Use Descartes's Rule of Signs to determine the possible number...

Question

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=5x³−3x²+3x−1

Accepted Answer

Hey everyone here we are asked to identify the possible number of positive and negative real zeros. Using the cards rule of signs for the function F of X is equal to eight X cubed minus six X squared plus nine x minus three. Now to start with the number of possible positive real zeros, we need to make sure that the X. Input of our function is positive which it is in our given function. So we don't need to make any changes. But next we need to note the number of sign changes within the function when we have a negative earth sorry positive input. So we can see that we start out with positive A. X cubed. Then we go to negative six X squared. So this is one sign change. And then we go from negative to positive nine X. Which will be our second sign change. And then we go from positive to negative three which will be our third sign change. So we have a total of three side changes. And with this we can now note that our number of positive real zeros will be a maximum of three or a minimum of one. And now we can move on to finding our number of negative real zeros. And so we need to make sure that the input this x value is negative. So we take F of negative X. And inputting negative X. We have eight times the quantity of negative X cubed minus six times the quantity of negative X squared plus nine times the quantity of negative x minus three. And now again we need to note the number of sign changes. So in this case we have negative X cubed which makes this first value here, this first term negative and then we have negative X squared which just gives us negative six X. So we go from a negative to a negative. So there's no sign change here and then we have not negative X. Which gives us another negative term. So there's not a sign change here either. And then we go from negative to negative three. So there's not a sign change here either. So we don't have any sign changes for a negative input. So now we can note that the number of negative real zeros will also just be zero. So with this we're left with answer choice B where we have three or one positive real zeros and no negative real zeros. Thanks for watching. I hope you found this video helpful. And I'll see you again next time.

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=5x³−3x²+3x−1

Key Concepts

Descartes's Rule of Signs

Polynomial Functions and Their Zeros

Evaluating f(-x) to Find Negative Zeros

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=5x3−3x2+3x−1

Key Concepts

Descartes's Rule of Signs

Polynomial Functions and Their Zeros

Evaluating f(-x) to Find Negative Zeros

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x)=5x³−3x²+3x−1