1. **Problem statement:** (i) Find the quadratic polynomial $p(x)$ given that $(2x+1)$ is a factor and the remainders when $p(x)$ is divided by $(x-1)$ and $(x-2)$ are $-6$ and $-5$ respectively. (ii) Find the zeros of $p(x)$. 2. **Given:** - $(2x+1)$ is a factor of $p(x)$, so $p\left(-\frac{1}{2}\right) = 0$. - $p(1) = -6$ and $p(2) = -5$. 3. **Form of $p(x)$:** Since $(2x+1)$ is a factor, write $p(x) = (2x+1)(ax + b)$ where $a,b$ are constants to find. 4. **Use the remainder conditions:** Calculate $p(1)$: $$p(1) = (2(1)+1)(a(1) + b) = 3(a + b) = -6$$ So, $3(a + b) = -6 \Rightarrow a + b = -2$. Calculate $p(2)$: $$p(2) = (2(2)+1)(a(2) + b) = 5(2a + b) = -5$$ So, $5(2a + b) = -5 \Rightarrow 2a + b = -1$. 5. **Solve the system:** From $a + b = -2$ and $2a + b = -1$: Subtract first from second: $$(2a + b) - (a + b) = -1 - (-2) \Rightarrow a = 1$$ Then $b = -2 - a = -2 - 1 = -3$. 6. **Write $p(x)$:** $$p(x) = (2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$$ 7. **Find zeros of $p(x)$:** Solve $2x^2 - 5x - 3 = 0$ using the quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times (-3)}}{2 \times 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4}$$ So zeros are: $$x = \frac{5 + 7}{4} = 3, \quad x = \frac{5 - 7}{4} = -\frac{1}{2}$$ --- 8. **Solve $2x^2 + 3x - 4 = 0$ by completing the square:** 9. **Divide entire equation by 2:** $$x^2 + \frac{3}{2}x - 2 = 0$$ 10. **Move constant to right side:** $$x^2 + \frac{3}{2}x = 2$$ 11. **Complete the square:** Take half of coefficient of $x$, square it: $$\left(\frac{3}{4}\right)^2 = \frac{9}{16}$$ Add to both sides: $$x^2 + \frac{3}{2}x + \frac{9}{16} = 2 + \frac{9}{16} = \frac{32}{16} + \frac{9}{16} = \frac{41}{16}$$ 12. **Rewrite left side as square:** $$\left(x + \frac{3}{4}\right)^2 = \frac{41}{16}$$ 13. **Take square root:** $$x + \frac{3}{4} = \pm \frac{\sqrt{41}}{4}$$ 14. **Solve for $x$:** $$x = -\frac{3}{4} \pm \frac{\sqrt{41}}{4} = \frac{-3 \pm \sqrt{41}}{4}$$ **Final answers:** (i) $p(x) = 2x^2 - 5x - 3$ (ii) Zeros of $p(x)$ are $x = 3$ and $x = -\frac{1}{2}$ (iii) Solutions to $2x^2 + 3x - 4 = 0$ by completing the square are $x = \frac{-3 \pm \sqrt{41}}{4}$