1. **Problem 1: Identify which is NOT a factor of the polynomial $P(x) = x^3 - \pi x^2 - x + \pi$.** We test each candidate factor by evaluating $P(x)$ at the root of the factor. If $P(r) = 0$, then $(x - r)$ is a factor. - For $x - 1$, test $x=1$: $$P(1) = 1^3 - \pi \cdot 1^2 - 1 + \pi = 1 - \pi - 1 + \pi = 0$$ So, $x - 1$ is a factor. - For $x + \pi$, test $x = -\pi$: $$P(-\pi) = (-\pi)^3 - \pi (-\pi)^2 - (-\pi) + \pi = -\pi^3 - \pi \pi^2 + \pi + \pi = -\pi^3 - \pi^3 + 2\pi = -2\pi^3 + 2\pi$$ Since $-2\pi^3 + 2\pi \neq 0$, $x + \pi$ is NOT a factor. - For $x + 1$, test $x = -1$: $$P(-1) = (-1)^3 - \pi (-1)^2 - (-1) + \pi = -1 - \pi + 1 + \pi = 0$$ So, $x + 1$ is a factor. - For $x - \pi$, test $x = \pi$: $$P(\pi) = \pi^3 - \pi \pi^2 - \pi + \pi = \pi^3 - \pi^3 - \pi + \pi = 0$$ So, $x - \pi$ is a factor. **Answer:** $x + \pi$ is NOT a factor. 2. **Problem 2: Find the average rate of change of $f(x) = 3 - 2x$ over $x_1 = -2$ to $x_2 = 7$.** The average rate of change formula is: $$\text{Average rate} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ Calculate: $$f(-2) = 3 - 2(-2) = 3 + 4 = 7$$ $$f(7) = 3 - 2(7) = 3 - 14 = -11$$ So, $$\frac{-11 - 7}{7 - (-2)} = \frac{-18}{9} = -2$$ **Answer:** The average rate of change is $-2$. 3. **Problem 3: Find the sum of the series $\sum_{k=1}^{15} k(8 - k)$.** First, expand the term inside the sum: $$k(8 - k) = 8k - k^2$$ So, $$\sum_{k=1}^{15} (8k - k^2) = 8 \sum_{k=1}^{15} k - \sum_{k=1}^{15} k^2$$ Use formulas for sums: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ Calculate each: $$\sum_{k=1}^{15} k = \frac{15 \times 16}{2} = 120$$ $$\sum_{k=1}^{15} k^2 = \frac{15 \times 16 \times 31}{6} = \frac{7440}{6} = 1240$$ Now substitute: $$8 \times 120 - 1240 = 960 - 1240 = -280$$ **Answer:** The sum of the series is $-280$.