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
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#
Rational Roots Theorem
By the rational roots theorem, any rational zeros of
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.
Descartes' Rule of Signs
The signs of the coefficients of
The signs of the coefficients of
Turning points
Let us first find where the turning points of
#f'(x) = 16x^3+9x^2-4x+5#
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 1296+4096-14580-172800-51840 = -233828#
Since
So
Therefore
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
There are no rational zeros.
graph{4x^4+3x^3-2x^2+5x-12 [-5, 5, -20, 20]}
Where do we go from here?
to be continued...
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
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++ ...
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#