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

Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=x^3+2x^2+x-10

Accepted Answer

Hello, everyone. We are asked for the given function by the various potential scenarios for the count of positive negative and non real complex zeros. The function we are given is F of X equals 12 X cubed plus three X squared plus 71 X minus 110. We have four answer choices with varying amounts of positive real zeros, negative real zeros and non-real complex zeros. The first thing I want to identify with the function we are given is how many zeros we should have total. We can find this by looking at the degree of the function. In this case, our highest exponent is a cube. So this is degree three, which means we should have a total when we're done of three zeros. So starting with the positive zeros to find our positive real zeros, we count the number of sign changes in the F of X function looking at my F of X function. My first term is a positive, second term is a positive, third term is a positive fourth term is a negative. This means there is one sign change. So at this point, we have a possibility of one positive real zero. Next thinking about my negative real zeroes to find negatives, we want to count how many sign changes in the F of negative X function. So we're gonna plug negative X in for X. So F of negative X would equal multiplied by negative X cubed plus three, multiplied by negative X squared plus 71 multiplied by negative X minus 110. So a negative cubed would still be a negative. So negative X cubed multiplied by 12 would give us negative 12 X cubed negative X squared and negative squared is a positive. So this would remain a positive three X squared, negative X multiplied by 71 would give us a negative 71 X and then minus 110. So again, let's go look at the signs, we start with a negative, then we have a positive and then our third term is a negative. Fourth term is a negative. So the sign changes from term one to term two and from term two to term three. So it changes twice. So we have two sign changes and recall that when we have an amount of zeros that is two or higher, we subtract a factor of two to account for any complex numbers that could be there. So we could have two negative real zeros or zero, negative real zeros. I'm gonna write all of my scenarios down right here. So we can find how many complex zeros there are So positives, we said there is one negatives, we said it is two or zero. So now to find complex recall that we need a total of three zeros regardless of the scenario. So if we have one positive two negative, that totals three, so we'd have zero complex zeros. However, we have one positive and zero negatives. In that case, we would have two non-real complex zeros. So our scenarios are one positive real zero, two or zero, negative real zeros and zero or two non-real complex zeros. And this matches with answer choice C have a nice day.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=x^3+2x^2+x-10

Key Concepts

Fundamental Theorem of Algebra

Descarte's Rule of Signs

Complex Zeros