How do you find all the zeros of #f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19#?

1 Answer
Jul 31, 2016

Answer:

Use Durand-Kerner method to find numerical approximations:

#x_1 ~~ -4.61495#

#x_(2,3) ~~ 1.34883+-0.412784i#

#x_(4,5) ~~ -0.374682+-0.74117i#

Explanation:

#f(x) = -3x^5-8x^4+25x^3-8x^2+x-19#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-19# and #q# a divisor of the coefficient #-3# of the leading term.

So the only possible rational zeros are:

#+-1/3, +-1, +-19/3, +-19#

Evaluating #f(x)# for each of these values we find none works. So #f(x)# has no rational zeros.

#color(white)()#
In common with most quintics, the zeros of this #f(x)# are not expressible in terms of #n#th roots. There are no algebraic solutions in terms of elementary functions, including trigonometric ones.

About the best we can do is use a numeric method such as Durand-Kerner to find approximations to the zeros:

Suppose the #5# zeros are: #p, q, r, s, t#.

Choose initial approximations for these zeros as follows:

#p_0 = (0.4+0.9i)^0#

#q_0 = (0.4+0.9i)^1#

#r_0 = (0.4+0.9i)^2#

#s_0 = (0.4+0.9i)^3#

#t_0 = (0.4+0.9i)^4#

Then iterate using the formulas:

#p_(i+1) = p_i-(f(p_i))/((p_i-q_i)(p_i-r_i)(p_i-s_i)(p_i-t_i))#

#q_(i+1) = q_i-(f(q_i))/((q_i-p_(i+1))(q_i-r_i)(q_i-s_i)(q_i-t_i))#

#r_(i+1) = r_i-(f(r_i))/((r_i-p_(i+1))(r_i-q_(i+1))(r_i-s_i)(r_i-t_i))#

#s_(i+1) = s_i-(f(s_i))/((s_i-p_(i+1))(s_i-q_(i+1))(s_i-r_(i+1))(s_i-t_i))#

#t_(i+1) = t_i-(f(t_i))/((t_i-p_(i+1))(t_i-q_(i+1))(t_i-r_(i+1))(t_i-s_(i+1))#

Keep iterating until the values are stable to the desired accuracy.

With this method, I found the following approximations for the zeros:

#x_1 ~~ -4.61495#

#x_(2,3) ~~ 1.34883+-0.412784i#

#x_(4,5) ~~ -0.374682+-0.74117i#

Here's a sample C++ program that implements the algorithm for the example quintic:

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