1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$$x ^ { 2 } + 1 + \frac { 1 } { 4 x ^ { 2 } }$$

$\frac{\left(2x+\frac{1}{x}\right)^{2}}{4}$

To add or subtract expressions, expand them to make their denominators the same. Multiply $x^{2}+1$ times $\frac{4x^{2}}{4x^{2}}$.

$$\frac{\left(x^{2}+1\right)\times 4x^{2}}{4x^{2}}+\frac{1}{4x^{2}}$$

Since $\frac{\left(x^{2}+1\right)\times 4x^{2}}{4x^{2}}$ and $\frac{1}{4x^{2}}$ have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}+1\right)\times 4x^{2}+1}{4x^{2}}$$

Do the multiplications in $\left(x^{2}+1\right)\times 4x^{2}+1$.

$$\frac{4x^{4}+4x^{2}+1}{4x^{2}}$$

$\frac{\left(2x^{2}+1\right)^{2}}{4x^{2}}$