# 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 #?

##### 1 Answer

#### Answer:

This quartic has

#### Explanation:

By the rational root theorem, any *rational* zeros of

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#

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:

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, -8#

That leaves

Trying each of the possible rational zeros in turn, we find that *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+-0.86572i#

#x_(3,4) ~~ -0.60384+-0.51093i#

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