What are the possible number of positive, negative, and complex zeros of #f(x) = –3x^4 – 5x^3 – x^2 – 8x + 4#?

Question

Lincoln Anderson · Accepted 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...

#f(x) = -3x^4-5x^3-x^2-8x+4#

Since there is one change of sign, #f(x)# has one positive zero.

#f(-x) = -3x^4+5x^3-x^2+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) = -12x^3-15x^2-2x-8#

Newton's method can be used to find approximate solutions.

Pick an initial approximation #a_0#.

Iterate using the formula:

#a_(i+1) = a_i - f(a_i)/(f'(a_i))#

Putting this into a spreadsheet and starting with #a_0 = 1# and #a_0 = -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 #-3x^2+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:

#-3x^2+0.325x-4.343#
#Delta = b^2-4ac = 0.325^2-(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.

Elias Ackerman · Answer

undefined The possible number of positive, negative, and complex zeros of $f(x) = -3x^4 - 5x^3 - x^2 - 8x + 4$ can be determined using the rules of Descartes' Rule of Signs and the Fundamental Theorem of Algebra.

1. **Descartes' Rule of Signs**:
   - The number of positive real zeros is either equal to the number of sign changes in the coefficients of $f(x)$ or less than that by an even integer.
   - The number of negative real zeros is either equal to the number of sign changes in $f(-x)$ or less than that by an even integer.

For $f(x)$, there are 3 sign changes, so there are either 3 positive real zeros or 1 positive real zero (3-2=1). There are 2 sign changes in $f(-x)$, so there are either 2 negative real zeros or 0 negative real zeros (2-2=0).

2. **Fundamental Theorem of Algebra**:
   - The number of complex zeros (including both real and imaginary parts) is equal to the degree of the polynomial, which is 4 in this case.

Combining the information from Descartes' Rule of Signs and the Fundamental Theorem of Algebra, we can conclude:
- There are 3 or 1 positive real zeros.
- There are 2 or 0 negative real zeros.
- There are 4 complex zeros.

Please note that these are the possible numbers of zeros, and the actual number of zeros may vary based on the specific roots of the polynomial.