SOLUTION: Example: Please help me solve this equation: {{{ x^4+2x^3+3x^2+2x+1=0 }}}.

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Question 113507: Example: Please help me solve this equation: +x%5E4%2B2x%5E3%2B3x%5E2%2B2x%2B1=0+.
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve this equation: +x%5E4%2B2x%5E3%2B3x%5E2%2B2x%2B1=0+.
----------------
The only possible rational roots are +1 and -1
But neither f(1) nor f(-1) is zero.
So, the equation has no rational solutions.
==============
Cheers,
Stan H.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

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


Example:  Please help me solve this equation: 

+x%5E4%2B2x%5E3%2B3x%5E2%2B2x%2B1=0+


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

(__x² + __x + __)(__x² + __ + __}

We know that x%5E4 factors as x%5E2 times x%5E2,
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:

+%28x%5E2+%2B+Ax+%2B+1%29%28x%5E2+%2B+Bx+%2B+1%29+=+0  

x%5E4+%2B+Bx%5E3+%2B+x%5E2+%2B+Ax%5E3+%2B+ABx%5E2+%2B+Ax+%2B+x%5E2+%2B+Bx+%2B+1+=+0

x%5E4+%2B+%28A%2BB%29x%5E3+%2B+%282%2BAB%29x%5E2+%2B+%28A%2BB%29x+%2B+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%5E4%2B2x%5E3%2B3x%5E2%2B2x%2B1=0+

we must have  

A%2BB+=+2 and 2%2BAB+=+3, or AB+=+1

Solving the first for B, B+=+2-A

Sunstituting into AB+=+1

A%282-A%29+=+1
2A-A%5E2+=+1
0+=+A%5E2-2A%2B1
0+=+%28A-1%29%28A-1%29
So A+=+1, and since B+=+2-A
B+=+2-1+=+1

So we have now factored te left side of

+x%5E4%2B2x%5E3%2B3x%5E2%2B2x%2B1=0+

as

%28x%5E2%2Bx%2B1%29%28x%5E2%2Bx%2B1%29+=+0

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

x%5E2%2Bx%2B1=0

Solve that by the quadratic formula and get

x = -1%2F2%2Bsqrt%283%29%2F2i and x = -1%2F2-sqrt%283%29%2F2i

Each solution has multiplicity 2.

Edwin