Algebra
10.4 Solving Equations
in Factored
Form
Quadratic Equations in Standard Form
You have already learned the standard form of
a quadratic equation:
ax² + bx + c = 0 where a ≠ 0
Quadratic Equations in Factored Form
Consider the equation: x² + 8x + 15 = 0
Here is the same equation, in factored form:
(x + 5)(x + 3) = 0
FOIL to be sure
x²+3x +5x +15 x² + 8x + 15 = 0
Two Ways to Solve: x² + 8x + 15 = 0
Quadratic Formula Factored form
-8 ± √ - 4(1)(15)
8² (x + 5)(x + 3) = 0
x=
2(1) This means that for the
equation to be true, either:
-8 ± √ 4
x=
2 x+5=0 or x+3=0
-8 ± 2 -5 -5 -3 -3
x=
-6 -10
=
x = -5 x = -3
2 2 2
x = -3, -5 x = -3, -5
ZERO PRODUCT PROPERTY
If a product equals zero, then one of the
factors must be 0.
If ab = 0, then either a = 0
b=0
or both = 0
Apply the ZERO PRODUCT PROPERTY……
We said that the solutions to the equation
x² + 8x + 15 = 0 or (x +5)(x + 3) are -3 and -5.
Substitute each solution into the equation in factored
form:
Substitute -3 for x (-3 + 5)(-3 + 3) = 0
(2) (0) = 0 TRUE
Substitute -5 for x (-5 + 5)(-5 + 3) = 0
(0) (-2) = 0 TRUE
Substitute into original equation: (-3)² + 8(-3) + 15 = 0 True
(-5)² + 8(-5) + 15 = 0 True
Try these
x + 7 = 0 or x – 3 = 0
(x + 7) (x – 3) = 0 x = -7 or x = 3
x - 4 = 0 or x – 5 = 0
(x - 4)(x – 5) = 0
x = 4 or x = 5
2x – 5 = 0 or x – 3 = 0
(2x – 5)(x – 3) = 0
+5 +5 +3 +3
2x = 5 x=3
2 2
5
x= x=3
2
Shortcut:
When the factors are in the
form (x +__ ) or (x - __ ) the
(x + 7) (x – 3) = 0 solutions will always be the
opposite of the factors, in
this case -7 and +3.
Ask yourself: Ask yourself:
what value of what value of
x will make x will make
this binomial this binomial
= 0? = 0?
-7 +3
Shortcut:
(3x + 2) (x – 1) = 0
Ask yourself: what value of x
will make this binomial = 0?
The other
solution is +1
(3 -2 + 2) = 0
3
The denominator would cancel with 3 when multiplied
The numerator would cancel with +2 when added
One solution is 2
-
3
Try these
(2x + 1) (x – 3) = 0 x = -½ or x = 3
4
(3x - 4)(2x + 5) = 0 x= or x = - 5
3 2
(3x – 5)(2x + 3) = 0 5 3
x= or x = -
3 2
A Few together from the HW
P. 600 #29, 43
Homework
pg. 600 # 19 - 46, 69-77