1. **Problem statement:** Given two polynomials: $$P(x) = 3x^3 + 5x - 12 - x^2 + x^3 - 6x^2 - 2x$$ $$Q(x) = -2x^3 + x - 3x^2 + 12 + 8x^2 - 10x - 2x^3$$ We need to: a) Simplify and arrange both polynomials in ascending powers of $x$. b) Calculate $H(x) = P(x) + Q(x)$ and $G(x) = 2P(x) - Q(x)$. c) Find the leading coefficient and constant term of $H(x)$ and $G(x)$. d) Evaluate $H(-1)$, $H(2)$, $G(-\frac{1}{2})$, and $G(0)$. e) Find the roots of $H(x)$. --- 2. **Simplify and arrange $P(x)$:** Combine like terms: $$P(x) = (3x^3 + x^3) + (-x^2 - 6x^2) + (5x - 2x) - 12$$ $$= 4x^3 - 7x^2 + 3x - 12$$ Arrange in ascending powers: $$P(x) = -12 + 3x - 7x^2 + 4x^3$$ 3. **Simplify and arrange $Q(x)$:** Combine like terms: $$Q(x) = (-2x^3 - 2x^3) + (-3x^2 + 8x^2) + (x - 10x) + 12$$ $$= -4x^3 + 5x^2 - 9x + 12$$ Arrange in ascending powers: $$Q(x) = 12 - 9x + 5x^2 - 4x^3$$ 4. **Calculate $H(x) = P(x) + Q(x)$:** $$H(x) = (-12 + 3x - 7x^2 + 4x^3) + (12 - 9x + 5x^2 - 4x^3)$$ Combine like terms: $$H(x) = (-12 + 12) + (3x - 9x) + (-7x^2 + 5x^2) + (4x^3 - 4x^3)$$ $$= 0 - 6x - 2x^2 + 0 = -6x - 2x^2$$ Arrange ascending: $$H(x) = -6x - 2x^2$$ 5. **Calculate $G(x) = 2P(x) - Q(x)$:** First, compute $2P(x)$: $$2P(x) = 2(-12 + 3x - 7x^2 + 4x^3) = -24 + 6x - 14x^2 + 8x^3$$ Then subtract $Q(x)$: $$G(x) = (-24 + 6x - 14x^2 + 8x^3) - (12 - 9x + 5x^2 - 4x^3)$$ $$= (-24 - 12) + (6x + 9x) + (-14x^2 - 5x^2) + (8x^3 + 4x^3)$$ $$= -36 + 15x - 19x^2 + 12x^3$$ Arrange ascending: $$G(x) = -36 + 15x - 19x^2 + 12x^3$$ 6. **Find leading coefficient and constant term:** - For $H(x) = -6x - 2x^2$, leading term is $-2x^2$ so leading coefficient is $-2$, constant term is $0$ (no constant). - For $G(x) = -36 + 15x - 19x^2 + 12x^3$, leading term is $12x^3$ so leading coefficient is $12$, constant term is $-36$. 7. **Evaluate values:** - $H(-1) = -6(-1) - 2(-1)^2 = 6 - 2 = 4$ - $H(2) = -6(2) - 2(2)^2 = -12 - 8 = -20$ - $G(-\frac{1}{2}) = -36 + 15(-\frac{1}{2}) - 19(-\frac{1}{2})^2 + 12(-\frac{1}{2})^3$ Calculate stepwise: $$15(-\frac{1}{2}) = -7.5$$ $$(-\frac{1}{2})^2 = \frac{1}{4}$$ $$-19 \times \frac{1}{4} = -4.75$$ $$(-\frac{1}{2})^3 = -\frac{1}{8}$$ $$12 \times -\frac{1}{8} = -1.5$$ Sum: $$-36 - 7.5 - 4.75 - 1.5 = -49.75$$ - $G(0) = -36 + 15(0) - 19(0)^2 + 12(0)^3 = -36$ 8. **Find roots of $H(x) = -6x - 2x^2 = 0$:** Rewrite: $$-2x^2 - 6x = 0$$ Divide both sides by $-2$: $$\cancel{-2}x^2 + \cancel{-2}3x = 0 \Rightarrow x^2 + 3x = 0$$ Factor: $$x(x + 3) = 0$$ Roots: $$x = 0 \quad \text{or} \quad x = -3$$