Question 310·Hard·Math

The even powers make this a quadratic in $x^2$. The factors $x^2-1$ and $x^2-k$ have the right first and last terms; their middle terms add to $-(k+1)x^2$. Factor:

The zero-product rule says that a product is zero if at least one factor is zero. Since $p(x)=0$, set each factor equal to zero:

Add $1$ to the first equation:

Take both square roots, the positive and negative numbers whose squares are $1$:

These two roots are present for every value of $k$.

Add $k$ to the other equation:

A negative $k$ gives no real roots because a real number's square can't be negative. A positive $k$ gives two opposite roots: they either both repeat $-1$ and $1$, or both are new. Only a zero right-hand side gives exactly one new root, $x=0$.

Set $k=0$ to make that second branch give $x=0$. For a graph check, type $k=0$ and the given formula for $p(x)$ into Desmos. Click its $x$-intercepts: they are $(-1,0)$, $(0,0)$, and $(1,0)$. The graph touches the axis at $0$, so don't miss that root. There are exactly three distinct real solutions when $k=0$. Choice B.