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Factoring Methods
When factoring a polynomial, there are three steps you want to follow.
1. Check if there’s a Greatest Common Factor (GCF)
2. Check if the polynomial can be factored by grouping (4 term
polynomial)
3. Check if the polynomial is a trinomial or a Special Case
Greatest Common Factor
A Common Factor is a number that can be divided into each term of the
polynomial evenly. The Greatest Common Factor (G.C.F.) is the largest common
factor by which you can evenly divide each term of the polynomial. Keep in
mind that 1 is not a GCF, but it might be needed in certain examples when
factoring by grouping.
Example:
Factor
3x2 + 15x
The G.C.F. is 3x
1. Write the G.C.F. before of the parentheses
2. Divide both terms by the G.C.F.
3. Write the quotient inside of the parentheses
3𝑥 2 15𝑥
+
3𝑥
3𝑥
3x(x + 5)
Other examples:
Factor
m2 + mn
12x3 + 16x2 – 8x
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m(m + n)
4x(3x2 + 4x – 2)
Factoring
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Factoring by Grouping
This factoring method applies to four term polynomials. The steps are as
follows:
1. Check if there’s a G.C.F.
2. If there’s no G.C.F. factor as follows:
Example:
Factor
2ab + 12a + 3b + 18
2ab + 12a|+ 3b + 18
Make two groups of two terms and take G.C.F. of
each pair separately
2a(b + 6) + 3(b + 6)
Factor (b + 6) which is the GCF from each group
(b + 6)
(b + 6)
(b + 6)(2a + 3)
and divide each term by GCF
Group the remaining terms in another
parentheses
Example: (Intermediate Algebra)
Factor
6xy – 5 – 15x +2y
6xy - 5|-15x + 2y
1(6xy – 5) – 1(15 -2y)
The parentheses don’t match
6xy +2y|– 15x -5
Re-arrange the terms
2y(3x + 1) -5(3x +1)
(3x + 1)(2y – 5)
Other examples:
Factor by Grouping
5q2 – 4pq – 5q +4p
(5q – 4p)(q – 1)
16x2 + 4xy2 + 8xy +2y3
2(4x + y2)(2x + y)
Intermediate Algebra Examples
4 + xy -2y – 2x
(2 – y)(2 – x)
3m2n3 + 15m2n – 2m2n2 – 10m2
m2 (n2 + 5)(3n – 2)
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Factoring
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Factoring Trinomial of the Form ax2 +bx +c where a = 1
To factor trinomials of this form, the trinomial has to be in descending order.
1. Find factors of c, where the sum of the factors adds up to b
Example:
Factor x2 - 5x +6
a=1 b=-5 c=6
Find factors of 6 that have a sum of -5
Factors
6
6·1
3·2
Sum of factors
-5
-6-1
-3-2
The factors of x2 - 5x +6 are
(x - 3)(x – 2)
Example:
Factor x2 – x – 2
Find factors of -2 that have a sum of -1
Factors
-2
2·1
Sum of factors
-1
-2+1
(x – 2)(x + 1)
Other examples:
Factor
a2 – 9ab + 18b2
(a – 3b)(a – 6b)
10t – 24 + t2
(t – 2)(t + 12)
3x2 +9x – 30
3(x + 5)(x -2)
6x3 + 54x2 + 120x
6x(x + 4)(x + 5)
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Factoring
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Factoring Trinomial of the form ax2 +bx +c where a  1
Determining Signs when Factoring Trinomial of the form ax2 +bx +c where
a1
This is a helpful hint to decide which signs should be placed on each factor
when dealing with trinomials of the form ax2 + bx + c:

When the last sign is (+), both factors take the sign of the middle term
ax2 + bx + c (___ + ___)(___ + ___)
ax2 – bx + c (___ - ___)(___ - ___)

When the last sing is (-), one factor is (+) and the other is (-)
ax2 – bx – c (___ + ___)(___ - ___)
ax2 + bx - c
(___ + ___)(___ - ___)
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Factoring
5
Scissors Method
Example:
Factor
3x2 + 10x – 8
1. Check if the trinomial has a G.C.F.
2. The trinomial has to be in descending order and the leading
coefficient should be positive.
3. List all the factors of a and c
4. Keep in mind that if the last term of the trinomial is negative, the
factors should be one (+) and one (-).
5. List the factors of a and c vertically and multiply crisscross
(scissors) as follows:
Factors of 3
3·1
Factors of 8
8·1
4·2
(3
2)
(1
4)
-2
1 times 2 negative sign
+12 3 times 4 positive sign
10
(3x - 2)(x + 4)
Factor
24x2 + 41x + 12
Keep in mind that the last term of the trinomial is positive, so the factors
should have the sign of the middle term.
Factors of 24
24·1
12·2
8·3
6·4
Factors of 12
12·1
6·2
4·3
(6
3)
12
(4
4)
24
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Factoring
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You can clearly that this combination won’t work, because there is a
G.C.F. for (4 and 4)
(8
3)
+9
Positive sign
(3
4)
+32
Positive sign
41
(8x + 3)(3x + 4)
Factor
8x2 – 14xy + 3y2
Factors of 8
8·1
4·2
Factors of 3
3·1
(4
3)
-6
(2
1)
-4
-10
Switch one of the factors
(4
1)
-2
Negative sign
(2
3)
-12
Negative sign
-14
(4x – 1y)(2x – 3y)
Other examples:
Factor
25n2 – 5n - 6
(5n + 2)(5n – 3)
12x3 – 34x2 + 24x
2x(3x - 4)(2x – 3)
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Factoring
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Factoring Trinomial of the form ax2 +bx +c where a  1
Trial and Error Method
Example:
Factor
12x2 - 5x – 2
1. Check if the trinomial has a G.C.F.
2. The trinomial has to be in descending order and the leading
coefficient should be positive.
3. Find products of a and c.
4. Choose inner and outer terms. Use various combinations of the
factors from Step 3 until the necessary middle term is found.
(____x ____) (____x ____)
Factors of 12 are 12·1
6·2
4·3
We try 4·3 for the first two terms
( 4x
)( 3x
)
Factor of -2 are -2·1 or
2·-1. We try both possibilities
(4x
- 2 )(3x + 1) Notice that 4x – 2 have a common factor, so this
combination is wrong.
8x
(4x
- 1 )(3x + 2) 8x – 3x = 5x. Wrong middle term
-3x
Interchange the sings
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Factoring
8
-8x
(4x
+ 1 )(3x - 2) -8x + 3x = -5x. Correct middle term
3x
Therefore the factors of 12x2 - 5x – 2 are (4x + 1 )(3x - 2)
Other examples:
18m2 – 19mn – 12n2
(9m + 4n)(2m – 3n)
-3x2 + 16x + 12
-(3x + 2)(x – 6)
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Factoring
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Special cases
Perfect Square Trinomials
A perfect square trinomial has the following form:
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
To identify a perfect square trinomial, check the following:
1. The first term and the last term are perfect squares.
2. The coefficient of the middle term is twice the square root of the
first term multiplied by the square root of the last term.
3. The last sign has to be positive.
Example:
Factor
4m2 + 20m + 25
(2m)2
+ (5)2
Middle term 2(2m)(5) is 20m
4m2 + 20m + 25 is a Perfect Square Trinomial.
The factored form is:
(2m + 5)2
Example:
Factor
p2 - 8p + 64
(p)2
+ (8)2
Middle term 2(p)(8) is not 8p
Not a Perfect Square Trinomial
Other Examples:
Factor
144a2 – 216ab + 81b2
(12a - 9b)2
49x2 + 14xy + y2
(7x + y)2
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Factoring
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Difference of Perfect Squares
The Difference of Perfect Squares has the following form:
a2 – b2 = (a + b)(a – b)
To identify a difference of perfect squares
1. The first term and the last term are perfect squares
2. The terms have opposite signs
Example:
Factor
x2 – 16
Both terms are perfect squares and they have opposite
signs
(x + 4)(x -4)
Example:
Factor
81a2 – 25b2
Both terms are perfect squares and they have opposite
signs
(9a + 5b)(9a – 5b)
Other examples:
4x2 – 25y2
(2x + 5y)(2x – 5y)
36a2b2 – 1
(6ab + 1)(6ab – 1)
2y2 – 18
2(y + 3)(y – 3)
3x3 – 48x
3x(x + 4)(x – 4)
Intermediate Algebra Examples:
a4 - 16
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(a2 + 4)(a + 2)(a – 2)
Factoring
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Sum and Difference of Perfect Cubes
(Intermediate Algebra)
The Sum and Difference of perfect Cubes have the following general forms:
a3 – b3 = (a - b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 - ab + b2)
To identify a difference of perfect squares
1. Both terms have to be perfect cubes
2. Follow the general form
Example:
Factor
x3 + 27
(x + 3)(x2 – 3x + 32)
(x + 3)(x2 – 3x + 9)
Example:
Factor
8x3 – 125
(2x – 5)[(2x)2 + 10x + (5)2]
(2x – 5)(2x2 + 10x + 25)
Other examples
8p3 -1
(2p – 1)( 4p2 + 2p +1)
2x3 + 2000
2(x + 10)(x2 – 10x + 100)
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Factoring
12
Factoring a Polynomial Using Substitution
(Intermediate Algebra)
Example:
Factor
(y + z)2 - 81
Let t = y + z
t2 – 81
Substitute t in the original binomial
(t + 9)(t – 9)
Factor the difference of perfect squares
(y + z + 9)(y + z – 9)
Replace t with y + z
Example:
Factor
2(x + 3)2 + 5(x + 3) – 12 Let t = x + 3
2t2 + 5t – 12
Factor using scissors or trial and error
(2t – 3)(t + 4)
Replace t with x + 3
[2(x + 3) – 3][(x + 3) + 4]
(2x + 6 – 3)(x + 3 +4)
Simplify
(2x + 3)(x + 7)
Example:
Factor
(a – 4)3 – b3
Let t = a – 4
t3 – b3
Factor as difference of cubes
(t – b)(t2 + tb + b2)
Replace t with a – 4
(a – 4 – b)[(a – 4)2 + (a – 4)b + b2]
(a – 4 – b)(a2 – 8a + 16 – ab – 4b + b2)
Other examples:
(x – 2y)2 – 4
(x – 2y + 2)(x – 2y -2)
4(p + 2) + m(p + 2)
(p + 2)(4 + m)
(p + 8q)2 – 10(p + 8q) + 25
(p + 8q – 5)2
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Factoring