1. Solving Quadratic Equations by Factoring
5.7
The Principle of Zero Products
Factoring to Solve Equations

2. A quadratic equation is an equation equivalent to one of the form
Second degree equations like 9t2  4 = 0 and x2 + 6x + 9 = 0 are called quadratic equations.
Quadratic Equation
A quadratic equation is an equation equivalent to one of the form
ax2 + bx + c = 0,
where a, b, and c are constants, with a ≠ 0.

3. The Principle of Zero Products
An equation AB = 0 is true if and only if A = 0 or B = 0, or both.
(A product is 0 if and only if at least one factor is 0.)

4. Solve: (x + 4)(x  3) = 0 Solution In order for a product to be 0, at least one factor must be 0. Therefore, either x + 4 = 0 or x  3 = 0 We solve each equation: x + 4 = 0 or x  3 = 0 x = 4 or x = 3 Both 4 and 3 should be checked in the original equation.

5. Check: For 4: For 3:
(x + 4)(x  3) = 0 (x + 4)(x  3) = 0
(4 + 4)(4  3) (3 + 4)(3  3)
0(7) (0)
0 = = 0
True True
The solutions are 4 and 3.

6. Solve: 4(3x + 1)(x  4) = 0
Solution Since the factor 4 is constant, the only way for (3x + 1)(x  4) to be 0 is for one of the other factors to be That is,
3x + 1 = or x  4 = 0
3x =  or x = 4
x = 4.
Check: For 1/3: For 4:
4(3x + 1)(x  4) = (3x + 1)(x  4) = 0
4( ) + 1)(  4) = (3(4) + 1)(4  4) = 0
4(0)( ) = (13)(0) = 0
0 = = 0
The solutions are 1/3 and 4.

7. Solve: 3y(y  7) = 0 Solution 3  y(y  7) = 0 y = 0 or y  7 = 0 y = 0 or y = 7 The solutions are 0 and 7. The check is left to the student.

8. Factoring to Solve Equations
By factoring and using the principle of zero products, we can now solve a variety of quadratic equations.

9. Solve: x2 + 9x + 14 = 0 Solution This equation requires us to factor the polynomial since there are no like terms to combine and there is a squared term. Then we use the principle of zero products: x2 + 9x + 14 = 0 (x + 7)(x + 2) = 0 x + 7 = 0 or x + 2 = 0 x = 7 or x = 2.

10. Check: For 7: For 2:
x2 + 9x + 14 = 0 x2 + 9x + 14 = 0
(7)2 + 9(7) (2)2 + 9(2)
49  
 
0 = = 0
True True
The solutions are 7 and 2.

11. Solve: x2 + 9x = 0
Solution Although there is no constant term, because of the x2-term, the equation is still quadratic. Try factoring:
x2 + 9x = 0
x(x + 9) = 0
x = or x + 9 = 0
x = or x = 9
The solutions are 0 and 9. The check is left to the student.

12. Caution! We must have 0 on one side of the equation before the principle of zero products can be used. Get all nonzero terms on one side and 0 on the other.

13. Solve: x2  12x = 36 Solution We first add 36 to get 0 on one side: x2  12x = 36 x2  12x + 36 =  (x  6)(x  6) = 0 x  6 = 0 or x  6 = 0 x = 6 or x = 6 There is only one solution, 6.

14. Solve: 9x2 = 49 Solution 9x2 = 49 9x2  49 = 0 (3x  7)(3x + 7) = 0 3x  7 = 0 or 3x + 7 = 0 3x = 7 or 3x = 7 The solutions are

15. Solve: 14x2 + 9x + 2 = 10x + 6 Solution Be careful with an equation like this! Since we need 0 on one side, we subtract 10x and 6 from the right side. 14x2 + 9x + 2 = 10x x2 + 9x  10x + 2  6 = 0 14x2  x  4 = 0 (7x  4)(2x + 1) = 0 7x  4 = 0 or 2x + 1 = 0 7x = 4 or 2x = 1 x = 4/7 or x = 1/2 The solutions are 4/7 and 1/2.

16. Find the x-intercepts for the graph of the equation shown
Find the x-intercepts for the graph of the equation shown. (The grid is intentionally not included) Solution To find the intercepts, we let y = 0 and solve for x. 0 = x2 + 2x  8 0 = (x + 4)(x  2) x + 4 = 0 or x  2 = 0 x = 4 or x = 2 The x-intercepts are (4, 0) and (2, 0).
y = x2 + 2x  8