Question

the function $f(x)=x^{6}-5x^{4}+4x^{2}$ has the graph shown below. though the fundamental theorem of algebra implies that this function will have 6 roots, there are only 5 roots shown in the graph. does this mean the fundamental theorem of algebra is incorrect? explain why or why not and include the factored form of the function in your explanation.

Explanation:

Step1: Factor the function

Let \(y = x^{2}\), then \(f(x)=x^{6}-5x^{4}+4x^{2}=y^{3}-5y^{2} + 4y\). Factoring out \(y\) first, we get \(y(y^{2}-5y + 4)\). Then factor \(y^{2}-5y + 4=(y - 1)(y - 4)\). Substituting \(y=x^{2}\) back, we have \(f(x)=x^{2}(x^{2}-1)(x^{2}-4)=x^{2}(x - 1)(x + 1)(x-2)(x + 2)\).

Step2: Analyze the roots

The roots of the function \(f(x)\) are found by setting \(f(x)=0\). So \(x^{2}(x - 1)(x + 1)(x-2)(x + 2)=0\). The roots are \(x = 0\) (with multiplicity 2), \(x=1\), \(x=-1\), \(x = 2\), \(x=-2\). The Fundamental Theorem of Algebra states that a non - zero polynomial of degree \(n\) with complex coefficients has exactly \(n\) roots (counting multiplicities) in the complex - number system. Here, the degree of \(f(x)\) is 6. The root \(x = 0\) has a multiplicity of 2, so in total, there are 6 roots when counting multiplicities, which is consistent with the theorem.

Answer:

No, the Fundamental Theorem of Algebra is not incorrect. The function \(f(x)=x^{2}(x - 1)(x + 1)(x-2)(x + 2)\) has a root \(x = 0\) with multiplicity 2, and roots \(x=1\), \(x=-1\), \(x = 2\), \(x=-2\). When counting multiplicities, there are 6 roots as the theorem predicts.