Question 580002
I can factor out an {{{ x^2 }}}
{{{ x^2*( x^4 - 14x^2 + 49 ) }}}
Now I can make the substitution {{{ x^2 = z }}}
{{{ z*( z^2 - 14z + 49 ) }}}
To factor this completely, I 
complete the square on the 2nd factor:
{{{ z^2 - 14z + (14/2)^2 = -49 + (14/2)^2 }}}
{{{ z^2 - 14z + 49 = -49 + 49 }}}
{{{ ( z - 7 )^2 = 0 }}}
{{{ z - 7 = 0 }}}
This is called a double root, so the factorization is:
{{{ z*( z - 7 )^2 }}}
Now substitute {{{ z = x^2 }}}
{{{ x^2*( x^2 - 7 )^2 }}}
This can be factored 1 more step:
{{{ x^2*( x + sqrt(7) )^2( x - sqrt(7) )^2 }}}
Hope I got it
Here's a plot of the equation:
{{{ graph( 500, 500, -4, 4, -2, 16, x^6 - 14x^4 + 49x^2 ) }}}