<s>[INST] How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$ 
 [/INST] You could factorise it, in the manner of $(x-2)^4=0$.  I saw those factors immediately.
One process is to note that $16$ has divisors, and one can try various combinations of this such that the sum gives eight.
Possibilities include $2, 2, 2, 2$ and $4, 4, 1, -1$.  However, one can not produce the second set to give +16, so trying $(x-2)(x-2)(x-2)(x-2)$ is more likely than $(x-4)(x-4)(x-1)(x+1)$.
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