1. **Problem 1:** Given polynomial $$P(x) = 3x^3 + ax^2 + bx - 4$$ leaves remainder -7 when divided by $$x + 1$$ and remainder 23 when divided by $$x - 2$$. By the Remainder Theorem: - Remainder when divided by $$x + 1$$ is $$P(-1) = -7$$ - Remainder when divided by $$x - 2$$ is $$P(2) = 23$$ Calculate: $$P(-1) = 3(-1)^3 + a(-1)^2 + b(-1) - 4 = -3 + a - b - 4 = a - b - 7$$ Set equal to remainder: $$a - b - 7 = -7$$ $$a - b = 0 ag{1}$$ $$P(2) = 3(2)^3 + a(2)^2 + b(2) - 4 = 24 + 4a + 2b - 4 = 20 + 4a + 2b$$ Set equal to remainder: $$20 + 4a + 2b = 23$$ $$4a + 2b = 3 ag{2}$$ From (1): $$a = b$$ Substitute into (2): $$4b + 2b = 3 \\ 6b = 3 \\ b = \frac{3}{6} = \frac{1}{2}$$ Thus, $$b = \frac{1}{2}$$ 2. **Problem 2:** Find $$k$$ if $$2x - 1$$ is a factor of $$P(x) = 2x^3 + kx^2 + ?$$ The problem is incomplete for full calculation; assuming we find $$k$$ by setting $$P\left(\frac{1}{2}\right) = 0$$ (since root corresponds to factor): 3. **Problem 3:** Identify which is NOT a factor of $$f(x) = x^3 - ?$$ with options: - (x + 4) - (x + 3) - (x + 1) - (x - 4) Since no explicit polynomial is given, cannot determine without more info. 4. **Problem 4:** Which polynomial in factored form matches graph with zeros at $$x = -4, 1, 3$$ and dips below x-axis between roots? Answer: $$f(x) = (x + 4)(x - 1)(x - 3)$$ 5. **Problem 5:** Find quotient when $$P(x) = x^5 - 1$$ is divided by $$x - 1$$. By formula for difference of powers: $$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$$ Hence quotient is: $$x^4 + x^3 + x^2 + x + 1$$ 6. **Additional request:** Calculate remainder when $$P(x) = 3x^3 + ax^2 + bx - 4$$ (from Problem 1) is divided by $$2x - 1$$. Apply remainder theorem for root of $$2x - 1 = 0 \\ x = \frac{1}{2}$$: Recall $$a = b$$ and $$b = \frac{1}{2}$$, so $$a = \frac{1}{2}$$. Calculate remainder: $$P\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^3 + \frac{1}{2}\left(\frac{1}{2}\right)^2 + \frac{1}{2}\left(\frac{1}{2}\right) - 4$$ $$= 3\times \frac{1}{8} + \frac{1}{2} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{2} - 4$$ $$= \frac{3}{8} + \frac{1}{8} + \frac{1}{4} - 4$$ $$= \frac{3}{8} + \frac{1}{8} + \frac{2}{8} - 4 = \frac{6}{8} - 4 = \frac{3}{4} - 4 = -\frac{13}{4}$$ **Final answers:** - $$b = \frac{1}{2}$$ - Quotient for problem 5 is $$x^4 + x^3 + x^2 + x + 1$$ - Polynomial for problem 4 is $$f(x) = (x + 4)(x - 1)(x - 3)$$ - Remainder when divided by $$2x - 1$$ is $$-\frac{13}{4}$$