1. Chapter 5 Section 7
Polynomial Equations and Their Applications.

2. Standard Form of a Quadratic Equation.
ax2 + bx + c = 0
where a, b, c are real numbers with a ≠ 0
Also called a second-degree polynomial equation in x

3. What can you conclude?
5(x) = 0
4p = 0

4. The Zero-Product Principle
If the product of two algebraic expression is zero, then at least one of the factors is equal to zero.
Example:
(x – 5)(x + 2) = 0
Either x – 5 = 0 or x + 2 = 0
So . . .

5. Solve by Factoring
Solve: 2x2 – 5x = 12

6. What is the difference Factoring a Polynomial Factor: 2x2 – 5x - 12
Solving a Quadratic Equation
Solve: 2x2 – 5x – 12 = 0

7. Solve the Quadratic Equation
5x2 = 20x
x2 + 4x = 8x - 12
(x – 7)(x + 5) = - 20

8. The Pythagorean Theorem
The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.
a2 + b2 = c2
a and b are legs a
c is hypotenuse b c

9. Page 393 number 82
A piece of wire measuring 20 feet is attached to a telephone pole as a guy wire. The distance along the ground from the bottom of the pole to the end of the wire is 4 feet greater than the height where the wire is attached to the pole. How far up the pole does the guy wire reach?
Draw a picture to describe the situation.

10. Modeling Motion Show Motion problem Page 386, ex 4
You throw a ball straight up from a rooftop 384 feet high with an initial speed of 32 feet per second. The function
s(t) = -16t2 + 32t + 384
describes the ball’s height above the ground, s(t), in feet, ‘t’ seconds after you throw it. The ball misses the rooftop on its way down and eventually strikes toe ground. How long will it take for the ball to hit the ground?

11. Summary Standard form of a Quadratic Equation Zero Product Principle
Solve Quadratic Equations
Polynomial Equations
Model Motion
Pythagorean Theorem