Prove that the function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots.

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∻Solution by Steps∻ ‖ step 1 ⋮ We start by factoring the polynomial $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$. ‖ ‖ step 2 ⋮ Using the Asksia-LL calculator, we find that $f(x)$ factors as $(x - 1)^4$. ‖ ‖ step 3 ⋮ Since $(x - 1)^4$ is a perfect fourth power, it is always non-negative and equals zero only when $x = 1$. ‖ ‖ step 4 ⋮ Therefore, the function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots other than $x = 1$. ‖ ∻Answer∻ ⚹ The function $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots other than $x = 1$. ⚹ ∻Key Concept∻ ⚹Factoring polynomials⚹ ∻Explanation∻ ⚹By factoring the polynomial $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ as $(x - 1)^4$, we see that it is always non-negative and equals zero only at $x = 1$. Therefore, it has no other real roots.⚹◊What method can be used to prove that a polynomial has no real roots⍭ Generate me a similar question◊

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