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. $ƒ (x) = 7 x^{5} + 6 x^{4} + 2 x^{3} + 9 x^{2} + x + 5$

Accepted Answer

Welcome back. I am so glad you're here for the numbers of positive negative and non-real complex zeros. We are asked to find out all possibilities. Given this polynomial function F of X equals 13 X to the fifth plus eight, X to the fourth plus seven X cubed plus 17 X squared plus five, X plus 15. 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 choice C positive real zeros zero, negative real zeros 53 or one non-real complex zeros 02 or four. And answer choice D positive real zeros two or zero, negative real zeros three or one non-real complex zeros zero or four. All right, to figure out how many zeros we've got, we recall from previous lessons, Descartes rule of signs. So we're going to look for variations of signs in our polynomial as long as everything is in proper degree order. And here it is, we start with the highest degree which is five and then it goes down in descending order until we have no XS. And noting that our highest degree is five, that means we're gonna have a maximum of five of any type of zeros. And so what we do is we look for sign changes of the coefficients. So from 13 X to the fifth, that's positive. Well, eight X to the fourth is also positive seven X cubed is also positive 17 X squared is also positive five X is also positive and 15 also positive. There are no sign changes. So for the positive real zeroes, there are zero possibilities because there are zero sign changes. Now let's look at the negative real zeros from Descartes rule of signs. For the negative real zeros, we are now going to plug in F of negative X and see how many sign changes we have. So we'll plug in negative X and then we'll simplify, we'll start off with 13 multiplied by negative X raised to the fifth plus eight, multiplied by negative X to the fourth plus seven multiplied by negative X cubed plus 17, multiplied by negative X squared plus five, multiplied by negative X and plus 15. Now simplifying that negative X to the fifth will be a negative X to the fifth multiplying that by the 13, that will be a negative 13 X to the fifth, that negative X to the fourth will be a positive X to the fourth. So this will still be plus eight X to the fourth negative X cubed will be a negative X cubed multiplying that by a positive seven will give us a negative seven X cubed, negative X squared is positive X squared. So this will remain plus 17 X squared. Five multiplied by negative X is a negative five X and then keeping our plus 15. So now sign changes our negative 13 X to the fifth is negative. Going to a positive eight X to the fourth. There's one from positive X to the fourth to a negative seven X cubed. That's two from negative seven X cube to positive 17 X squared. There's three from positive 17 X squared to a negative five X. There's 1/4 sign change and from negative five X to a positive 15, there's 1/5 sign change. So the maximum possibility of negative real zeros is five. But recalling Descartes rule of signs, we also know that it could be less than that by a positive even integer, positive even integer here is two. So two less than five is three. We could also have three but we can get another positive even integer out of that. We can take two less than three, which is one. So the negative real zeros possibilities are five or three or one. Now let's talk about the non-real complex zeros or the non-real complex zeros. There are two things to note they need to occur in pairs because of that plus or minus with the I S. So those will always occur in pairs and they need to add up when you take the positive real zeros plus the negative real zeros plus the non-real complex zeros, they're going to add up to the degree of this polynomial which we know is five. So for the positive real zeros, that's always zero. For this problem, the non-real zeros changes. So if the non-real zeros if or excuse me, not non-real, the negative real zeros, that one changes. If there are five of the negative real zeros, how many non-real complex zeros will there be if they have to add up to five? Or in other words, zero plus five plus what equals five? Well, in that case, the non-real complex zeros would be zero. But what if our negative real zeros? What if there's three of them? Well, zero plus three plus what gives us five? That's two. And that makes sense because they have to occur in pairs. What if for the negative real zeros there's one of them. Well, zero plus one plus what gives us five? That's two or excuse me, it's four. And that makes sense because they need to occur in pairs. So the non-real complex zeros could be zero or two or four. And that's how many possible zeros we have. We look at our answer choices and this matches with answer choice. C 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. $ƒ (x) = 7 x^{5} + 6 x^{4} + 2 x^{3} + 9 x^{2} + x + 5$

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)=7x5+6x4+2x3+9x2+x+5ƒ(x)=7x^5+6x^4+2x^3+9x^2+x+5

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. $ƒ (x) = 7 x^{5} + 6 x^{4} + 2 x^{3} + 9 x^{2} + x + 5$