1. **State the problem:** We are given two polynomials: $$P(x) = -3x^3 + 2x^2 - 1$$ $$Q(x) = 5 + x + 2x^4$$ We need to find the derivatives of the product $(P \times Q)(x)$ and the sum $(P + Q)(x)$. 2. **Recall the formulas:** - Derivative of a product: $\frac{d}{dx}[P(x)Q(x)] = P'(x)Q(x) + P(x)Q'(x)$ - Derivative of a sum: $\frac{d}{dx}[P(x) + Q(x)] = P'(x) + Q'(x)$ 3. **Find the derivatives of $P(x)$ and $Q(x)$:** $$P'(x) = \frac{d}{dx}(-3x^3 + 2x^2 - 1) = -9x^2 + 4x$$ $$Q'(x) = \frac{d}{dx}(5 + x + 2x^4) = 0 + 1 + 8x^3 = 1 + 8x^3$$ 4. **Compute the derivative of the product:** $$\frac{d}{dx}[P(x)Q(x)] = P'(x)Q(x) + P(x)Q'(x)$$ Substitute: $$= (-9x^2 + 4x)(5 + x + 2x^4) + (-3x^3 + 2x^2 - 1)(1 + 8x^3)$$ 5. **Expand each term:** First term: $$(-9x^2)(5) = -45x^2$$ $$(-9x^2)(x) = -9x^3$$ $$(-9x^2)(2x^4) = -18x^6$$ $$(4x)(5) = 20x$$ $$(4x)(x) = 4x^2$$ $$(4x)(2x^4) = 8x^5$$ Sum first term: $$-45x^2 - 9x^3 - 18x^6 + 20x + 4x^2 + 8x^5$$ Simplify: $$-18x^6 + 8x^5 - 9x^3 + (-45x^2 + 4x^2) + 20x = -18x^6 + 8x^5 - 9x^3 - 41x^2 + 20x$$ Second term: $$(-3x^3)(1) = -3x^3$$ $$(-3x^3)(8x^3) = -24x^6$$ $$(2x^2)(1) = 2x^2$$ $$(2x^2)(8x^3) = 16x^5$$ $$(-1)(1) = -1$$ $$(-1)(8x^3) = -8x^3$$ Sum second term: $$-3x^3 - 24x^6 + 2x^2 + 16x^5 - 1 - 8x^3$$ Simplify: $$-24x^6 + 16x^5 + (-3x^3 - 8x^3) + 2x^2 - 1 = -24x^6 + 16x^5 - 11x^3 + 2x^2 - 1$$ 6. **Add both terms:** $$(-18x^6 + 8x^5 - 9x^3 - 41x^2 + 20x) + (-24x^6 + 16x^5 - 11x^3 + 2x^2 - 1)$$ Combine like terms: $$(-18x^6 - 24x^6) + (8x^5 + 16x^5) + (-9x^3 - 11x^3) + (-41x^2 + 2x^2) + 20x - 1$$ $$= -42x^6 + 24x^5 - 20x^3 - 39x^2 + 20x - 1$$ 7. **Compute the derivative of the sum:** $$\frac{d}{dx}[P(x) + Q(x)] = P'(x) + Q'(x) = (-9x^2 + 4x) + (1 + 8x^3)$$ Simplify: $$8x^3 - 9x^2 + 4x + 1$$ **Final answers:** - $$\frac{d}{dx}[P(x)Q(x)] = -42x^6 + 24x^5 - 20x^3 - 39x^2 + 20x - 1$$ - $$\frac{d}{dx}[P(x) + Q(x)] = 8x^3 - 9x^2 + 4x + 1$$