What are all the possible rational zeros for #f(x)=4x^4+3x^3-2x^2+5x-12# and how do you find all zeros?

2 Answers
May 5, 2017

Answer:

The "possible" rational zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12#

The actual zeros are irrational and complex.

Explanation:

Given:

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

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-12# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12#

That's rather a lot of possibilities to try, so let us find out more about the nature of these zeros.

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Descartes' Rule of Signs

The signs of the coefficients of #f(x)# are in the pattern #+ + - + -#. With #3# changes of sign, that means that #f(x)# has #3# or #1# real positive zero.

The signs of the coefficients of #f(-x)# are in the pattern #+ - - - -#. With #1# change of sign, that means that #f(x)# has exactly #1# real negative zero.

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Turning points

Let us first find where the turning points of #f(x)# are to help guide our search. Turning points occur where the derivative #f'(x)# of #f(x)# is zero.

#f'(x) = 16x^3+9x^2-4x+5#

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=16#, #b=9#, #c=-4# and #d=5#, so we find:

#Delta = 1296+4096-14580-172800-51840 = -233828#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

So #f'(x)# has only one real zero and #f(x)# has only one turning point.

Therefore #f(x)# has one real negative zero and one real positive zero.

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Back to the rational possibilities...

Trying some of the possible rational zeros, we find:

#f(-2) = 10#

#f(-3/2) = -13.875#

#f(1) = -2#

#f(3/2) = 21.375#

Hence we can deduce that there is an irrational zero in #(-2, -3/2)# and one in #(1, 3/2)#.

There are no rational zeros.

graph{4x^4+3x^3-2x^2+5x-12 [-5, 5, -20, 20]}

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Where do we go from here?

to be continued...

May 6, 2017

Answer:

Use a numerical method (Durand-Kerner) to find approximations to the irrational zeros...

Explanation:

Given:

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

In my first answer, we found that the zeros are not rational.

We can find approximations to the zeros using an algorithm known as Durand-Kerner.

Suppose the #4# zeros are #p, q, r, s#.

Choose initial approximations:

#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#

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))#

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

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

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

This can be done using a computer program, such as this one written in C++ ...

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After a few iterations, the approximations settle down to give:

#p ~~ 1.0707#

#q ~~ -1.85404#

#r ~~ 0.0166701-1.22921i#

#s ~~ 0.0166701+1.22921i#