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)=4x^3-x^2+2x-7

Accepted Answer

Hello, everyone. We are asked for the given function to find 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 eight X cubed minus seven X squared plus 15 X minus 13. We have four answer choices all with slightly varying amounts of positive real zeros, negative real zeros and non real complex zeros. To start, we're gonna look at positive real zeroes. Recall that we can find the number of positive real zeros by counting how many times the sign changes on the coefficient in the F of X function. So looking at our F of X function, our first term is a positive second is a negative, third is a positive fourth is a negative. So how many times does the sign change? Well, it changes once from positive to negative, once from negative to positive and once from positive back to negative. So we have three sign changes. So this means we have three possible positive real zeros. But we also have to take into account that we have to subtract a multiple of two. So we could have three or we could have one because two of those might be imaginary or complex zeros. We don't really know yet. Next to find our negative real zeroes, we want to find the amount of sign changes for the F of negative X function. So using the same function, I'm gonna put a negative X in for X. So we'd have eight multiplied by negative X cubed minus seven multiplied by negative X squared plus 15, multiplied by negative X minus 13. Negative X cubed will result in a negative X negative X multiplied by eight is still gonna be a negative. So we have negative eight X cubed looking at the negative X squared, a negative squared is a positive. So that will not change the sign on the seven. So that's still minus seven X squared for 15 multiplied by negative X. That will change the sign on the 15 to a negative 15 X and minus 13 does not change. So looking at our signs, our first term is a positive. Oops, I'm sorry, it's a negative. Our second term is a negative. Third term is a negative. Fourth term is a negative. So counting our sign changes, there are zero sign changes. So this means we have possibly zero, negative real zeros. So there are none. So making a list again, we have positives, you could have three or one negative, we have zero. So next, we need to think about complex. Looking at the degree of the function, it has a degree of three, which means our answer should have three zeros. So if we had three positive zeros and zero negative zeros, this means there are no zeros left to be complex. So we'd have zero complex zeros. However, if we have one positive zero, no negative zeros, we need to find two more zeros. So then there'd be a possibility of two complex zeros because the amount of zeros we have has to total to the degree of the function which is three. So this is our answer. We would have three or one positive real zeros, zero, negative real zeros and zero or two complex real zeros. And that matches answer choice A have a nice day.

Struggling with College Algebra?

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

Zeros of Polynomial Functions practice set