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 0 b. positive: 1; negative: 3 or 1; complex: 2 or 0 c. positive: 3 or 1; negative: 1; complex: 2 or 0 d. positive: 4 or 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0

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...

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:

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:

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.

answer: yes, he is correct

step-by-step explanation: 10x-1=59

                                              1-1=0     59+1=60

60/10=x=6

not solvable

step-by-step explanation:

we need more information, or at least an