Factoring (foil)
Completing the Square
Quadratic equation
(Click one to begin)
2
2x
+ 10x + 8=-4
SET THE EQAUTION
EQUAL TO ZERO
2x2+10x+8=-4
To set the equation equal to zero
simply add four to both sides of the
equation.
Click for answer
2
2x
+ 10x + 12= 0
TAKE OUT THE GREATEST
COMMON FACTOR (GCF)
2
2x
+ 10x + 12=0
To complete the next step of this method you
take out the Greatest Common Factor (GCF) or
largest number that is a factor of every term in
the equation. For this equation the GCF is 2,
because it is the largest factor of ALL the terms.
Click for answer
2
2(x
+ 5x + 6)= 0
SET UP TWO SETS OF
PARENTHESIS ( ) ( )…
(
… Easy enough
)(
)
Step 3
continued
The next thing that you need to do to solve the equation is learn
how to “FOIL”. To set up your “FOIL” equation you place the
letters of the word FOIL to the corresponding places shown below.
(
F
L
O
I
)(
F
L
I
O
)
CONTINUE FOILING
The next step in “FOILing” the equation is to take the first term in
the trinomial and take half of the exponent and put the variable and
exponent in the “F” places in the equation. The trinomial in this
equation is (x2 + 5x + 6), so the first term would be x2. Half of the
exponent (2) would be 1, so you would put x1 (x) in the “F” places.
Click for answer
F
L
(x
O
F
L
) (x
I
O
)
I
WHAT OPERATION
TO USE?
After you have the first term in each of your Parenthesis you need to
put an operation, either addition or subtraction, in the equation. You
need to know whether or not the operations will be the same, and what
they will be, addition or subtraction. To do this you look at the second
sign, or operation, in the trinomial. If it is positive (addition) the signs
will be the same (whatever the first sign is), if it is negative
(subtraction) the signs are different. In our example problem the
trinomial is (x2 + 5x + 6), so the second sign is addition. We now
know that the signs will be the same. The first sign is positive, so we
know that both sets of parenthesis will have addition for a sign.
Click for answer
(x+
) (x+
)
FIND THE
SECOND TERM
Now that you have the first term, and operation of both sets of
parenthesis, you need to find the second term for each set. To do
this you have to list the factors of the last term in the trinomial. In
our equation the last term in the trinomial is 6. (x2 + 5x + 6).
Click for answer
6
1
6
2
3
WHAT TO DO WITH
THE FACTORS?
If the second sign in your trinomial is addition, you find the pair of factors that add
to get the second term in the trinomial. If the second sign is subtraction, you find the
pair of factors that subtract to get the second term. The second term in our problem is
addition, so you would have to find the factor pair of six, that adds to get five. If
there are no factor pairs that add, or subtract to equal B, the equation is PRIME.
(x2 + 5x + 6)
Click for answer
6
1
+
6=7
2
+
3=5
3 and 2 are the factor pair
you want, because they
add to equal 5.
Leading coefficient
other than one
(Continued)
4y  6 y  2  0
2
If the leading coefficient of an equation is a value other than one, you have
to find the factors of (a) (first coefficient), and (c) (third coefficient). You
multiply the factors of (a) by the factors of (c), and find a combination that
either add or subtract to equal (b) (depending on the sign). To determine
where the factors a placed in the two sets of parenthesis you find the greater
product of the two sets. You take the last value of the pair of factors with
the greatest product, and place it in the last spot in the parenthesis that has a
corresponding sign with that in the first spot in the equation. Click for answer
Factors of 4
4
2
1
4
2
2
1
2
Factors of 2
(1)
(2) = 2
(4)
(1) = 4
2 and 4 add to equal 6
(4y+2)(y+1)
y  -1,-
1
2
FINISHING UP
This factor pair is going to be the last thing you put into your set of
parenthesis to solve the problem. Once you have found your factor
pair you have to decide which value goes into which parenthesis. To
do this you have to look at the signs that you put in your parenthesis in
step 5. If the signs are the same, place the factor pairs in the “L” spots.
If the signs are different, place the greatest factor of the pair in the
parenthesis that contains the same sign as the first of the trinomial. If
the first sign of the trinomial is (–) and your factor pair is (6,3) then
you place the 6 in the parenthesis with the (–) sign. In our problem the
signs are both addition, so you place one value in each parenthesis.
Click for answer
2(x + 3)(x + 2)
Finding x’s values
Now that you have your new equation 2(x+3)(x+2) you need to
find out what x equals. To do this you set the equation equal to
zero, and x will then equal the opposite of whatever is being
added to it, or subtracted from it, in each parenthesis.
2 (x + 3) (x + 2) = 0
Click for answer
x= -3, -2
This is your final answer
To check you answer you go use the “FOIL” set up, and multiply the
value under each letter in the first parenthesis, by the corresponding
value of the same letter in the second parenthesis.
F
L
F
L
(x + 3) (x + 2)
F: (x)(x)= x2
O: (x)(2)= 2x
I: (3)(x)= 3x
L: (3)(2)= 6
O
I
I
O
Once you have these values, you add them all
together. If the equation you get corresponds to
the original equation your answer is correct.
Click for answer
2
ax
+ bx + c = 0
2
x –
7x – 10= 0
MOVE THE
CONSTANT
2
x – 7x
– 10 = 0
The first step in solving an equation using this method is to move the
constant to the right side of the equal sign. The terms with variables
stay on the left side of the equal sign. In this equation, you move the
constant to the right side of the equal sign by adding it (10) to both
sides.
Click for answer
2
x
– 7x = 10
LEADING
COEFFICIENT
In some equations, the leading coefficient will be a value other
than one. In the equation 2x2 +4x=8, the leading coefficient is 2.
If you have an equation like this, you divide every term by the
leading coefficient.
2
2x
+ 4x = 8
(Divide every term by 2)
2
x
+ 2x = 4
Click for answer
USING “B”
x2 – 7x = 10
The next step in solving this equation involves using the formula for
completing the square, which is, ax2 + bx + c= 0. The next thing that you
would do is to take (1/2 b)2 and add it to both sides of the equation. In
our equation b is –7. Half of 7 is (7/2). (7/2)2 is (49/4)
Click for answer
49
49
x  7x 
 10 
4
4
2
SIMPLIFY
49
49
x  7x 
 10 
4
4
2
The next step you take is to simplify the right side of the equation. To
do this in this equation, you would add 10 to the quantity of (49/4). In
order to add, the two values must have the same denominator. You
could change 10 to (40/4). This would give you common
denominators so you could then simply add the terms.
49 40 49
x  7x 


4
4
4
2
49 89
x  7x 

4
4
2
Click for answer
REWRITE
49 89
x  7x 

4
4
2
Once you have simplified the right side of the equation, the next thing to
do is rewrite the left side of the equation as a binomial squared. To do
this, you take the quantity of x plus or minus (depending on what the first
sign is) half of b and square that quantity. The first sign in this equation
is subtraction, so we would subtract half of b from x and square the
quantity.
Click for answer
7

x  
2

2
SQUARE ROOT
2
7  89

x   
2
4

After you rewrite the left side of the equation, and the equation looks
similar to the one shown above, the next thing to do is take the square
root of both sides of the equation. The left side of the equation is a
quantity squared, so the square root of it is simply that quantity. For the
right side of the equation, you take the square root of the numerator and
the denominator and keep it as a fraction. Since the square root of a term
is a positive and negative value, the left side will equal + or - the right.
Click for answer
7  89
x 
2
2
SOLVE
7
89
x 
2
2
The final step in solving the equation using completing the square is to solve for x. To
do this, the first thing you need to do, is get x alone on one side of the equation. Then
you would move the other values on the same side as x to the other side of the equation.
In this problem you would ADD (7/2) to both sides of the equation. You would then
have x alone and could simplify to get your final answer. In some equations the number
inside the square root sign will be negative. If your equation has an answer like that,
you could leave it as it is, or write “no rational roots”. You could write this because the
square root of a negative number is not rational.
Click for answer
x
7
89

2
2
(SIMPLIFY)
7  89
x
2
 b  b  4(a )(c)
2(a)
2
4x  (3)x  2  0
2
Learning the
formula
ax  bx  c  0
2
4x  (3)x  2  0
2
Using the Quadratic equation (shown above) you can set up your
equation to work in the quadratic formula (shown below).The
example equation is set up to work in the quadratic formula.
Click for answer
 b  b  4(a )( c)
2a
2
x=
Setting up
the formula
4x  (3)x  2  0
2
Now that the equation is set up in the form of the correct form you
can substitute the values of your equation for the corresponding
variables in the quadratic formula (shown below)
4x  (3)x  2  0
2
Click for answer
a=4 c=2
b=(-3)
 b  b  4( a )( c)
2a
2
x=
=
x=
3  (3)2  4(4)( 2)
2(4)
Simplify
3  (3)  4(4)( 2)
x=
2(4)
2
After you have put all the values in the equation into the quadratic
formula the next step is to simplify (using order of operations).
Click for answer
3  9  4(4)( 2)
2(4)
=
3  9  32
8
=
x=
3   23
8
Finishing up
x=
3   23
8
In this equation the final answer contains the square root of a negative
number. The answer to this would be not be rational, so therefore your
final answer could be “no rational roots” or leave it as it is. This is only
the case in problems where you have a square root of a negative number.
If your answer contains the square root of a positive number, you could
simplify (if it contains a perfect square) or leave it as is.
x=
4 8
3
Can be reduced to:
(Click for answer)
42 2
3
Practice
problems
Menu
Click for problem 1
Solve by factoring
y  7 y  6  6
2
ANSWERS
(choose one)
y  3,4
y  3,4
y  3,4
y  3,4
y  7 y  6  6
2
y  7 y  12  0
( y  3)( y  4)  0
2
y  3,4
Solve by using the
quadratic formula
2x  5 x  3  0
2
3
x  ,1
2
3
x  ,1
2
1
x  -3,
2
-3
x  ,1
2
2x  5 x  3  0
2
 5  52  4(2)(3)
x
2(2)
 5  25  24
x
4
 5 1
x
4
5 1
x

4 4
3
x
,1
2
Solve by completing
the square
z  24z  63  0
2
3
z  21,
2
z  21,3
z  21
, 3
21
z  ,3
3
z  24z  63  0
2
z  24z  63
2
z  24z  144  81
2
z  12
2
  81
z  12  9
z  12  9
z  21,3
Solve by
factoring
2w  10 w  8  0
2
w  -1,-4
1
w  ,1
4
Prime
-1
w  ,3
4
2w  10 w  8  0
2
2w
2
 5w  4  0
2w  1w  4  0
w  1
, 4
Solve by using the
quadratic formula
x  3x  1
2
Answer
3  13
x
2
x 2  3x  1
3  9  4(1)( 1)
2(1)
3  9  ( 4)
2
3
13
2
Solve by completing
the square
x  x 1 0
2
Answer
 1 5
x
2
x2  x - 1  0
x  x 1
2
1 5
x x 
4 4
2
2
 1 5
x -  
 2 4
1  5
x 
2
2
 1 5
x
2
1
5
x 
2 2
Solve by factoring
3x2  7x  2  0
1
x  3,
2
1
x  ,2
3
1
x   ,2
3
Prime
3x  7x  2  0
2
(x  2)(3x  1)  0
1
x   ,2
3
Solve by using
quadratic formula
4g  9g  8  0
2
Answer
- 9  - 47
g
8
(no rational roots)
4g  9g  8  0
2
- 9  81 - 4(4)(8)
2(4)
- 9  81 - 128
8
- 9  - 47
8
Solve by
completing the
square
3r  3r  4  0
2
Answer
3   39
r
6
(no rational roots)
3r 2  3r  4  0
3r 2  3r  4
r2  r  
4
3
1
13
r r 
4
12
2
1
13


r -   
12
 2
2
r
1
  13

2
2 3
r
3   39
6
x  3x  1  0
2
x  3,1
Prime
x  3,1
1
x  ,1
3
The answer to this equation is PRIME (b), because
no factors of 1, add to equal 3.