In Exercises 33–38, use Descartes's Rule of Signs to determine...

Question

In Exercises 33–38, use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=2x⁴−5x³−x²−6x+4

Accepted Answer

Hello. Today we're going to be using. Discard this rule of science on our given function to identify the positive number of positive and negative real zeros. So the function is F of X is equal to six X to the fourth minus X cubed minus four X squared minus two X plus nine. Now we need to go ahead and identify the number of sign changes that are in F of X. And that's going to give us the amount of positive zeros for the function. So our leading coefficient is positive six. And the next and it leads into the next term which is negative X. So that's one sign change right there because we go from positive to negative. Then for the 2nd and 3rd term we go from negative to negative. So there's no sign change there. Then from the third to fourth term we have negative two negative. So there's no sign change. And then from the fourth to fifth term we have negative to positive and that's another sign change. So for F of X there's a total of two sign changes. And that means that there's going to be two or positive zeros. So we found the amount of possible positive zeros. But now we need to find the amount of possible negative zeros. And in order to do that we need to look at f of negative X for a given function. So we're going to replace every X with negative X. And that's going to give us six times negative X to the fourth power minus negative X cubed minus four times negative X squared minus two times negative X. And then plus nine. And now we need to go ahead and simplify. So we have negative X rays to an even power and that's going to produce a positive value. So we have six X to the fourth on the next term we have negative x rays to an odd power and that's going to produce a negative value. But keep in mind we have this negative sign on the outside. So we get plus X cubed negative X squared is negative X times negative X. And that produces a positive value. So negative four times X squared is going to give us negative four X squared negative two times negative X is going to give us a positive value which is plus two X. And then finally we have plus nine at the end. Now let's go ahead and check the sign changes for the first term which is positive. Six to the next term which is positive X cubed. There's no sign change because we go from positive to positive for the 2nd and 3rd term, we go from positive to negative. So that's one sign change right there. And then from the third to the fourth term we go from negative to positive. So that's another sign change. And then finally from the fourth to the fifth term we have positive to positive. So there's no sign change and that means that for f of negative X. There's a total of two sign changes and that means that there's going to be two or zero negative zeros. So we just figured out the possible amount of positive and negative zeros. There's two or zero positive zeros and there's two or zero negative zeros. With that being said, that's gonna be the amount of zeros or possible zeros forgiving function. And the answer to this problem is going to be a So I hope this video helps you in understanding how to use this card to rule of signs, and I'll go ahead and see you all in the next video.

In Exercises 33–38, use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=2x⁴−5x³−x²−6x+4

Key Concepts

Descartes's Rule of Signs

Polynomial Functions and Their Zeros

Evaluating f(-x) for Negative Zeros

In Exercises 33–38, use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=2x4−5x3−x2−6x+4

Key Concepts

Descartes's Rule of Signs

Polynomial Functions and Their Zeros

Evaluating f(-x) for Negative Zeros

In Exercises 33–38, use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=2x⁴−5x³−x²−6x+4