```Solving Equations by Factoring
Zero-Factor Property
Equation
Quadratic equations can be written in the form
ax2 + bx + c = 0
where a, b, and c are real numbers with a  0.
Standard form for a quadratic equation
is in descending order equal to zero.
Standard Form
p  81  18 p
p  18 p  81  0
x  9 x  18
x  9 x  18  0
y  25
y  25  0
2
2
2
2
2
2
Zero-Factor Property
If a and b are real numbers
and if ab =0, then
a = 0 or
b = 0.
BACK
Solve the equation (x + 2)(2x - 1)=0
• By the zero factor property we know...
• Since the product is equal to zero then one
of the factors must be zero.
( x  2)  0
x  2
1
x  {2, }
2
OR (2 x  1)  0
2x  1
2x 1

2 2
1
x
2
( x  5)( x  2)  0
x  5  0 OR x  2  0 Solution Set
x5
x  2 x  {2, 5}
x(5 x  3)  0
x0
x0
Solution Set
3
x  { , 0}
5
OR
5x  3  0
5 x  3
5
3
x
5
5
3
x
5
Solving a Quadratic Equation by Factoring
Step 1 Write the equation in standard
form.
Step 2 Factor completely.
Step 3 Use the zero-factor property.
Set each factor with a variable equal
to zero.
Step 4 Solve each equation produced
in step 3.
Solve.
x  9 x  18
2
x  9 x  18  0
2
( x  6)( x  3)  0
x  {6, 3}
Solve.
m  3m  10  0
2
( m  5)( m  2 )  0
m  {5,  2}
Solve.
x  7x  0
x ( x  7)  0
2
x0
x7  0
x  {0, 7}
Number Of Solutions
• The degree of a polynomial
is equal to the number of
solutions.
x  2 x  3x
3
2
Three solutions!!!
x  2 x  3x
2
x( x  2 x  3)
3
2
x (x + 1)(x – 3) = 0
Set each of the three factors equal to 0.
x=0
x+1=0
x = -1
x–3=0
x=3
Solve the resulting equations.
Write the solution set.
x = {0, -1, 3}
Solve.
r  2r  8
2
r  2r  8  0
( r  2)( r  4 )  0
r  {2,4}
2
x2 – 9x + 20 = 0
(x – 4)(x – 5) = 0
x–4 = 0
x=4
x–5=0
x=5
x = {5, 4}
• Standard form
• Factor
• Set each factor = 0
• Solve
• Write the solution
set
4x2 – 49 = 0
Example
(2x + 7)(2x – 7) = 0
• Standard form
2x + 7 = 0
7
x
2
2x – 7 = 0
7
x
2
•
•
•
•
Factor
Set each factor = 0
Solve
Write the solution set
 7 7
x   , 
 2 2
How would you solve the following equation?
x2 – 36 = 0
Factor the polynomial.
(x - 6)(x + 6) = 0
(x – 6) = 0 or (x + 6) = 0.
Therefore
x–6=0
+ 6 +6
x=6
or
x+6=0
- 6 -6
x = -6
This equation has two solutions or zeros:
x = 6 or x = -6.
Solve the following equations.
1. x2 – 25 = 0  x  5 x  5  0 5,5
2. x2 + 7x – 8 = 0  x  8 x  1  0 1,8
3. x2 – 12x + 36 = 0  x  6 x  6  0 6
4. c2 – 8c = 0
c ( c  8)  0
0,8
1. Get a value of zero on one side of the
equation.
2. Factor the polynomial if possible.
3. Apply the zero product property by
setting each factor equal to zero.
4. Solve for the variable.
Solving Equations by Factoring
Zero-Factor Property
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