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

1 Answer
Jan 7, 2016

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) = -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:

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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.