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

Accepted Answer

Hey, everyone here we are asked to determine all of the possibilities for the number of positive negative and non-real complex zeros. For our given function F of X is equal to two X to the sixth power minus 16 X to the fifth power plus 27 X Q minus five X squared plus 11 X plus one. Here we have four answer choice options. Answer choice. A positive real zeros four or two or zero, negative real zeros two or zero non-real complex zeros zero or two or four. Answer B positive real zeros two or zero, negative real zeros, one non-real complex zeros two or four. Answer C positive real zeros, one negative real zeros three or one non-real complex zeros one or three. And answer D positive real zeros two or zero, negative real zeros three or one and non-real complex zeros zero or four. So here looking at our given function, we can first determine the amount of possible positive real zeros by finding these sine variations of F of X. So here going from the first to the second term, we see we go from a positive to a negative. And so this is already one sign variation. Then from the second to the third term, we go from a negative to a positive. So we have a second sign variation. Then from the third to the fourth term, we go from a positive to a negative, which is our third sign variation. And then we go from the negative to a positive, from the fourth to the fifth term, which is our fourth sign variation. And lastly, we go from a positive to a positive. So there is no additional sign variation. And this tells us that the variation of F of X is equal to four. Well, this variation of four tells us that we could have four positive real zeros. Well, if we take this four and subtract two, we get two, which tells us we could also have just two, two positive real zeros. So we could have four or two. But if we take the four again and subtract four, we see we could also have zero positive real zeros. So we could have 42 or zero, positive real zeros. So now we can move on to finding the amount of possible negative real zeros. And we do this by checking the sine variation of F of negative X. So substituting in negative X into our original function, we see that our first term remains the same as two X to the six power since we have an keep in power. But then our second term since it is an odd power, the negative 16 X to the fifth power becomes positive 16 X to the fifth power. Then from our third term, we also have an odd power. So this positive 27 X cubed becomes negative 27 X. Then from our next term, we have negative five X squared which remains the same as again, it has an even power. Then from the next term we had positive 11 X which now becomes negative 11 X. And then our last term is constant. So it remains the same as just plus one. So now we can check the sine variation according to what we have for F of negative X. And we see from our first to the second term, we go from a positive to a positive. So there is no sign variation. Well, then we go from a positive to a negative. So there is one, one sign variation so far, then we go from a negative to a negative. So there is not an additional sign variation. Then we again go from a negative to another negative. And then lastly, we go from a negative to a positive. So we see we have a second sign variation. So here we can conclude or summarize that the variation when we have F of negative X is just two. So this tells us that one possibility is that we have two negative real zeros. But if we take this value of two, this possibility of two and just subtract two like we did previously, we get zero. So we see that we could either have two or we could have zero negative real zeros. And lastly for our complex roots, we have that our highest degree or our overall degree of the polynomial is six. So this tells us that we must have five complex zeros. So in summary, with all of our information that we gathered, we see that the possible number of positive real zeros of positive real zeros is 42 or zero. But the number of negative real zeros we found to be two or zero. And then lastly the number of possible non-real non real complex zeros, we have 02 or four. So with all of our information summarized the number of all of our possible combinations, we are left with answer choice. A where again, for the positive real zeros, they could be four or two or zero, negative real zeros could be two or zero and non-real complex zeros can be zero or two or four. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = 5 x^{6} - 6 x^{5} + 7 x^{3} - 4 x^{2} + x + 2$

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)=5x6−6x5+7x3−4x2+x+2ƒ(x)=5x^6-6x^5+7x^3-4x^2+x+2

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