1. **State the problem:** Find the equation of the normal line to the graph of the function $$F(x) = (x^4 - 3x^2 + 2x)(x^3 - 2x + 3)$$ at the origin (point where $x=0$). 2. **Find $F(0)$ to confirm the point on the graph:** $$F(0) = (0^4 - 3\cdot0^2 + 2\cdot0)(0^3 - 2\cdot0 + 3) = 0 \cdot 3 = 0$$ So the point is $(0,0)$. 3. **Find the derivative $F'(x)$ using the product rule:** Let $$u = x^4 - 3x^2 + 2x, \quad v = x^3 - 2x + 3$$ Then $$u' = 4x^3 - 6x + 2, \quad v' = 3x^2 - 2$$ By the product rule, $$F'(x) = u'v + uv' = (4x^3 - 6x + 2)(x^3 - 2x + 3) + (x^4 - 3x^2 + 2x)(3x^2 - 2)$$ 4. **Evaluate $F'(0)$ to find the slope of the tangent at $x=0$:** $$F'(0) = (4\cdot0 - 6\cdot0 + 2)(0 - 0 + 3) + (0 - 0 + 0)(0 - 2) = 2 \times 3 + 0 = 6$$ 5. **Find the slope of the normal line:** The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: $$m_{normal} = -\frac{1}{6}$$ 6. **Write the equation of the normal line passing through $(0,0)$:** Using point-slope form, $$y - 0 = -\frac{1}{6}(x - 0)$$ Simplifies to $$y = -\frac{1}{6}x$$ **Final answer:** The equation of the normal line to the graph of $F(x)$ at the origin is $$y = -\frac{1}{6}x$$