Question 113507
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Solution by Edwin
`
As Stanbon says, it has
no rational solutions, but
it can have irrational or 
complex solutions,as we'll
see!!</font>

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Example:  Please help me solve this equation: 

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


We try to see if we can factor that this 
as the product of two quadratics:

(__x² + __x + __)(__x² + __ + __}

We know that {{{x^4}}} factors as {{{x^2}}} times {{{x^2}}},
so we choose 1's for the coefficients of x².

1 factors as either 1×1 or (-1)×(-1).

Since all the terms are positive we chose +1's to go in the
last blanks. So we try to see if we can find A and B such that
the left side factors this way:

{{{ (x^2 + Ax + 1)(x^2 + Bx + 1) = 0}}}  

{{{x^4 + Bx^3 + x^2 + Ax^3 + ABx^2 + Ax + x^2 + Bx + 1 = 0}}}

{{{x^4 + (A+B)x^3 + (2+AB)x^2 + (A+B)x + 1 = 0}}}

For this to be equivalent to the original equation, the
coefficients of like powers must be equal

Since the original equation is

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

we must have  

{{{A+B = 2}}} and {{{2+AB = 3}}}, or {{{AB = 1}}}

Solving the first for {{{B}}}, {{{B = 2-A}}}

Sunstituting into {{{AB = 1}}}

{{{A(2-A) = 1}}}
{{{2A-A^2 = 1}}}
{{{0 = A^2-2A+1}}}
{{{0 = (A-1)(A-1)}}}
So {{{A = 1}}}, and since {{{B = 2-A}}}
{{{B = 2-1 = 1}}}

So we have now factored te left side of

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

as

{{{(x^2+x+1)(x^2+x+1) = 0}}}

We set each factor = 0, but since
the factors are the same, we only need
set one of them = 0:

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

Solve that by the quadratic formula and get

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

Each solution has multiplicity 2.

Edwin</pre>