FACTORING
Factoring is similar to breaking up a number into its multiples. For example, 10=5*2. The multiples
are ‘5’ and ‘2’. In a polynomial it is the same way, however, the procedure is somewhat more
complicated since variables, not just numbers, are involved. There are different ways of factoring an
equation depending on the complexity of the polynomial.
Factoring out the greatest common factor
The first thing to do when factoring is to look at all terms and break up each term into its multiples:
Examples:
Factor: 8 x 3 + 4 x 2 + 10 x
(
)
x 8 x 2 + 4 x + 10 →
In this polynomial the only variable in common to
all is x.
2x(4x+2x+5) →
The two is also common to all terms. Therefore,
this is as far as the polynomial can be factored.
Factor: 6 x 3 + 8 x 2 + 16 x
6 x 3 + 8 x 2 + 16 x
(
2 x 3x 2 + 4 x + 8
)
→
The common variable is x, and the smallest
exponent is 1. The common multiple is 2.
Therefore, the greatest common factor is 2x.
You can also look at it this way:
(2 x )(3x 2 ) = 6 x 3
(2 x )(4 x ) = 8 x 2
(2 x )(8) = 16 x
Factoring by grouping
There will be occasions when there are no common factors for all terms, but there will be terms that
have a variable in common while other terms have a different variable in common. In this case, we can
factor by grouping together these common terms.
Examples:
Factor: ax + ay + bx + by
ax + ay + bx + by
ax + bx + ay + by
x(a + b ) + y (a + b )
→
Notice how there are two variables, x and y. Group
together the x’s and the y’s and then
factor out the common factor in each group.
Factor: 4 − 2 x 2 + x 2 y 3 − 2 y 3
4 − 2x 2 + x 2 y 3 − 2 y 3 →
(
)
(
)
→
(
)
(
)
→
2 2 − x2 + y3 x2 − 2
2 2 − x2 − y3 2 − x2
(2 − x )(2 − y )
2
There are two variables present, x and y.
Start by grouping two terms at a time.
Notice how the two polynomials inside the
parentheses are similar. To make them the
same change the sign in the second
polynomial.
Now that they are the same, factor out the
common term.
Now it is completely factored out.
→
3
Factoring a difference of squares
For a difference of squares, a 2 − b 2 , the factors will be (a − b )(a + b ) .
Examples:
x 2 − y 2 = (x − y )(x + y )
( ) =(1 − x )(1 + x ) =(1 − x )(1 + x ) (1 + x )
1 − x 4 =(1) − x 2
2
2
2
2
2
Factoring the sum and difference of two cubes
For the addition or subtraction of two cubes, the following formulas apply:
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a 3 − b3
Examples:
2
+ a +bb 2
)
)
x3 + 8
3
3
= ( x ) + ( 2)
(
= ( x + 2) ( x
→ Manipulate to be in “ a 3 + b3 ” form.
=+
( x 2 ) x 2 − ( x )( 2 ) + 22
2
− 2x + 4
)
8 y 3 − 27
=
)
→ follow formula.
→ Manipulate to be in “ a 3 − b3 ” form.
→
( 2 y ) − ( 3)
2
=
( 2 y − 3) ( ( 2 y ) + ( 2 y )( 3) + 32 )
=
3
3
follow formula.
( 2 y − 3) ( 4 y 2 + 6 y + 9 )
Factoring trinomials
Factoring trinomials is based on finding the two integers whose sum and product meet the given
requirements.
Example:
Find two integers whose sum is –11 and whose product is 30.
Step 1 – The product is positive, the sum is negative, therefore, both integers must be
negative.
( − )( − ) =+ and ( − ) + ( − ) =−
Step 2 – Possible pairs of factors that will give a positive answer:
30
1. ( −1)( −30 ) =
30
3. ( −3)( −10 ) =
30
2. ( −2 )( −15 ) =
30
4. ( −5 )( −6 ) =
Step 3 – The pair whose sum is –11 is –5 and –6. Therefore, the critical integers are –5
and –6, ( −5 − 6 =−11) .
Factoring trinomials of the form x 2 + bx + c
To factor trinomials of the form x 2 + bx + c follow the same principle described above. Find two
integers whose sum equals the middle term and whose product equals the last term.
Examples:
Factor: x 2 − x − 6
Step 1 – Find two integers whose sum is equal to –1 and whose product is equal to –6.
The product is negative; therefore, we need integers that have different sign.
The sum is negative, so the bigger integer is negative.
Step 2 – Possible pairs of factors that will give us –6:
1. (− 3)(2 ) = −6
3. (− 1)(6 ) = −6
2. (3)(− 2 ) = −6
4. (1)(− 6 ) = −6
Step 3 – The pair whose sum is –1 is –3 and 2, (− 3 + 2 = −1) .
The critical integers are –3 and 2.
Step 4 – The factors for x 2 − x − 6 are ( x − 3)( x + 2) .
Factor: x 2 + 16 x + 60
Step 1 – Find two integers whose sum is 16 and whose product is 60. Both integers must
be positive since their sum and product are positive.
Step 2 – Possible pairs of factors:
1. (1)( 60 ) = 60
3. ( 3)( 20 ) = 60
5. ( 5 )(12 ) = 60
2. ( 2 )( 30 ) = 60
4. ( 4 )(15 ) = 60
6. ( 6 )(10 ) = 60
Step 3 – The pair whose sum is 16 is (6, 10). Therefore, the critical integers are 6, 10.
Step 4 – The factors for x 2 + 16 x + 60 are ( x + 6)( x + 10) .
Factoring trinomials of the form ax2 + bx + c
In this type of trinomials the coefficient of x 2 is not 1, therefore we must consider the product of a ⋅ c .
Examples:
Factor: 2 x 2 − 7 x − 4
−8 . The
Step 1 – Find two integers whose sum is –7 and whose product is ( 2 )( −4 ) =
integers must be of different signs since their product is a negative number. The
larger integer is negative since their sum is negative.
Step 2 – Possible pairs of factors:
−8
1. ( −1)( 8 ) =
−8
3. ( −2 )( 4 ) =
−8
2. (1)( −8 ) =
−8
4. ( 2 )( −4 ) =
Step 3 – The pair whose sum is –7 is 1 and –8. The critical integers are 1 and –8.
Step 4 – Use the critical integers to break the first degree term into two parts:
2 x2 − 7 x − 4 = 2 x2 − 8x + x − 4
Step 5 – Factor by grouping the first two and last two terms:
2 x 2 − 8 x + x − 4 = 2 x ( x − 4 ) + 1( x − 4 )
There are now two terms in the expression. The binomial factor in each of these
two terms should be the same; otherwise, there is an error. In this case, the
common binomial factor is ( x − 4 ) .
Step 6 – Factor out the common binomial factor: 2 x ( x − 4) + 1( x − 4 ) = ( x − 4 )( 2 x + 1)
Factor: 6 x 2 − 23 x + 20
Step 1 – Find two integers whose sum is –23 and whose product is 6(20) = 120. Since
the product is positive and the sum is negative, the two integers are negative.
Step 2 – Possible pairs of factors:
120
1. ( −1)( −120 ) =
120
4. ( −4 )( −30 ) =
120
7. ( −8 )( −15 ) =
120
2. ( −2 )( −60 ) =
120
5. ( −5 )( −24 ) =
120
8. ( −10 )( −12 ) =
120
3. ( −3)( −40 ) =
60
6. ( −6 )( −20 ) =
Step 3 – The sum of –8 and –15 is –23. Therefore, the critical integers are –8 and –15.
Step 4 – Use the critical integers to break the first degree term into two parts:
6 x 2 − 23 x + 20 = 6 x 2 − 15 x − 8 x + 20
Step 5 – Factor, separately, the first two and last two terms:
6 x 2 − 15 x − 8 x + 20 = 3 x ( 2 x − 5) − 4( 2 x − 5)
Step 6 – Factor out the common binomial factor.
3 x ( 2 x − 5) − 4 ( 2 x − 5) = ( 2 x − 5)( 3 x − 4)
FACTORING – EXERCISES
1. 3 x 2 − 10 x − 8
2. 2 x 2 (x − 3) − 5 x( x − 3) − 3( x − 3)
3. x 3 + 2 x 2 − 9 x − 18
4. x 2 − 10 x + 25 − y 2
5. x 6 − y 6
6. a 3 −
1
8
7. ( x + 5) 2 + 4( x + 5) + 4
8. 1 − 2t 3 + t 6
9. 3 x 4 − 18 x 3 + 27 x 2
10. 100 x 2 − 100 x − 600
11. 8 x 2 − 2 x − 15
12. x 2 + 3 xy + 2 y 2
13. 2 x 2 ( x + 2) + 13x( x + 2) + 15( x + 2)
14. x 2 + x − 12
15. x 2 y + 3 y + 2 x 2 + 6
16. 18a 2 − 50
17. x 2 + 8 x + 15
18. 16 x 3 − 20 x 2 − 50 x
19. 9 x 2 + 33 x + 30
20. 12 x 2 + 2 x − 2
FACTORING – ANSWER TO EXERCISES
2. 2 x 2 ( x − 3) − 5 x ( x − 3) − 3 ( x − 3)
1. 3 x 2 − 10 x − 8
3 x 2 − 12 x + 2 x − 8
( x − 3)(2 x 2 − 5 x − 3)
3x ( x − 4 ) + 2 ( x − 4 )
( x − 3)(2 x 2 − 6 x + x − 3)
( 3x + 2 )( x − 4 )
( x − 3)(2 x( x − 3) + 1( x − 3))
( x − 3)( 2 x + 1)( x − 3) = ( x − 3) ( 2 x + 1)
2
4. x 2 − 10 x + 25 − y 2
3. x 3 + 2 x 2 − 9 x − 18
x 2 ( x + 2) − 9( x + 2)
( x 2 − 10 x + 25) − y 2
( x 2 − 9)( x + 2)
x 2 − 9 = ( x + 3)( x − 3)
( x − 3)( x + 3)( x + 2)
( x − 5) 2 − y 2
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
y
where a =
( x − 5) and b =
( x − 5 − y )( x − 5 + y )
5. x 6 − y 6
6. a 3 −
x 6 − y 6=
(x ) −( y )
3
2
3
2
Difference of cubes:
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
3
where
=
a x=
and b y 3
(
x −y = x −y
6
6
3
3
)( x
3
+y
3
)
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a −b
3
2
+ a +bb
(
a 3 − b3 = ( a − b ) a 2 + a +bb 2
a = a and b =
)
1
2
2
1  2

1 1 
=
 a −   a + ( a )   +   
2 

2 2 
Sum & Difference of Cubes:
3
1
8
2
)
)
= ( x + y )( x 2 − xy + y 2 )( x − y )( x 2 + xy + y 2 )
= ( x + y )( x − y )( x 2 + xy + y 2 )( x 2 − xy + y 2 )
1 
a 1

=  a −  a 2 + + 
2 
2 4

7. ( x + 5 ) + 4 ( x + 5 ) + 4 =
2
8. 1 − 2t 3 + t 6 =
(t
3
( x + 7 )( x + 7 ) = ( x + 7 )
)(
) ( t − 1) ( t
2
−1 t3 −1 =
2
)
+ t +1
2
2
2
9. 3 x 4 − 18 x 3 + 27 x=
3 x 2 ( x 2 − 6 x + 9=
) 3 x 2 ( x − 3)
2
10. 100 x 2 − 100 x − 600= 100 ( x − 3)( x + 2 )
11. 8 x 2 − 2 x − 15 =
( 4 x + 5)( 2 x − 3)
12. x 2 + 3 xy + 2 y 2 =( x + 2 y )( x + y )
13. 2 x 2 ( x + 2 ) + 13 x ( x + 2 ) + 15 ( x + 2 ) = ( x + 2 )( 2 x + 3)( x + 5 )
14. x 2 + x − 12 =
( x + 4 )( x − 3)
15. x 2 y + 3 y + 2 x 2 + 6=
(x
2
)
+ 3 ( y + 2)
16. 18a 2 − 50= 2 ( 3a + 5 )( 3a − 5 )
17. x 2 + 8 x + 15 = ( x + 5 )( x + 3)
18. 16 x 3 − 20 x 2 − 50 x= 2 x ( 2 x − 5 )( 4 x + 5 )
19. 9 x 2 + 33x + 30 = 3 ( x + 2 )( 3 x + 5 )
2 2 ( 3 x − 1)( 2 x + 1)
20. 12 x 2 + 2 x −=