Question

let $p(x)=4x^{4}-4kx^{2}$, where $k$ is a real number. jasper claims that the polynomial $p(x)$ can have three distinct real zeros, $x = -sqrt{k},x = 0$, and $x=sqrt{k}$. drag each case of $k$ to the correct column to identify whether the claim is true or false in each case.

true	false

$k < 0$ $k = 0$ $k>0$

Explanation:

Step1: Factor the polynomial

Let \(y = x^{2}\), then \(p(x)=4x^{4}-4kx^{2}=4y^{2}-4ky = 4y(y - k)\). Substituting back \(y=x^{2}\), we get \(p(x)=4x^{2}(x^{2}-k)\).

Step2: Analyze the case when \(k<0\)

If \(k < 0\), then \(x^{2}-k>0\) for all real \(x\) except \(x = 0\). The equation \(p(x)=4x^{2}(x^{2}-k)=0\) has only one real - zero \(x = 0\) since \(x^{2}-k=0\) has no real solutions (\(x^{2}=k\) and \(k<0\)). So the claim is false.

Step3: Analyze the case when \(k = 0\)

If \(k = 0\), then \(p(x)=4x^{4}\), and the equation \(p(x)=0\) has only one real - zero \(x = 0\) (with multiplicity 4). So the claim is false.

Step4: Analyze the case when \(k>0\)

If \(k>0\), then \(x^{2}-k = 0\) gives \(x=\pm\sqrt{k}\) and \(4x^{2}=0\) gives \(x = 0\). The polynomial \(p(x)=4x^{2}(x^{2}-k)\) has three distinct real zeros \(x=-\sqrt{k},x = 0,x=\sqrt{k}\). So the claim is true.

Answer:

True: \(k>0\)
False: \(k < 0\), \(k = 0\)