1. **State the problem:** We have a polynomial $p(t)$ of degree 6 with factors $(t - 10)$, $(2t - 2)$, $(t + 10)$, and $(20t + 2)$. We need to determine which graph among A, B, C, or D could represent $y = p(t)$. 2. **Analyze the factors:** Each factor corresponds to a root of $p(t)$. - From $(t - 10)$, root is $t = 10$. - From $(2t - 2)$, root is $t = 1$ (since $2t - 2 = 0 \Rightarrow t = 1$). - From $(t + 10)$, root is $t = -10$. - From $(20t + 2)$, root is $t = -\frac{1}{10}$ (since $20t + 2 = 0 \Rightarrow t = -\frac{1}{10}$). 3. **Degree and roots:** The polynomial is degree 6 but only 4 linear factors are given. This means there are two more factors (possibly repeated roots or other factors). 4. **Behavior at roots:** Since these are linear factors, the polynomial crosses the $t$-axis at these roots unless the root has even multiplicity (which would cause the graph to touch but not cross). 5. **End behavior:** The polynomial is degree 6 (even degree). The leading coefficient's sign determines end behavior. - If leading coefficient is positive, both ends go up. - If negative, both ends go down. 6. **Check graphs:** - Graph A: even degree, both ends up, crosses $t$-axis multiple times. - Graph B: even degree, left end down, right end up (not possible for even degree). - Graph C: even degree, both ends down (possible if leading coefficient negative). - Graph D: downward opening with single hump (degree 2 behavior, not degree 6). 7. **Roots on graphs:** The roots are at $-10$, $-\frac{1}{10}$, $1$, and $10$. The graph must cross the $t$-axis near these points. 8. **Conclusion:** Graph A matches the behavior of an even-degree polynomial with positive leading coefficient and multiple roots crossing the axis near the given roots. **Final answer:** Graph A