Download Review of factoring

Prepared by Frieda Nilforoushan
Review of factoring
Factoring means put an expression in multiplication from that is ( )( )( )( )….., such that non-of the
expressions inside parentheses could break down further.
Example: (
)(
)(
)
There are four methods of factoring:




Pull-out the greatest common factors ( G.C.F.) of all terms.
Factoring by Grouping.
Factoring second degree tri-nomials.
Factoring, using special formulas.
Pull-out the greatest common factors (G.C.F.) of all terms
Example: Factor completely.
Solution:
12 x3  4 x 2 y  16 x 4
12 x3  4 x 2 y  16 x 4
There are 3 terms, (remember, terms are separated by + or -).
4 x2 (
The GCF of 12, 4 and 16 is 4. x 2 is common to all terms. Y is not
)
common to all terms. So, the GCF of all 3 terms is 4x 2 . Pull out
and put it behind a set of parentheses, i.e. 4 x 2 (
) . Now
here is the rule for writing the terms inside parentheses ( of
course you can guess the terms, however, I don’t recommend
it): Divide each term of the expression by the GCF i.e.,
12 x3
 3x ,
4 x2
4 x 2 (3x  y  4 x 2 )
4 x2 y
y
4x2
16 x 4
,
 4 x 2
2
4x
Now, the expression is factored completely.
Note 1: You can always check your factoring by multiplications:
4 x2 (3x  y  4 x 2 )  12 x3  4 x 2 y 16 x 4
Note 2: You can pull out the positive or negative of the common factor that is;
4 x2 (3x  y  4 x 2 )
Example: Factor completely
Solution:
r12  r10
r12  r10
There are two terms. r10 , is the common factor of both terms.
r10 (r 2  1)
Pull it out, put it behind a set of parentheses. Then divide each
r10 (
term by the common factor to find the inside terms:
)
r12
r10
2

r
,
1
r10
r10
r10 (r 2  1)
Example: Factor completely.
The expression is now factored completely.
8 p5q 2  16 p6 q3  12 p 4 q7
Solutions: 8 p5 q 2  16 p6 q3  12 p 4 q7 , We have three terms.
4 p 4 q 2 (2 p  4 p 2 q  3q5 ) ,
Example: Factor completely
Solution:
is the greatest common factor.
1 9 3 2
x  x
4
4
1 9 3 2
x  x
4
4
We have two terms.
1 2
x (
4
The greatest common factor is
)
.
1 2 7
x ( x  3)
4
Note: Just remember, in any method of factoring, you must pull-out the GCF of all terms first. Then try
other methods of factoring. You will see the examples later.
Factoring by Grouping
This method is used when we have even number of terms; four or more.
Example: Factor completely
Solution:
XP  4P  XY  4Y
X P  4P  XY  4Y
We have 4 terms that have nothing in common.
(
)
(
)
However, if we group the first two terms, and the last
two, each group has something in common that we can
be pulled-out .
(
)(
)
Now, we have two terms: P( X  4), Y ( X  4) . These
two terms have ( X  4) in common. Pull-out ( X  4)
and put it behind a set of parentheses. Then write the
terms inside the parentheses, using the technique I
showed you in the first method above i.e:
(
)(
Example: Factor completely
Solution:
P( X  4)
 P,
( X  4)
)
Y ( X  4)
Y
( X  4)
18P2  12PT  3 XP  2 XT
36P2  24PT  6 XP  4 XT
We have 4 terms and they all have 2 in common. So, we
2(18P2  12PT  3 XP  2 XT ) must pull-out 2 (GCF) first. Then, group the first two
terms, and the last two terms inside parentheses. Now,
pull-out the common factors from each group.
2[6P(3P  2T )  X (3P  2T )] Now, inside brackets we have two terms that have
2(3P  2T )(6P  X )
Example: Factor completely
Solution:
A2  6 A  AB  6B
A ( A  6)  B( A  6)
Notice that you need to factor –B from the second
( A  6)( A  B)
group in order to see the common factor.
Example: Factor completely
Solution:
(3P  2T ) in common. Pull it out.
1  M  MN  N
1  M  MN  N
Group the first two and the last two terms.
1(1  M )  N (M  1)
You can always pull-out 1 if terms have nothing in
common.
1(1  M )  N (1  M )
(1  M )(1  N )
Factoring trinomials
I learned this method of factoring for second degree polynomials in High School. It is a quick and
accurate way to factor second degree trinomials.
Example: Factor the second degree trinomial 6 x 2  7 x  2
Solution:
1. Always pull-out the GCF of all terms, if there is one.
2. Multiply the first and the last number, 6.2=12 , this is the product
of two numbers. The coefficient of x (middle term) would be the sum of
two numbers.
3. Use trial and error to find two numbers whose product is 12 and sum is 7,
and put them in (x+….)(x+….). In this case ( x  3)( x  4) .
6 x2  7 x  2
Product=12
1.12
2.6
3.4
Sum=7
3+4=7
( x  3)( x  4)
4. Must divide each number by the coefficient of x2 i.e.
5. You must simplify each fraction and then move each denominator over to x.
3
4
( x  )( x  )
6
6
1
2
( x  )( x  )
2
3
(2 x  1)(3x  2)
Example: Factor completely 20 x 2  18x  4
Solution:
1. Always pull-out the GCF of all terms if there is one.
2(10 x 2  9 x  2)
2. Now we need to factor 10 x 2  9 x  2 . We must find two numbers whose
product is 20 (multiply the first and the last number, 10.2=20), and sum is 9
( the coefficient of x term).
3. Use trial and error to find two numbers whose product is 20 and sum is 9,
Product=20
1.20
2.10
4.5
and put them in (x+….)(x+….). In this case ( x  4)( x  5) .
2( x  4)( x  5)
4. Must divide each number by 10, the coefficient of
x2 .
5. You must simplify each fraction, and then move each denominator
over to x.
2( x 
4
5
)( x  )
10
10
2
1
2( x  )( x  )
5
2
2(5x  2)(2 x  1)
Sum=9
4+5=9
Example: Factor completely 12 x 2  17 x  5
Solution:
12 x2  17 x  5
1. Always pull-out the common factor of all terms if there is one.
2. Now, we need to factor 12 x 2  17 x  5 . Multiply the first and the last
number, 12.(-5)=-60 , this is the product of two numbers. The coefficient
of x, would be the sum of two numbers.
3. Use trial and error to find two numbers whose product is -60 and sum is
-17, and put them in (x+….)(x+….). In this case
P= - 60
1.-60
2.-30
3.-20
S= - 17
3+(-20)
=-17
( x  3)( x  20)
4. Must divide each number by the coefficient of x2 in this
(x 
3
20
)( x  )
12
12
Case 12
5. You must simplify each fraction and then move each denominator
1
5
( x  )( x  )
4
3
(4 x  1)(3x  5)
over to x.
Factoring using special formulas
Recall that (a  b)(a  b)  a 2  b2 . If you switch sides, we get: a 2  b2  (a  b)(a  b) . As you see
the right hand side is the factored form of the left hand side.
Steps to factor difference of two squares a 2  b2 :
1. Open up two sets of parentheses
a 2  b2  (
2. Find square roots of a 2 and b 2 .
Square root of a 2 = a
)(
)
Square root of b2  b
3. Add a and b and put them in one set of
Parentheses, and subtract them, and put them in the other.
2
Example: Factor completely x  16 .
Solution:
x2  16  ( )(
)
Sq. root of x 2  x , Sq. root of 16  4
a 2  b2  (a  b )(a  b )
x2  16  ( x  4)( x  4)
Example: Factor completely a 2b2  1 .
Solution:
a 2b2  1  ( )(
)
Sq. root of a 2b2  ab , Sq. root of 1  1
a 2b2  1  (ab  1)(ab  1)
Example: Factor completely 128  2y 4 .
Solution:
128  2 y 4  2( 64  y 2 )
 2(
)(
Must pull-out the GCF first.
)
4
2
Sq. root of 64  8 , Sq. root of y  y
 2(8  y 2 )(8  y 2 )
Example: Factor completely x  y .
8
Solution:
x8  y8  ( )(
8
)
Sq. root of x8  x 4 , Sq. root of y  y
8
x8  y8  ( x4  y 4 )( x4  y 4 )
4
Sum of two squares (
) is prime and not
Not factorable. But (
) is.
( x8  y8 )  ( x4  y 4 )( x2  y 2 )( x2  y 2 )
( x8  y8 )  ( x4  y 4 )( x2  y 2 )( x  y )( x  y)