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Solving Equations by Factoring
Definition of Quadratic Equations
Zero-Factor Property
Strategy for Solving Quadratics
Standard Form Quadratic
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.
BACK
Examples of Quadratic Equations
p
p 18
81
2


18
9
2

 x
x
25
2

y
0
81
18
2


 p
p
Standard Form
0
18
9
2


 x
x
0
25
2


y
BACK
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.
0
)
2
( 

x
2


x
OR (2 1) 0
x  
1
2 
x
2
1
2
2

x 2
1

x
}
2
1
,
2
{

x
BACK
Solve the equation. Check your answers.
0
)
2
)(
5
( 

 x
x
5

x
OR 0
2 

x
2


x { 2, 5}
x  
0
5 

x Solution Set
BACK
Solve each equation. Check your answers.
0
)
3
5
( 

x
x
0

x OR
0
3
5 

x
5
3


x
}
0
,
5
3
{


x
0

x
Solution Set
3
5 

x
5
3
5
5 

x
BACK
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.
BACK
Solve.
18
9
2

 x
x
0
18
9
2


 x
x
0
)
3
)(
6
( 

 x
x
}
3
,
6
{

x BACK
Solve.
0
7
2

 x
x
0
)
7
( 

x
x
}
7
,
0
{

x
BACK
Number Of Solutions
•The degree of a polynomial
is equal to the number of
solutions.
x
x
x 3
2 2
3


Three solutions!!!
Example
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}
BACK
Solve the following equations.
1. x2
– 25 = 0
2. x2
+ 7x – 8 = 0
3. x2
– 12x + 36 = 0
4. c2
– 8c = 0
  
x x
  
5 5 0
  
x x
  
8 1 0
  
x x
  
6 6 0
c c
( )
 
8 0
 
5
,
5 
 
8
,
1 
 
6
 
8
,
0
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.

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Solving Quadratic Equation by Factoring.ppt

  • 1. Solving Equations by Factoring Definition of Quadratic Equations Zero-Factor Property Strategy for Solving Quadratics
  • 2. Standard Form Quadratic 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. BACK
  • 3. Examples of Quadratic Equations p p 18 81 2   18 9 2   x x 25 2  y 0 81 18 2    p p Standard Form 0 18 9 2    x x 0 25 2   y BACK
  • 4. Zero-Factor Property If a and b are real numbers and if ab =0, then a = 0 or b = 0. BACK
  • 5. 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. 0 ) 2 (   x 2   x OR (2 1) 0 x   1 2  x 2 1 2 2  x 2 1  x } 2 1 , 2 {  x BACK
  • 6. Solve the equation. Check your answers. 0 ) 2 )( 5 (    x x 5  x OR 0 2   x 2   x { 2, 5} x   0 5   x Solution Set BACK
  • 7. Solve each equation. Check your answers. 0 ) 3 5 (   x x 0  x OR 0 3 5   x 5 3   x } 0 , 5 3 {   x 0  x Solution Set 3 5   x 5 3 5 5   x BACK
  • 8. 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. BACK
  • 11. Number Of Solutions •The degree of a polynomial is equal to the number of solutions. x x x 3 2 2 3   Three solutions!!!
  • 12. Example 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} BACK
  • 13. Solve the following equations. 1. x2 – 25 = 0 2. x2 + 7x – 8 = 0 3. x2 – 12x + 36 = 0 4. c2 – 8c = 0    x x    5 5 0    x x    8 1 0    x x    6 6 0 c c ( )   8 0   5 , 5    8 , 1    6   8 , 0
  • 14. 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.