1. **State the problem:** Resolve the following into partial fractions: $$\frac{x}{x^2+x+4}$$ and $$\frac{x-2}{x^2+4x+4}$$ 2. **Recall the formula and rules:** Partial fraction decomposition is used to express a rational function as a sum of simpler fractions. For quadratic denominators that cannot be factored into real linear factors, the partial fraction form is $$\frac{Ax+B}{quadratic}$$. 3. **First expression:** $$\frac{x}{x^2+x+4}$$ The denominator $$x^2+x+4$$ does not factor nicely (discriminant $$\Delta = 1^2 - 4\times1\times4 = 1-16 = -15 < 0$$), so it is irreducible over the reals. Thus, the partial fraction form is: $$\frac{x}{x^2+x+4} = \frac{Ax+B}{x^2+x+4}$$ Multiply both sides by the denominator: $$x = (Ax+B)$$ Since the denominators are the same, this implies: $$x = Ax + B$$ Equate coefficients: Coefficient of $$x$$: $$1 = A$$ Constant term: $$0 = B$$ So, $$A=1$$ and $$B=0$$. Therefore: $$\frac{x}{x^2+x+4} = \frac{1\cdot x + 0}{x^2+x+4} = \frac{x}{x^2+x+4}$$ This means the expression is already in simplest partial fraction form. 4. **Second expression:** $$\frac{x-2}{x^2+4x+4}$$ Factor the denominator: $$x^2+4x+4 = (x+2)^2$$ Since the denominator is a repeated linear factor, the partial fraction decomposition is: $$\frac{x-2}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}$$ Multiply both sides by $$ (x+2)^2 $$: $$x-2 = A(x+2) + B$$ Expand: $$x-2 = Ax + 2A + B$$ Equate coefficients: Coefficient of $$x$$: $$1 = A$$ Constant term: $$-2 = 2A + B$$ Substitute $$A=1$$: $$-2 = 2(1) + B \Rightarrow B = -2 - 2 = -4$$ Therefore: $$\frac{x-2}{(x+2)^2} = \frac{1}{x+2} - \frac{4}{(x+2)^2}$$ **Final answers:** $$\frac{x}{x^2+x+4}$$ (already simplest form) $$\frac{x-2}{x^2+4x+4} = \frac{1}{x+2} - \frac{4}{(x+2)^2}$$