1. **Problem statement:** A quadratic polynomial $p(x)$ has $(2x + 1)$ as a factor. When $p(x)$ is divided by $(x - 1)$ and $(x - 2)$, the remainders are $-6$ and $-5$ respectively. Find: i) $p(x)$ ii) The zeros of $p(x)$ 2. **Formula and rules:** Since $(2x + 1)$ is a factor, $p(x)$ can be written as $p(x) = (2x + 1)(ax + b)$ where $a$ and $b$ are constants to be determined. 3. **Using the remainder conditions:** - When divided by $(x - 1)$, remainder is $p(1) = -6$ - When divided by $(x - 2)$, remainder is $p(2) = -5$ 4. **Express $p(x)$:** $$p(x) = (2x + 1)(ax + b) = 2a x^2 + (2b + a) x + b$$ 5. **Evaluate $p(1)$ and $p(2)$:** $$p(1) = 2a(1)^2 + (2b + a)(1) + b = 2a + 2b + a + b = 3a + 3b = -6$$ $$p(2) = 2a(2)^2 + (2b + a)(2) + b = 8a + 4b + 2a + b = 10a + 5b = -5$$ 6. **Solve the system:** From $3a + 3b = -6$, divide both sides by 3: $$a + b = -2$$ From $10a + 5b = -5$, divide both sides by 5: $$2a + b = -1$$ 7. **Subtract equations:** $$ (2a + b) - (a + b) = -1 - (-2) \\ a = 1$$ Substitute $a=1$ into $a + b = -2$: $$1 + b = -2 \\ b = -3$$ 8. **Write $p(x)$ explicitly:** $$p(x) = (2x + 1)(1 imes x - 3) = (2x + 1)(x - 3)$$ Expanding: $$p(x) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$$ 9. **Find zeros of $p(x)$:** Solve $2x^2 - 5x - 3 = 0$ using quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times (-3)}}{2 \times 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4}$$ $$x = \frac{5 \pm 7}{4}$$ So zeros are: $$x = \frac{5 + 7}{4} = 3$$ $$x = \frac{5 - 7}{4} = -\frac{1}{2}$$ **Final answers:** - i) $p(x) = 2x^2 - 5x - 3$ - ii) Zeros are $x = 3$ and $x = -\frac{1}{2}$