1. The problem asks which of the given expressions is NOT a factor of the polynomial $f(x) = x^3 - 13x - 12$. 2. Recall that if $(x - r)$ is a factor of $f(x)$, then $f(r) = 0$. We test each candidate by evaluating $f(-4)$, $f(-3)$, $f(-1)$, and $f(4)$. 3. Calculate: $$f(-4) = (-4)^3 - 13(-4) - 12 = -64 + 52 - 12 = -24 \neq 0$$ $$f(-3) = (-3)^3 - 13(-3) - 12 = -27 + 39 - 12 = 0$$ $$f(-1) = (-1)^3 - 13(-1) - 12 = -1 + 13 - 12 = 0$$ $$f(4) = 4^3 - 13(4) - 12 = 64 - 52 - 12 = 0$$ 4. From the above, $(x+3)$, $(x+1)$, and $(x-4)$ are factors, but $(x+4)$ is NOT a factor. --- 5. Next problem: Find the possible maximum number of negative real roots of $P(x) = x^3 + 5x^2 - 4x - 20$. 6. Use Descartes' Rule of Signs for $P(-x)$: $$P(-x) = (-x)^3 + 5(-x)^2 - 4(-x) - 20 = -x^3 + 5x^2 + 4x - 20$$ Sign changes in coefficients are: from $-x^3$ (negative) to $+5x^2$ (positive): 1 change, $+5x^2$ to $+4x$ no change, $+4x$ to $-20$ negative: 1 change. 7. Total sign changes in $P(-x)$ are 2, so the possible number of negative real roots is 2 or 0 (decreasing by multiples of 2). 8. The maximum possible number of negative real roots is thus 2. --- 9. Finally, for $P(x) = 3x^3 + a x^2 + b x - 1$, given $P(-1) = -7$ and $P(2) = 23$, find the remainder when divided by $2x - 1$. 10. Express $P(-1)$ and $P(2)$: $$P(-1) = 3(-1)^3 + a(-1)^2 + b(-1) - 1 = -3 + a - b - 1 = a - b - 4 = -7$$ $$P(2) = 3(2)^3 + a(2)^2 + b(2) - 1 = 24 + 4a + 2b - 1 = 4a + 2b + 23 = 23$$ From $P(-1) = -7$: $$a - b - 4 = -7 \Rightarrow a - b = -3$$ From $P(2) = 23$: $$4a + 2b + 23 = 23 \Rightarrow 4a + 2b = 0 \Rightarrow 2a + b = 0 \Rightarrow b = -2a$$ 11. Substitute $b = -2a$ into $a - b = -3$: $$a - (-2a) = -3 \Rightarrow a + 2a = -3 \Rightarrow 3a = -3 \Rightarrow a = -1$$ Then $b = -2(-1) = 2$. 12. To find the remainder when $P(x)$ is divided by $2x - 1$, note the divisor root is: $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ 13. The remainder when dividing by a linear divisor is the polynomial's value at that root: $$R = P\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^3 + (-1)\left(\frac{1}{2}\right)^2 + 2 \left(\frac{1}{2}\right) - 1 = 3 \cdot \frac{1}{8} - \frac{1}{4} + 1 - 1 = \frac{3}{8} - \frac{1}{4} + 0 = \frac{3}{8} - \frac{2}{8} = \frac{1}{8}$$ 14. Final answers: - Not a factor of $f(x)$: $(x + 4)$. - Maximum possible negative real roots for $P(x)$: 2. - Remainder when dividing $P(x)$ by $2x - 1$: $\frac{1}{8}$.