1. **State the problem:** We have a polynomial $$p(x) = 10x^3 + ax^2 - 10x + b$$ where $$a$$ and $$b$$ are integers. - It is divisible by $$2x + 1$$. - When divided by $$x + 1$$, the remainder is $$-24$$. We need to find: (a) The values of $$a$$ and $$b$$. (b) An expression for $$p(x)$$ as the product of three linear factors. (c) The remainder when $$p(x)$$ is divided by $$x$$. --- 2. **Use the divisibility condition:** Since $$p(x)$$ is divisible by $$2x + 1$$, the root of $$2x + 1 = 0$$ is $$x = -\frac{1}{2}$$. By the Factor Theorem, $$p\left(-\frac{1}{2}\right) = 0$$. Calculate: $$p\left(-\frac{1}{2}\right) = 10\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 - 10\left(-\frac{1}{2}\right) + b = 0$$ Simplify powers: $$= 10\left(-\frac{1}{8}\right) + a\left(\frac{1}{4}\right) + 5 + b = 0$$ Calculate: $$-\frac{10}{8} + \frac{a}{4} + 5 + b = 0$$ Simplify $$-\frac{10}{8} = -\frac{5}{4}$$: $$-\frac{5}{4} + \frac{a}{4} + 5 + b = 0$$ Multiply entire equation by 4 to clear denominators: $$-5 + a + 20 + 4b = 0$$ Simplify: $$a + 4b + 15 = 0$$ Rearranged: $$a + 4b = -15 \quad \quad (1)$$ --- 3. **Use the remainder condition:** When $$p(x)$$ is divided by $$x + 1$$, remainder is $$-24$$. By the Remainder Theorem: $$p(-1) = -24$$ Calculate: $$p(-1) = 10(-1)^3 + a(-1)^2 - 10(-1) + b = -24$$ Simplify powers: $$= 10(-1) + a(1) + 10 + b = -24$$ Calculate: $$-10 + a + 10 + b = -24$$ Simplify: $$a + b = -24 \quad \quad (2)$$ --- 4. **Solve the system of equations:** From (1): $$a + 4b = -15$$ From (2): $$a + b = -24$$ Subtract (2) from (1): $$ (a + 4b) - (a + b) = -15 - (-24) $$ $$ a + 4b - a - b = 9 $$ $$ 3b = 9 $$ $$ b = 3 $$ Substitute $$b=3$$ into (2): $$ a + 3 = -24 $$ $$ a = -27 $$ --- 5. **Answer for (a):** $$a = -27, \quad b = 3$$ --- 6. **Find the factorization for (b):** We know $$2x + 1$$ is a factor. Divide $$p(x) = 10x^3 - 27x^2 - 10x + 3$$ by $$2x + 1$$. Use polynomial long division or synthetic division: Set divisor root: $$x = -\frac{1}{2}$$. Divide: Coefficients: 10, -27, -10, 3 Synthetic division with root $$-\frac{1}{2}$$: - Bring down 10 - Multiply: $$10 \times -\frac{1}{2} = -5$$ - Add: $$-27 + (-5) = -32$$ - Multiply: $$-32 \times -\frac{1}{2} = 16$$ - Add: $$-10 + 16 = 6$$ - Multiply: $$6 \times -\frac{1}{2} = -3$$ - Add: $$3 + (-3) = 0$$ (remainder 0, confirms factor) Quotient polynomial: $$10x^2 - 32x + 6$$ Factor out 2: $$2(5x^2 - 16x + 3)$$ Factor quadratic: Find two numbers that multiply to $$5 \times 3 = 15$$ and add to $$-16$$: $$-15$$ and $$-1$$. Rewrite: $$5x^2 - 15x - x + 3 = 0$$ Group: $$(5x^2 - 15x) - (x - 3) = 0$$ Factor: $$5x(x - 3) - 1(x - 3) = (5x - 1)(x - 3)$$ So, $$p(x) = (2x + 1)(5x - 1)(x - 3)$$ --- 7. **Answer for (b):** $$p(x) = (2x + 1)(5x - 1)(x - 3)$$ --- 8. **Find remainder when $$p(x)$$ divided by $$x$$ for (c):** Remainder when dividing by $$x$$ is $$p(0)$$. Calculate: $$p(0) = 10(0)^3 - 27(0)^2 - 10(0) + 3 = 3$$ --- 9. **Answer for (c):** Remainder is $$3$$. --- **Final answers:** - (a) $$a = -27$$, $$b = 3$$ - (b) $$p(x) = (2x + 1)(5x - 1)(x - 3)$$ - (c) Remainder when divided by $$x$$ is $$3$$.