This week in Precalculus 11, we learned about solving quadratic equations by factoring.

**Solving**: finding a variable’s value

**Quadratic equation**: an equation with a degree of 2

**Degree**: sum of the variables’ exponents in a term

**Factoring**: separating an expression into its components

**Variable**: an unknown number represented by a letter

The zero-product property is one strategy to solve quadratic equations. If the product of two numbers is 0, than one or both of the numbers have a value of 0. To solve a quadratic equation using this strategy, we first move all terms to one side of the equation, so the other has a value of 0.

Example:

– 6x = 27

– 6x – 27 = 0

We then factor the expression.

(x – 9)(x + 3) = 0

Then we make one of the components have a value of zero and solve for the variable.

x – 9 = 0

x = 0 + 9

x = 9

OR

x + 3 = 0

x = 0 – 3

x = -3

We then verify the solutions by replacing x with the solutions in the original equation.

– 6x = 27

– 6(9) = 27

81 – 54 = 27

27 = 27

OR

– 6x = 27

– 6(-3) = 27

9 + 18 = 27

27 = 27

This is an interesting new way to solve equations, especially since I did not know a variable could have more than one solution.