This week in Pre-calc  11, we have spent a lot of time reviewing the basics of factoring one new thing I learned and will go over this lesson is solving quadratic equations and using zero product property to determine what the variable is.

Understanding Quadratic Equations

Before we get into the Zero Product Property, let’s refresh our understanding of quadratic equations. A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable (usually denoted as ‘x’) is 2. Its standard form is:

ax^2+bx+c=0

A, b, and c are constants, and a cannot be zero.

How can you identify a quadratic equation? Well, look at the leading X if it’s holding an exponent of 2 then it will be quadratic.

We have to make both sides equal at that’s by making one side equal 0 throughout the process of factoring.

After solving for the variable you must end up with two solutions.

When factoring you should end up with two factors

Make sure to follow the zero product property:

The Zero Product Property

The Zero Product Property is a fundamental concept in algebra, particularly when dealing with quadratic equations. It states that if the product of two or more factors is zero, then at least one of those factors must be zero.

Mathematically, this can be represented as follows:

If a×b= 0, then either $a=0$or b$=0$ (or both).

Solving Quadratic Equations Using the Zero Product Property

Now that we understand the Zero Product Property, let’s apply it to solve quadratic equations step by step:

Step 1: Write the quadratic equation in standard form.

For example, let’s say we have the equation:

X^2 − 5x + 6=0

Step 2: Factor the quadratic equation, if possible.

Sometimes, the quadratic equation can be factored into two binomial factors. For our example, we can factor:

X^2 − 5x + 6

$(x-2)(x-3)$

Step 3: Apply the Zero Product Property.

Set each factor equal to zero:

$x-2=0\text{and}x-3=0$

Step 4: Solve for ‘x’ in each equation.

For the first equation, $-2=0$, add 2 to both sides to isolate x, yielding $x=2$.

For the second equation, $-3=0$, add 3 to both sides to isolate x, giving us $x=3$.

Step 5: Verify the solutions.

Plug the solutions back into the original equation to ensure they satisfy it. In our example, substituting $x=2$ and $x=3$ into

X^2− 5x + 6=0 confirms that both values are indeed solutions.