors of eight and QS of the factors of three and we have the same multiple choice. So we can test negative one again see if that comes up with anything. So doing this synthetic division, bring down the three multiply negative one multiplied by three is negative three, add one plus negative three is negative two, multiply negative one multiplied by negative two is a positive two, add six plus two is positive eight, multiply negative one multiplied by eight is negative eight, add eight plus negative eight is zero. Yes, the Q remainder theorem negative one. We've got another one as one of our zeros. So now we have got a three X squared minus two X plus eight as our polynomial that's all equal to zero. Well, now that it's quadratic, we could see if we could factor it. This one is not easily factor. So we can pull out our good friend the quadratic formula to solve for the remainder of our zeros. We recall from previous lessons, the quadratic formula is negative B plus or minus the square root of B squared minus four, multiplied by a multiplied by C all divided by two A. And in this case, our A coefficient of our X squared term is three B, the coefficient of our X term is negative two and C our constant term is eight. So now we can plug these in. So negative B that's a negative negative two, that's a positive two plus or minus the square root of B squared. That's negative two squared. That's a positive four minus four, multiplied by A A is three multiplied by C C is eight all divided by two multiplied by A A is three. So two multiplied by A is six. Now let's simplify inside of the radical. Doing the multiplication negative four, multiplied by three is negative 12, multiplied by eight is a negative 96. So inside of the radical, we have four minus 96 outside of the radical. In front of that, we have two plus or minus. And that's all in the numerator divided by six. So we can simplify in the radical further. So let's bring over in the numerator two plus or minus. And in the radical four minus 96 will give us a negative 92 all still divided by six. Now taking a look in that radical a little bit more, we have the square root of negative which is the same as breaking that down into its factors a little bit. We have a negative one multiplied by four, multiplied by 23. We know that the square root of negative one is I and the square root of four is two. So we have a two I that we can pull out that's multiplied by the square root of 23 that 23 cannot come outside of the radical. So in our numerator, we have two plus or minus two I square at 23 all divided by six. Now let's do that division by six. Simplify this a little bit. So two divided by six, that's 2/6 which simplifies to one third. And that is plus or minus two, I divided by six. That's one third I, and that's still multiplied by the square root of 23. And we want to get this into the standard format of A plus B I for our complex number. So we'll bring that square root of 23 back to the numerator of one third. So this will be, we'll have one third plus or minus the square it of 23 3rd. I and those are all of our zeroes here. So looking at our answer choices, we have a zero as a zero. And then we have negative one with a multiplicity of two and then we have one third plus and minus the square root of 23 3rd. I, and this matches with answer choice. A well, Dan, we'll catch you on the next one.

Struggling with College Algebra?

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=2x^5+11x^4+16x^3+15x^2+36x

Zeros of Polynomial Functions practice set