Let $y=x^2$ so that $p(x)=0$ becomes $y^2-(k+1)y+k=0$. Try to factor this quadratic in $y$; look for two numbers that multiply to $k$ and add to $-(k+1)$.

Once you have something like $(x^2-1)(x^2-k)=0$, think about what values of $x$ each equation $x^2=1$ and $x^2=k$ can produce, depending on whether $k$ is negative, zero, or positive.

For each given value of $k$, determine how many distinct real $x$-values come from $x^2=1$ and from $x^2=k$, and then see which choice gives exactly three distinct real solutions in total.

Type p(x) = x^4 - (k + 1)x^2 + k into Desmos. When you press Enter, Desmos will prompt you to add a slider for k; create that slider.

Move the k slider exactly to each of the four values: $-1$, $0$, $2$, and $5$. For each value, look at the graph and count how many distinct $x$-intercepts (points where the graph crosses or just touches the $x$-axis) the curve has.

Among the four tested values of $k$, find the one for which the graph of $y=p(x)$ has exactly three distinct $x$-intercepts. That value of $k$ is the correct answer.

Notice that $p(x)=x^4-(k+1)x^2+k$ has only even powers of $x$. Let $y=x^2$. Then the equation $p(x)=0$ becomes

This is a quadratic equation in $y$.

Factor the quadratic in $y$:

$y^2-(k+1)y+k = (y-1)(y-k)$, because $(y-1)(y-k) = y^2-(k+1)y+k$.

Now replace $y$ with $x^2$:

So the solutions must satisfy either $x^2=1$ or $x^2=k$.

From $x^2=1$, we always get two real solutions: $x=1$ and $x=-1$.

From $x^2=k$:

• If $k<0$, there are no real solutions.
• If $k=0$, there is one real solution: $x=0$.
• If $k>0$, there are two real solutions: $x=\sqrt{k}$ and $x=-\sqrt{k}$.

The total number of distinct real solutions to $p(x)=0$ is the number from $x^2=1$ plus the number from $x^2=k$, taking care not to double-count any value.

Now apply the cases above to each answer choice:

• $k=-1$: $k<0$, so $x^2=k$ has no real solutions. Only $x^2=1$ contributes, giving $x=\pm1$ (2 distinct real solutions).
• $k=0$: from $x^2=1$ we get $x=\pm1$, and from $x^2=0$ we get $x=0$. Together, the distinct real solutions are $-1,0,1$ (3 distinct real solutions).
• $k=2$: $k>0$, so $x^2=2$ gives $x=\pm\sqrt{2}$ in addition to $x=\pm1$, for 4 distinct real solutions.
• $k=5$: similarly, $x=\pm1$ and $x=\pm\sqrt{5}$, again 4 distinct real solutions.

Only $k=0$ makes $p(x)=0$ have exactly three distinct real solutions, so the correct answer is $\boxed{0}$.