INTRODUCTION TO
FACTORING
POLYNOMIALS
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Janice Levasseur
Definitions
• Recall: Factors of a number are the numbers
that divide the original number evenly.
• Writing a number as a product of factors is
called a factorization of the number.
• The prime factorization of a number is the
factorization of that number written as a product
of prime numbers.
• Common factors are factors that two or more
numbers have in common.
• The Greatest Common Factor (GCF) is the
largest common factor.
Ex: Find the GCF(24, 40).
Prime factor each number:
24
2 12
2 6
2 3  24 = 2*2*2*3 = 23*3
 GCF(24,40) = 23 = 8
40
2 20
2 10
2 5  40 = 2*2*2*5 = 23*5
The Greatest Common Factor of terms of a
polynomial is the largest factor that the
original terms share
• Ex: What is the GCF(7x2, 3x)
7x2 = 7 * x * x
3x = 3 * x
The terms share a factor of x
 GCF(7x2, 3x) = x
Ex: Find the
5
3
2
GCF(6a ,3a ,2a )
• 6a5 = 2*3*a*a*a*a*a
• 3a3 = 3*a*a*a
• 2a2 = 2*a*a
The terms share two
factors of a
 GCF(6a5,3a3,2a2)= a2
Note: The exponent of the variable in the GCF is
the smallest exponent of that variable the terms
Definitions
• To factor an expression means to write
an equivalent expression that is a product
• To factor a polynomial means to write the
polynomial as a product of other
polynomials
• A factor that cannot be factored further is
said to be a prime factor (prime
polynomial)
• A polynomial is factored completely if it is
written as a product of prime polynomials
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
– Is the polynomial a difference of squares?
• a2 – b 2
– Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
– Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Ex: Factor 7x2 + 3x
Think of the Distributive Law: a(b+c) = ab + ac
 reverse it ab + ac = a(b + c)
Do the terms share a common factor?
What is the GCF(7x2, 3x)?
Recall: GCF(7x2, 3x) = x
Factor out
 7 x + 3 x = x(
What’s left?
2
x
+
x
 7x2 + 3x = x(7x + 3)
)
Ex: Factor 6a5 – 3a3 – 2a2
Recall: GCF(6a5,3a3,2a2)= a2
3
1
3a3 –
6a5 –
a2
a2
2a2 = a2( 6a3 - 3a - 2 )
a2
 6a5 – 3a3 – 2a2 = a2(6a3 – 3a – 2)
Ex: Factor x(a + b) – 2(a + b)
Always ask first if there is common factor the terms share . .
.
x(a + b) – 2(a + b) Each term has factor (a +
b)
 x(a + b) – 2(a + b) = (a + b)( x – 2 )
(a + b)
(a + b)
 x(a + b) – 2(a + b) = (a + b)(x – 2)
Ex: Factor a(x – 2) + 2(2 – x)
As with the previous example, is there a common factor
among the terms?
Well, kind of . . . x – 2 is close to 2 - x . . . Hum . . .
Recall: (-1)(x – 2) = - x + 2 = 2 – x
 a(x – 2) + 2(2 – x)
=
a(x – 2) + 2((-1)(x – 2))
= a(x – 2) + (– 2)(x – 2)
= a(x – 2) – 2(x – 2)
 a(x – 2) – 2(x – 2) = (x – 2)( a
(x – 2)
(x – 2)
– 2
)
Ex: Factor b(a – 7) – 3(7 – a)
Common factor among the terms?
Well, kind of . . . a – 7 is close to 7 - a
Recall: (-1)(a – 7) = - a + 7 = 7 – a
 b(a – 7) – 3(7 – a)
=
b(a – 7) – 3((-1)(a – 7))
= b(a – 7) + 3(a – 7)
= b(a – 7) +3(a – 7)
 b(a – 7) + 3(a – 7) = (a – 7)( b
(a – 7)
(a – 7)
+3
)
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
– Is the polynomial a difference of squares?
• a2 – b 2
– Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
– Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Factor by Grouping
•
1.
2.
3.
4.
5.
If the polynomial has four terms, consider
factor by grouping
Factor out the GCF from the first two terms
Factor out the GCF from the second two terms
(take the negative sign if minus separates the
first and second groups)
If factor by grouping is the correct approach,
there should be a common factor among the
groups
Factor out that GCF
Check by multiplying using FOIL
Ex: Factor 6a3 + 3a2 +4a + 2
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
GCF(6a3,
1 2)
3a
2
2
2a
=
3a2
 6a3 + 3a = 3a (2a + 1 )
2
2
3a
3a
Common factor among the 2nd two
terms?
GCF(4a, 2) = 2
2
1
 4a + 2 = 2( 2a + 1
2
2
Now put it all together . . .
)
6a3 + 3a2 +4a + 2 = 3a2(2a + 1) + 2(2a + 1)
Four terms  two terms. Is there a common
factor?
Each term has factor (2a + 1)
3a2(2a + 1) + 2(2a + 1) = (2a + 1)( 3a2 + 2 )
(2a + 1)
(2a + 1)
6a3 + 3a2 +4a + 2 = (2a + 1)(3a2 + 2)
Ex: Factor 4x2 + 3xy – 12y – 16x
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
4x
GCF(4x2,
3y
3xy)
=x
 4x2 + 3xy = x( 4x + 3y )
x
x
Common factor among the 2nd two
terms?
GCF(-12y, - 16x) = -4
3y
4x
 -12y – 16x = - 4( 3y + 4x )
-4
-4
Now put it all together . . .
4x2 + 3xy – 12y – 16x = x(4x + 3y) – 4(4x + 3y)
Four terms  two terms. Is there a common
factor?
Each term has factor (4x + 3y)
x(4x + 3y) – 4(4x + 3y) = (4x + 3y)( x – 4 )
(4x + 3y)
(4x + 3y)
4x2 + 3xy – 12y – 16x = (4x + 3y)(x – 4)
Ex: Factor 2ra + a2 – 2r – a
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
GCF(2ra, a2) = a
 2ra + a2 = a(2r + a )
a
a
Common factor among the 2nd two
terms?
GCF(-2r, - a) = -1
 -2r – a = - 1( 2r + a )
-1 -1
Now put it all together . . .
2ra + a2 –2r – a = a(2r + a) – 1(2r + a)
Four terms  two terms. Is there a common
factor?
Each term has factor (2r + a)
a(2r + a) – 1(2r + a) = (2r + a)( a – 1 )
(2r + a)
(2r + a)
2ra + a2 – 2r – a = (2r + a)(a – 1)
To factor a polynomial completely, ask
• Do the terms have a common factor
(GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
– Is the polynomial a difference of squares?
• a2 – b2
– Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two
binomials?
• Factored completely?
Special Polynomials
Is the polynomial a difference of squares?
• a2 – b2 = (a – b)(a + b)
Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 = (a + b)2
• a2 – 2ab + b2 = (a – b)2
Ex: Factor x2 – 4
Notice the terms are both perfect squares
and we have a difference  difference of
squares
x2 = (x)2
4 = (2)2
 x2 – 4 = (x)2 – (2)2 = (x – 2)(x + 2)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor 9p2 – 16
Notice the terms are both perfect squares
and we have a difference  difference of
squares
9p2 = (3p)2
16 = (4)2
 9a2 – 16 = (3p)2 – (4)2 = (3p – 4)(3p + 4)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor y6 – 25
Notice the terms are both perfect squares
and we have a difference  difference of
squares
y6 = (y3)2
25 = (5)2
 y6 – 25 = (y3)2 – (5)2 = (y3 – 5)(y3 + 5)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor 81 – x2y2
Notice the terms are both perfect squares
and we have a difference  difference of
squares
2
81 = (9)
x2y2 = (xy)2
 81 – x2y2 = (9)2 – (xy)2 = (9 – xy)(9 + xy)
a2 – b2 = (a – b)(a + b)
factors as
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
– Is the polynomial a difference of squares?
• a2 – b 2
– Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
– Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
FOIL Method of Factoring
• Recall FOIL
– (3x + 4)(4x + 5) = 12x2 + 15x + 16x + 20 = 12x2 + 31x + 20
 The product of the two binomials is a trinomial
 The constant term is the product of the L terms
 The coefficient of x, b, is the sum of the O & I
products
 The coefficient of x2, a, is the product of the F
terms
FOIL Method of Factoring
1. Factor out the GCF, if any
2. For the remaining trinomial, find the F
terms (__ x + )(__ x +
) = ax2
3. Find the L terms ( x + __ )( x + __ ) = c
4. Look for the outer and inner products to
sum to bx
5. Check the factorization by using FOIL to
multiply
Ex: Factor b2 + 6b + 5
1. there is no GCF
2. the lead coefficient is 1  (1b
3. Look for factors of 5
)(1b
)
1, 5 & 5, 1
(b + 1)(b + 5) or (b + 5)(b + 1)
4. outer-inner product?
(b + 1)(b + 5)  5b + b = 6b
or (b + 5)(b + 1)  b + 5b = 6b
Either one works  b2 + 6b + 5 = (b + 1)(b + 5)
5. check: (b + 1)(b + 5) = b2 + 5b + b + 5
= b2 + 6b + 5
Ex: Factor y2 + 6y – 55
1. there is no GCF
2. the lead coefficient is 1  (1y
)(1y
)
3. Look for factors of – 55 1, -55 & 5, - 11 & 11, - 5 & 55, - 1
(y + 1)(y – 55) or (y + 5)(y - 11) or ( y + 11)(y – 5) or (y + 55)(y –
1)
4. outer-inner product?
(y + 1)(y - 55)  -55y + y = - 54y(y + 5)(y - 11)  -11y + 5y = -6y
(y + 55)(y - 1)  -y + 55y =
54y
2
(y + 11)(y - 5)  -5y + 11y = 6y
 y + 6y - 55 = (y + 11)(y – 5)
5. check: (y + 11)(y – 5) = y2 – 5y + 11y - 55
= y2 + 6y – 55
Factor completely – 3 Terms
•
Always look for a common factor
–
–
•
•
immediately take it out to the front of the expression all common
factors
show what’s left inside ONE set of parenthesis
Identify the number of terms.
If there are three terms, and the leading coefficient is positive:
–
–
–
–
–
find all the factors of the first term, find all the factors of the last term
Within 2 sets of parentheses,
•
place the factors from the first term in the front of the parentheses
•
place the factors from the last term in the back of the parentheses
NEVER put common factors together in one parenthesis.
check the last sign,
•
if the sign is plus: use the SAME signs, the sign of the 2nd term
•
if the sign is minus: use different signs, one plus and one minus
“smile” to make sure you get the middle term
•
multiply the inner most terms together then multiply the outer most
terms together, and add the two products together.
Factor completely:
2
2x
– 5x – 7
• Factors of the first term: 1x & 2x
• Factors of the last term:
-1 & 7 or 1 & -7
• (2x – 7)(x + 1)
Factor Completely.
2
4x
+ 83x + 60
• Nothing common
• Factors of the first term: 1 & 4 or 2 & 2
• Factors of the last term: 1,6 2,30 3,20 4,15 5,12
6,10
• Since each pair of factors of the last has an even number,
we can not use 2 & 2 from the first term
•
(4x + 3)(1x + 20 )
Sign Pattern for the Binomials
Trinomial Sign Pattern
Binomial Sign Pattern
+
+
(
+
)(
+
)
-
+
(
-
)(
-
)
-
-
(
-
)(
+
)
+
-
(
+
)(
-
)
But as you can tell from the previous example, the
FOIL method of factoring requires a lot of trial and
error (and hence luck!) . . . Better way?
ac Method: factoring
1.
2.
3.
4.
2
ax
+ bx + c
Factor out the GCF, if any
For the remaining trinomial, multiply ac
Look for factors of ac that sum to b
Rewrite the bx term as a sum using the
factors found in step 3
5. Factor by grouping
6. Check by multiplying using FOIL
Ex: Factor 3
3x2 – 4
4x – 15
1. Is there a GCF? No
2. Multiply ac  a = 3 and c = – 15 3(-15) = 45
3. Factors of -45 that sum to – 4
1
– 45  – 44
3
– 15  – 12
5
–9 –4
Note: although there are more
factors of – 45, we don’t have to
check them since we found
what we were looking for!
4. Rewrite the middle term
3x2 – 4x – 15 = 3x2 – 9x + 5x – 15
Four-term polynomial . . . How should we
proceed to factor?
Factor by grouping . . . 3x2 – 9x + 5x – 15
Common factor among the 1st two terms? 3x
3
 3 x 2 – 9x = 3x( x – 3 )
3x
3x
Common factor among the 2nd two
terms?
5
3
 5 x – 15 = 5( x – 3 )
5
5
 3x2 – 9x + 5x – 15 = 3x(x – 3) + 5(x – 3)
= (x – 3)( 3x + 5 )
Ex: Factor 2
2t2 + 5t
5 – 12
1. Is there a GCF? No
2. Multiply ac  a = 2 and c = – 12 2(-12) = 24
3. Factors of -24 that sum to 5
1
– 24  – 23
Close but wrong sign so reverse it
2
– 12  – 10
-3 8 5
3
–8 –5
4. Rewrite the middle term
2t2 + 5t – 12 = 2t2 – 3t + 8t – 12
Four-term polynomial . . . Factor by grouping . .
.
2t2 – 3t + 8t – 12
Common factor among the 1st two terms? t
3
 2 t 2 – 3t = t( 2t –
t
t
3)
Common factor among the 2nd two
terms?
2
4
3
 8 t – 12 = 4( 2t – 3 )
4
4
 2t2 – 3t + 8t – 12 = t(2t – 3) + 4(2t – 3)
= (2t – 3)( t + 4 )
Ex: Factor 9
9x4 + 18
18x2 + 8
1. Is there a GCF? No
2. Multiply ac  a = 9 and c = 8
9(8) = 72
3. Factors of 72 that sum to 18
1 72  73
Bit big  think bigger factors
3 24  27
6 12  18 
4. Rewrite the middle term
9x4 + 18x2 + 8 = 9x4 + 6x2 + 12x2 + 8
Four-term polynomial . . . Factor by grouping . .
.
9x4 + 6x2 + 12x2 + 8
Common factor among the 1st two terms? 3x2
3x2
2
 9x4 + 6x2 = 3x2(3x2 + 2
3x2 3x2
)
Common factor among the 2nd two
terms?
3
4
3
 12x2 + 8 = 4( 3x2 + 2 )
4
4
 9x4 + 6x2 + 12x2 + 8 = 3x2(3x2 + 2) + 4(3x2 + 2)
= (3x2 + 2)( 3x2 + 4
)
y + 6y2 Pick one to be the
Ex: Factor 12
12x2 – 17
17xy
variable
two variables
1. Is there a GCF? No, but notice
2. Multiply ac  a = 12 and c = 6y2  12(6y2) = 72y2
3. Factors of 72y2 that sum to - 17y
-1y -72y  -73y
-6y -12y  -18y
-8y -9y  -17y 
Each factor need a y, both need to
be negative
Too big, think bigger factors
4. Rewrite the middle term
12x2 – 17xy + 6y2 = 12x2 – 8xy – 9xy + 6y2
Four-term polynomial . . . Factor by grouping . .
.
12x2 – 8xy – 9xy + 6y2
Common factor among the 1st two terms? 4x
3x
2y
 12x2 – 8xy = 4x( 3x – 2y )
4x
4x
Common factor among the 2nd two
terms?
3
- 3y
-2y
 – 9xy + 6y2 = - 3y( 3x – 2y )
-3y
-3y
 12x2 – 8xy – 9xy + 6y2 = 4x(3x – 2y) – 3y(3x – 2y)
= (3x – 2)( 4x – 3y )
Ex: Factor x3 + 3x2 – 4x – 12
1. Is there a GCF? No
2. Notice four terms  grouping
Common factor among the 1st two terms? x2
x
 x3 + 3x2 = x2( x
x2
+ 3 )
x2
Common factor among the 2nd two
terms?
3
 – 4x – 12 = – 4( x
-4
+3
)
-4
 x3 + 3x2 - 4x – 12 = x2(x + 3) – 4(x + 3)
= (x + 3)( x2 – 4 )
-4
Cont: we have (x + 3)(x2 – 4)
But are we done? No. We have to make sure
we factor completely.
Is (x + 3) prime?  can x + 3 be factored further?
No . . . It is prime
What about (x2 – 4)? Recognize it?
Difference of Squares
x2 = (x)2
4 = (2)2
 x2 – 4 = (x)2 – (2)2 = (x – 2)(x + 2)
Therefore x3 + 3x2 – 4x – 12 = (x + 3)(x2 – 4)
= (x + 3)(x – 2)(x + 2)
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
– Is the polynomial a difference of squares?
• a2 – b 2
– Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
– Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Special Polynomials
Is the polynomial a sum/difference of cubes?
• a3 + b3 = (a + b)(a2 - ab + b2)
• a3 – b3 = (a - b)(a2 + ab + b2)
Ex: Factor 8p3 – q3
Notice the terms are both perfect cubes
and we have a difference  difference of cubes
8p3 = (2p)3
q3 = (q)3
 8p3 – q3 = (2p)3 – (q)3 = (2p – q)((2p)2 + (2p)(q) + (q)2)
a3 – b3 = (a – b)(a2 + ab + b2)
factors as
= (2p – q)(4p2 + 2pq + q2)
Ex: Factor x3 + 27y9
Notice the terms are both perfect cubes
and we have a sum  sum of cubes
x3 = (x)3
27y9 = (3y3)3
 x3 + 27y9 = (x)3 + (3y3)3 = (x + 3y3)((x)2 - (x)(3y3) + (3y3)2)
a3 + b3 = (a + b)(a2 - ab + b2)
factors as
= (x + 3y3)(x2 – 3xy3 + 9y6)