How do you use the rational roots theorem to find all possible zeros of # P(x) = 3x^4 + x^3 + 2x^2 - 2#?

1 Answer
Aug 8, 2016

Answer:

This quartic has no rational zeros, but we can find approximations:

#x_1 ~~ 0.694795#

#x_2 ~~ -0.800431#

#x_(3,4) ~~ -0.113849+-1.08894i#

Explanation:

#P(x) = 3x^4+x^3+2x^2-2#

By the rational root theorem and rational zeros of #P(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-2# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-2/3, +-1, +-2#

None of these works, so #P(x)# has no rational zeros.

As a quartic, it is possible to solve #P(x)# algebraically, but typically for a quartic, it gets very messy. It is much simpler to find approximations to the zeros using a numerical method such as Durand-Kerner.

Here's a C++ program for this quartic...

enter image source here

Hence we can find approximations:

#x_1 ~~ 0.694795#

#x_2 ~~ -0.800431#

#x_(3,4) ~~ -0.113849+-1.08894i#