1. **State the problem:** Find the square root of the polynomial $$x^4 + x^3 - \frac{31}{4}x^2 - 4x + 16$$ using the division method. 2. **Recall the division method for square roots:** We want to find a polynomial $$P(x) = ax^2 + bx + c$$ such that $$P(x)^2 = x^4 + x^3 - \frac{31}{4}x^2 - 4x + 16$$. 3. **Set up the equation:** $$\left(ax^2 + bx + c\right)^2 = a^2x^4 + 2abx^3 + (b^2 + 2ac)x^2 + 2bcx + c^2$$ 4. **Match coefficients with the given polynomial:** - Coefficient of $$x^4$$: $$a^2 = 1 \implies a = 1$$ (taking positive root for simplicity) - Coefficient of $$x^3$$: $$2ab = 1 \implies 2 \cdot 1 \cdot b = 1 \implies b = \frac{1}{2}$$ - Coefficient of $$x^2$$: $$b^2 + 2ac = -\frac{31}{4}$$ Substitute $$b = \frac{1}{2}$$ and $$a=1$$: $$\left(\frac{1}{2}\right)^2 + 2 \cdot 1 \cdot c = -\frac{31}{4}$$ $$\frac{1}{4} + 2c = -\frac{31}{4}$$ $$2c = -\frac{31}{4} - \frac{1}{4} = -\frac{32}{4} = -8$$ $$c = -4$$ - Coefficient of $$x$$: $$2bc = -4$$ Substitute $$b = \frac{1}{2}$$ and $$c = -4$$: $$2 \cdot \frac{1}{2} \cdot (-4) = -4$$ $$-4 = -4$$ (checks out) - Constant term: $$c^2 = 16$$ $$(-4)^2 = 16$$ (checks out) 5. **Conclusion:** The square root polynomial is $$x^2 + \frac{1}{2}x - 4$$ 6. **Verification:** $$\left(x^2 + \frac{1}{2}x - 4\right)^2 = x^4 + x^3 - \frac{31}{4}x^2 - 4x + 16$$ **Final answer:** $$\boxed{x^2 + \frac{1}{2}x - 4}$$