Solving by Factoring

Today's lesson revisits the concept of solving quadratic equations. Last time, we used square roots to solving quadratics (September 30th). This time, we are using the Zero Product Property. This property states that the only way to multiply 2 (or more) factors together for a product of 0 is if at least one of the factors is already 0. (Zero times any number results in zero!)

1. Set the equation equal to zero.
2. Completely factor the polynomial expression.
3. Set each factor equal to zero and solve the equations.

Example: Solve r² + 12r - 15 = 30

Step 1 - Set equal to zero, here subtract 30 from both sides: r² + 12r - 45 = 0

Step 2 - Factor the polynomial: (r + 15) (r - 3) = 0

Step 3 - Set each factor equal to zero and solve: r + 15 = 0 or r - 3 = 0. So, r = -15 or r = 3.

Here, we find out that there are 4 vocabulary terms that are very similar in use/definition: solutions, roots, zeros, and x-intercepts. The methods are almost the exact same for finding any one of those terms. However, when we are presented with a function in function notation, step 1 - Set equal to zero simply becomes rewrite f(x) as 0.

Example: Find the zeros of f(x) = 5x² + 15x - 50.

Step 1 - Set equal to zero, here rewrite f(x) as 0: 0 = 5x² + 15x - 50

Step 2 - Factor the polynomial, here a gcf and then a quadratic:

First, 0 = 5 (x² + 3x - 10)

Then, 0 = 5 (x + 5) (x - 2)

Step 3 - Set each facgtor equal to zero and solve: 5 = 0 (not possible!) or x + 5 = 0 or x - 2 = 0. So, x = -5 or x = 2.

Homework: Page 79, #2 - 12 even and #22 - 30 even. Page 84, #10 - 24 even.

Remember: Quiz Thursday. Review session Thursday morning, 7:45am, in room 4303

---