Question 113774
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find all real and imaginary roots

{{{x^4+3x^3+x^2+4=0}}}

Possible rational roots, if it has any,
are ±1,±2,±4

DesCartes rule of signs tells us that
there are no positive roots, so that
narrows down the possible rational
roots to -1, -2, and -4

Try {{{x = -1}}}, or {{{x + 1 = 0}}}

So we try dividing synthetically 
by {{{x + 1}}}. But we must first put in a
zero term since the original
equation does not contain a term
in x.  So we rewrite the original
equation, showing all coefficients: 

{{{1x^4+3x^3+1x^2+0x+4=0}}}

-1|1  3  1  0  4
  |<u>  -1 -2  1 -1</u>
   1  2 -1  1  3

So we see that we don't get 0 as a
remainder, the bottom right number,
but rather 3, so -1 is not a root.

So we try {{{x = -2}}} to see if it is a root.
Dividing synthetically by {{{x + 2}}} 

-2|1  3  1  0  4
  |<u>  -2 -2  2 -4</u>
   1  1 -1  2  0

So {{{x = -2}}} is a root, since we get 0
as a remainder.  The 4 numbers to the
left of the zero represent the quotient
when we divided by {{{x + 2}}}, so we have now
factored the polynomial equation as

{{{x^4+3x^3+x^2+4=0}}}

{{{(x+2)(x^3 + x^2 - x + 2) = 0}}}

So we now try to find a rational root of

{{{x^3 + x^2 - x + 2 = 0}}}

DesCartes rule of signs tells us this
has possible roots ±1 and ±2.  But since
the original equation has no positive roots
this can only have rational roots -1 and -2.
We know -1 is not a root because it was not
a root of the original equation, so we can
only try {{{x = -2}}} again as a root of multiplicity
2.  {{{x = -2}}} is equivalent to {{{x + 2 = 0}}}. so we
divide the new polynomial also by {{{x + 2}}}:

-2| 1  1 -1  2
  |<u>   -2  2 -2</u>
    1 -1  1  0

Yes, we get 0 as a remainder, so {{{x = -2}}} is
a root of multiplicity 2. So now our
factorization of the original polynomial
is now:

{{{x^4 + 3x^3 + x^2 + 4 = 0}}}
{{{(x + 2)(x^3 + x^2 - x + 2) = 0}}}
{{{(x + 2)(x + 2)(x^2 - x + 1) = 0}}}

We set each factor = 0,

{{{x + 2 = 0}}} gives root {{{x = -2}}}
{{{x + 2 = 0}}} gives root {{{x = -2}}}, again

{{{x^2 - x + 1 = 0}}}
This does not factor, so must be solved by the 
quadratic formula:

Its roots are 

{{{(-(-1) +- sqrt((-1)^2 - 4(1)(1)))/(2(1))}}}

or, simplifying:

{{{(1 +- sqrt(1 - 4))/2}}}

{{{(1 +- sqrt(-3))/2}}}

{{{(1 +- i*sqrt(3))/2}}}

So there are four roots, counting multiplicities
of roots as though they were separate roots:

{{{x = -2}}}, {{{x = -2}}}, {{{x=(1 + i*sqrt(3))/2}}}, {{{x = (1 - i*sqrt(3))/2}}}


Edwin</pre>