How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8 P(x)=12x4+x3+4x2+7x+8?

1 Answer
Aug 2, 2016

This quartic has 44 Complex non-Real zeros.

Explanation:

P(x) = 12x^4+x^3+4x^2+7x+8P(x)=12x4+x3+4x2+7x+8

By the rational root theorem, any rational zeros of P(x)P(x) are expressible in the form p/qpq for integers p, qp,q with pp a divisor of the constant term 88 and qq a divisor of the coefficient 1212 of the leading term.

That means that the only possible rational zeros are:

+-1/12, +-1/6, +-1/4, +-1/3, +1/2, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8±112,±16,±14,±13,+12,±23,±1,±43,±2,±83,±4,±8

In addition note that all of the signs of the coefficients are positive, so there are no changes of sign. So by Descartes rule of signs there are no positive zeros.

Reversing the signs on the terms with odd degree, we get the pattern of signs: + - + - ++++. With 44 changes of sign, that means that there are 44, 22 or 00 negative Real zeros.

So the only possible rational zeros are:

-1/12, -1/6, -1/4, -1/3, -1/2, -2/3, -1, -4/3, -2, -8/3, -4, -8112,16,14,13,12,23,1,43,2,83,4,8

That leaves 00, 22 or 44 non-Real Complex zeros, occuring in Complex conjugate pairs.

Trying each of the possible rational zeros in turn, we find that P(x) > 0P(x)>0 for all of them. So P(x)P(x) has no rational zeros.

Though it is possible to find the zeros analytically, it gets very messy. It is substantially easier to find approximations to the zeros using a numerical method such as Durand-Kerner to find:

x_(1,2) ~~ 0.56217+-0.86572ix1,20.56217±0.86572i

x_(3,4) ~~ -0.60384+-0.51093ix3,40.60384±0.51093i

See https://socratic.org/s/awGUf8Qf for more information on solving quartics using Durand-Kerner.