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I think what you're trying to ask is how to factor
x4 + 7x2 + 6
The first thing to notice is the unusual exponents (4 and 2 instead of 2 and 1). What we can do to make it more familiar is to treat x2 as the variable rather than just x. So we have:
(x2)2 + 7x2 + 6
If that looks too complicated, it might be easier to set x2 equal to another variable, say y:
y2 + 7y + 6, where y = x2
Once it's in your preferred form, it's just like what you're used to. You need to find two numbers that multiply to 6 and add to 7. 2 and 3 doesn't work, but 1 and 6 does:
y2 + 7y + 6 = (y + 1)(y + 6)
If you used y, remember to put it back in terms of the original problem, which used x.
x4 + 7x2 + 6 = (x2 + 1)(x2 + 6)
It's impossible to factor further without getting into imaginary numbers.
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