can this equation be factored further?
y= x^4+2x^3+4x^2+8x+16
Not in the real number system. If you plot the function, you will see the minimum is at x=-1.1 (approx) and y is positive. At no x does the function equal zero, so there are no real roots, which means, no factors.
You can do some moving around..
x^4+2x^3+4x^2+8x+16
x^3(x + 2) + 4x(x + 2) + 16
(x+2)(x^3+ 4x) + 16
x(x +2)(x^2+4) + 16
I don't personally see that as progress.
Not having real roots doesn't mean it can't be factored: x^4+2x^2+1=(x^2+1)^2 has no real roots, but it does factor.
In theory every polynomial over R factors as polynomials of degree 1 or 2, but finding the factors can be time consuming in practice.
True.
In this case, the equation y = x^4 + 2x^3 + 4x^2 + 8x + 16 does not appear to have any further factorization in the real number system. To confirm this, you can plot the function and observe that it does not intersect the x-axis, meaning there are no real roots. Consequently, there are no linear factors (of the form x - a) that can be factored out.
However, it is important to note that not having real roots does not necessarily mean that a polynomial cannot be factored further. For example, the equation x^4 + 2x^2 + 1 = (x^2 + 1)^2 has no real roots, yet it can still be factored.
In theory, every polynomial over the real numbers can be factored as a product of linear (degree 1) or quadratic (degree 2) polynomials. However, finding these factors can be challenging and time-consuming in practice. In the case of the given equation, it does not seem to have any further factorization in the real number system.