1. **State the problem:** Find two non-equivalent polynomial functions $f(x)$ and $g(x)$ such that they have the same values at $x=0$ and $x=-3$, i.e., $f(0)=g(0)$ and $f(-3)=g(-3)$. 2. **Key idea:** Two functions are non-equivalent if they are different as polynomials. But they can share the same values at specific points. So we want $f(x)\neq g(x)$ but $f(0) = g(0)$ and $f(-3) = g(-3)$. 3. **Approach:** Let $h(x) = f(x) - g(x)$. Then $h(0)=0$ and $h(-3)=0$, so $h(x)$ has roots at $0$ and $-3$. 4. **Construct $h(x)$:** Since $h(x)$ has roots 0 and -3, $h(x)$ is divisible by $x(x+3)$. Let $h(x) = k x(x+3)$ for some nonzero constant $k$. 5. **Choose $g(x)$ arbitrarily:** For simplicity, let $g(x) = x^2$. Then we have $$f(x) = g(x) + h(x) = x^2 + k x(x+3) = x^2 + k(x^2 + 3x) = (1+k)x^2 + 3k x.$$ 6. **Check the conditions:** Evaluate at $x=0$, $f(0) = 0$, $g(0) = 0$, so $f(0)=g(0)$. Evaluate at $x = -3$, $f(-3) = (1+k)9 + 3k(-3) = 9 + 9k - 9k = 9$, $g(-3) = 9,$ so $f(-3) = g(-3)$. 7. **Non-equivalence:** For $k \ne 0$, $f(x) \neq g(x)$, so $f$ and $g$ are non-equivalent polynomial functions satisfying the conditions. **Example:** Take $k=1$, $$f(x) = 2x^2 + 3x, \quad g(x) = x^2.$$