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Factoring quadratics is an essential skill when dealing with quadratic expressions. Let's explore this by focusing on the quadratic expression in our problem: \( x^2 + 8x + 12 \). Factoring involves rewriting this expression as a product of two binomials. To achieve this, we need to find two numbers that both add up to the middle coefficient (which is 8 in this case) and multiply to give the constant term, 12.

• Look for factors of 12: 1 and 12, 2 and 6, 3 and 4.
• Only the pair 2 and 6 add up to 8.

Therefore, the quadratic \( x^2 + 8x + 12 \) can be factored into \( (x+2)(x+6) \). Factoring simplifies complex expressions, making them more manageable to work with.

Cross multiplication is a technique used to solve equations involving fractions. In our problem, we arrive at the equation \( \frac{x-1}{x+2} = x+6 \) after simplification. Cross multiplication helps eliminate the fraction, turning the equation into a more straightforward form.
Here's how cross multiplication works in this context:

• Multiply both sides by the denominator from the left side, \( x+2 \).

This results in: \[ x - 1 = (x+2)(x+6) \] Now, the equation no longer contains fractions, and you can proceed with solving a linear equation.

Solving quadratic equations is a critical task in algebra. After expanding the expression \( (x+2)(x+6) \), we have the quadratic equation \( x^2 + 8x + 12 \). Rearrange it to standard form: \( x^2 + 7x + 13 = 0 \).
There are several methods to solve quadratics:

• Factoring: This is effective when the quadratic can be easily factored, but not every quadratic is factorable.
• Completing the Square: This involves turning every quadratic into a perfect square trinomial.
• Quadratic Formula: Works for any quadratic equation and we'll cover it later.

In our scenario, the equation does not factor nicely, so we would proceed using the Quadratic Formula.

When a quadratic equation cannot be easily factored, the Quadratic Formula provides a reliable solution. This formula is a staple in algebra because it works universally for any standard quadratic equation of the form \( ax^2 + bx + c = 0 \).
The formula is as follows:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our equation, \( x^2 + 7x + 13 = 0 \), we identify the coefficients as \( a=1 \), \( b=7 \), and \( c=13 \). Substituting these values results in:\[ x = \frac{-7 \pm \sqrt{49 - 52}}{2} \]Since \( b^2 - 4ac = -3 \), realize the discriminant is negative, indicating complex solutions:\[ x = -3.5 \pm \sqrt{2.25} \]The Quadratic Formula allows us to find the solutions, even when they are not real numbers.