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

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Answer from Sia

Posted about 2 months ago

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 $f(x)$ can be factored as $(x - 1)^4$

step 3

Since $(x - 1)^4$ is a perfect fourth power, it is always non-negative for all real $x$

step 4

The expression $(x - 1)^4$ equals zero only when $x = 1$

step 5

Therefore, $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ has no real roots other than $x = 1$, and since it is a perfect fourth power, it does not cross the x-axis

Answer

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

Key Concept

Factoring polynomials

Explanation

By factoring the polynomial $f(x)$, we can determine that it is a perfect fourth power, which means it does not have any real roots other than $x = 1$, and it does not cross the x-axis.

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