What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 − 5x3 − x2 − 8x + 4? A. Positive: 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0B. Positive: 1; negative: 3 or 1; complex: 2 or 0C. Positive: 3 or 1; negative: 1; complex: 2 or 0D. Positive: 4 or 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0

Answer:Look at changes of signs to find this has 1 positive zero, 1 or 3 negative zeros and 0 or 2 non-Real Complex zeros.Then do some sums...Explanation:f(x)=−3x4−5x3−x2−8x+4 Since there is one change of sign, f(x) has one positive zero.f(−x)=−3x4+5x3−x2+8x+4 Since there are three changes of sign f(x) has between 1 and 3 negative zeros.Since f(x) has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so f(x) has exactly 1 or 3 negative zeros counting multiplicity, and 0 or 2 non-Real Complex zeros.f'(x)=−12x3−15x2−2x−8 Newton's method can be used to find approximate solutions.Pick an initial approximation a0 .Iterate using the formula:ai+1=ai−f(ai)f'(ai) Putting this into a spreadsheet and starting with a0=1 and a0=−2 , we find the following approximations within a few steps:x≈0.41998457522194 x≈−2.19460208831628 We can then divide f(x) by (x−0.42) and (x+2.195) to get an approximate quadratic −3x2+0.325x−4.343 as follows: Notice the remainder 0.013 of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.Check the discriminant of the approximate quotient polynomial:−3x2+0.325x−4.343 Δ=b2−4ac=0.3252−(4⋅−3⋅−4.343)=0.105625−52.116=−52.010375 Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly 2 non-Real Complex zeros, 1 positive zero and 1 negative one.