1. **State the problem:** We need to determine which of the given quadratic expressions has a factor of $x+2b$, where $b$ is a positive constant. 2. **Recall the factor theorem:** If $x+2b$ is a factor of a polynomial $P(x)$, then $P(-2b) = 0$. 3. **Apply the factor theorem to each polynomial:** - For $P_1(x) = 3x^2 + 7x + 14b$, evaluate $P_1(-2b)$: $$P_1(-2b) = 3(-2b)^2 + 7(-2b) + 14b = 3(4b^2) - 14b + 14b = 12b^2$$ Since $12b^2 \neq 0$ for $b > 0$, $x+2b$ is not a factor. - For $P_2(x) = 3x^2 + 28x + 14b$, evaluate $P_2(-2b)$: $$P_2(-2b) = 3(4b^2) + 28(-2b) + 14b = 12b^2 - 56b + 14b = 12b^2 - 42b$$ This is not zero for all $b > 0$, so no factor. - For $P_3(x) = 3x^2 + 42x + 14b$, evaluate $P_3(-2b)$: $$P_3(-2b) = 3(4b^2) + 42(-2b) + 14b = 12b^2 - 84b + 14b = 12b^2 - 70b$$ Not zero for all $b > 0$, so no factor. - For $P_4(x) = 3x^2 + 49x + 14b$, evaluate $P_4(-2b)$: $$P_4(-2b) = 3(4b^2) + 49(-2b) + 14b = 12b^2 - 98b + 14b = 12b^2 - 84b$$ Not zero for all $b > 0$, so no factor. 4. **Conclusion:** None of the given polynomials have $x+2b$ as a factor for all positive $b$. However, if we consider the possibility that $b$ satisfies the equation $12b^2 - 42b = 0$ (from $P_2$), then $b(12b - 42) = 0$ which gives $b=0$ or $b=3.5$. Since $b$ is positive, $b=3.5$ makes $P_2(-2b)=0$, so $x+2b$ is a factor of $P_2$ when $b=3.5$. Similarly, for $P_3$, $12b^2 - 70b=0$ gives $b=0$ or $b=\frac{70}{12} = 5.833...$. For $P_4$, $12b^2 - 84b=0$ gives $b=0$ or $b=7$. Therefore, $x+2b$ is a factor of $P_2$, $P_3$, and $P_4$ only for specific positive values of $b$. 5. **Final answer:** Among the options, $P_2$, $P_3$, and $P_4$ can have $x+2b$ as a factor for specific positive values of $b$, but $P_1$ cannot.