Determine the different possibilities for the numbers of posit...

Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = - 2 x^{5} + 10 x^{4} - 6 x^{3} + 8 x^{2} - x + 1$

Accepted Answer

Welcome back. I am so glad you're here. We are asked for the numbers of positive negative and non-real complex zeros. Find out all possibilities. And we are given this polynomial of F of X equals negative 13 X to the fifth plus five, X to the fourth minus six X cubed plus four, X squared minus nine X plus 36. Our answer choices are answer choice. A positive real zeros 42 or zero, negative real zeros, one non-real complex zeros or four answer choice B positive real zeros two or zero, negative real zeros, one non-real complex zeros two or four. Answer to ac positive real zeros zero, negative real zeros 53 or one non-real complex zeros 02 or four. In answer choice D positive real zeros 53 or one negative real zeros zero and non-real complex zeros 02 or four. All right. So to figure out all the different types of zeros, we have, we have to recall from previous lessons, Descartes rule of signs. And what we're going to do is we'll start off with the positive real zeroes. And we are going to see how many variations of signs we have. So as long as all of our X are in the correct degree order, so starting with the highest going down to the lowest. And in this case, they are, we're going to look for sign changes for their coefficients. So the first one is negative and then the next one is positive. Well, there's one and then from five X to the fourth, from two negative six X Q, there's another sign change. There's two from negative six X cubed to positive for X squared. There's another sign change three from positive four X squared to negative nine X. That's another one. There's four and from the negative nine X to the positive there's another sign change. So there are five variations of signs. So what does that tell us? Well, we look at our degree of this polynomial and it's five, that's the maximum amount of zeros that we're going to have. And we have five sign changes. So we have a possibility of five positive real zeros. However, it could be five or it could be less than that by a positive even integer. So a positive even integer less than five, that's two, less than five. So it could be three positive real zeros and we can have two, less than 32, less than three is one. So the number of positive real zeros could be five or three or one. Now let's work on the negative real zeros. The way we figure out the negative real zeros is we plug in F of negative X simplify and see how many sign changes we have with that. So we'll have a negative 13, negative X raised to the fifth plus five multiplied by negative X to the fourth minus six, multiplied by negative X cubed. Oops that cubed goes outside the parentheses plus four multiplied by a negative X squared minus nine multiplied by negative X. And then we do have the plus 36. Simplifying this, that negative X to the fifth will be a negative X to the fifth. Multiplying that by negative 13, we're going to have a positive 13 X to the fifth. Our negative X to the fourth will be a positive X to the fourth. So plus five X to the fourth, negative X cubed is negative X cubed multiplied by negative six. It's going to be a positive six X cubed negative X squared is a positive X squared. So this will be plus four X squared negative nine multiplied by negative X is a positive nine X and then our plus 36. Now how many sign changes do we have? Well, everything here is positive, we don't have any sign changes. So there are zero negative real zeros. That's how many are possible for the negative real zeros. Now let's work on the non real complex zeros. For the non-real complex zeros, we need to have two conditions non-real complex zeros. Must occur in pairs. That's the plus or minus with those I s. And they also need to add up to the degree when combined with the positive real zeros and the negative real zeros. So essentially, we're going to have the positive real zeros plus the negative real zeros plus the non-real complex zeros will add up to five. And we've got three different possibilities for the positive real zeros. The first one is going to be five. So we have five plus for the negative real zeros, it's always going to be zero. And then how, what are our non-real complex zeros? If we have five plus zero plus something equals five? Well, that's going to be zero. So one possibility is zero for our non-real complex zeros. But what if for our positive real zeros, we have the three, three plus zero plus what gives us five? That's two. So two is another possibility. And that makes sense because they need to come in pairs. And then for the last possibility, if our positive real zeroes are the one, how many non-real complex zeros would we have? Then one plus zero plus what equals five? That's four. And that makes sense because they would be able to come in pairs. So there are our answers. We look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = - 2 x^{5} + 10 x^{4} - 6 x^{3} + 8 x^{2} - x + 1$

Key Concepts

Fundamental Theorem of Algebra

Descartes' Rule of Signs

Complex Conjugate Root Theorem

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=−2x5+10x4−6x3+8x2−x+1ƒ(x)=-2x^5+10x^4-6x^3+8x^2-x+1

Key Concepts

Fundamental Theorem of Algebra

Descartes' Rule of Signs

Complex Conjugate Root Theorem

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = - 2 x^{5} + 10 x^{4} - 6 x^{3} + 8 x^{2} - x + 1$