**1. Basic Factoring**

To solve this quadratic equation by factoring, we can look for two numbers that multiply to give 8 and add up to 6. These numbers are 2 and 4.

Therefore, we can rewrite the equation as

Hence, the solutions to the equation \(x^2 + 6x + 8 = 0\) are \(x = -2\) and \(x = -4\).

**2. Basic Factoring with negative numbers**

To solve this quadratic equation by factoring, we are still looking for two numbers that multiply to give 8 but this time they add up to -6. These numbers are -2 and -4.

Therefore, we can rewrite the equation as

Hence, the solutions to the equation \(x^2 - 6x + 8 = 0\) are \(x = 2\) and \(x = 4\).

**2. Basic Factoring with a constant (Part A)**

We ignore the 3 and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).

So, we haveHence, the solutions to the equation \( 3(x^2 - 6x + 8) = 0\) are \(x = 2\) and \(x = 4\).

**2. Basic Factoring with a constant (Part B)**

Now as in the previous problem, we ignore the 4 and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).

So, we haveHence, the solutions to the equation \( 4x^2 - 24x + 32 = 0\) are \(x = 2\) and \(x = 4\).

**2. Factoring out the gcf **

Now as in the previous problems, we ignore the \( 4x \) term and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).

So, we haveHence, the solutions to the equation \( 4x^3 - 24x^2 + 32x = 0\) are \( x = 0, x = 2\) and \(x = 4\).

**2. Two minor modifcations to the quadratic **

Now as discussed above, we factor the quadratic as \((x - 2)(x - 4)\).

Setting each factor equal to zero, we haveHence, the solutions to the equation \( -x^2 + 6x - 8 = 0\) are also \(x = 2\) and \(x = 4\).

Then, we can proceed to solve the quadratic equation as usual. So, as before, we factor the quadratic as \((x - 2)(x - 4)\).

Setting each factor equal to zero, we haveHence, the solutions to the equation \( -x^2 + 6x - 8 = 0\) are also \(x = 2\) and \(x = 4\).

**2. Missing Constant Term **

Setting each factor equal to zero, we have:

Therefore, the solutions to the equation \(x^2 - 6x = 0\) are \(x = 0\) and \(x = 6\).

**2. Missing \( x \) Term **

**2. The \(ax^2 + bx + c \) method **

All quadratics can be written as \(ax^2 + bx + c = 0\).

When we compare equation \(3x^2 + 5 = 6x\) with \(ax^2 + bx + c = 0\), we getIn the equation \(3x^2 + 5 = 6x\), we first move the \( 6x \) to the left to get the equation \(3x^2 -6x + 5 = 0\). Then we get

**2. The Quadratic Formula (Your Last Resort) **

All quadratics can be solved using this formula: