91Ó°ÊÓ

Factoring quadratics is a method used to solve quadratic equations, which are equations of the form \(ax^{2} + bx + c = 0\). The goal is to express the quadratic equation as a product of two binomials. In the example given, the equation \(u^{2} + 5u + 6 = 0\) is factored. To do this, we need to find two numbers that:

• Multiply to give the constant term, which is 6.
• Add up to give the middle coefficient, which is 5.

Coincidentally, 2 and 3 fit these criteria since \(2 \times 3 = 6\) and \(2 + 3 = 5\). Thus, the equation can be expressed as \( (u + 2)(u + 3) = 0 \). Splitting the quadratic into binomials makes it easier to solve as we can use the Zero Product Property.

The substitution method helps in simplifying complex equations by temporarily replacing a part of the equation with a new variable. This method makes the equation easier to handle. Start by identifying the expression you want to substitute. In the exercise, \(x + 3\) is replaced by \(u\). This sets \(u = x + 3\), transforming the original equation \( (x+3)^{2} + 5(x+3) + 6 = 0 \) into a simpler form \( u^{2} + 5u + 6 = 0 \). This reduces the complexity of handling nested expressions. After solving the simpler equation, you substitute back to find the solution in terms of the original variable. So, for \(u = -2\), back-substitution gives \(x = -5\) as \(x + 3 = -2\). Similarly, for \(u = -3\), it results in \(x = -6\) as \(x + 3 = -3\).

Once solutions are found, it's crucial to check if they satisfy the original equation to ensure there are no errors. For the given problem, after finding \(x = -5\) and \(x = -6\), we substitute these values back into the original equation:

• For \(x = -5 \: (x + 3)^{2} + 5(x + 3) + 6 = 0\) turns into \( (-5 + 3)^{2} + 5(-5 + 3) + 6 = 0 \) which simplifies to \( 0 = 0 \).
• For \(x = -6 \: (x + 3)^{2} + 5(x + 3) + 6 = 0\) turns into \( (-6 + 3)^{2} + 5(-6 + 3) + 6 = 0 \) which also simplifies to \( 0 = 0 \).

This confirms that both solutions satisfy the original equation, validating their correctness. Checking solutions is a vital step to verify that all arithmetic and algebraic manipulations were done correctly.