Sect. 5.3
Common Factors & Factoring by Grouping
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Definitions
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Factor
Common Factor of 2 or more terms
Factoring a Monomial into two factors
Identifying Common Monomial Factors
Factoring Out Common Factors
Arranging 4 Term Polynomials into 2 Groups
Factoring Out Common Binomials
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What’s a Factor?
product = (factor)(factor)(factor) … (factor)
84 is a product that can be expressed by many different factorizations:
84 = 2(42) or 84 = 7(12) or 84 = 4(7)(3) or 84 = 2(2)(3)(7)
Only one example, 84 = 2(2)(3)(7), shows 84 as the product of prime integers.
Factoring is the reverse of multiplication.
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Factoring Monomials
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12x3 also can be expressed in many ways:
12x3 = 12(x3) 12x3 = 4x2(3x) 12x3 = 2x(6x2)
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Usually, we only look for two factors
Your turn – factor these monomials into two factors:
4a =
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=
x(x2) or x2(x)
14y2 =
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2(2a) or 4(a)
14(y2) or 14y(y) or 7(2y2) or 7y(2y) or y(14y) or …
43x5 =
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43(x5) or 43x(x4) or x3(43x2) or 43x2(x3) or …
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Common Factors
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Sometimes multi-termed polynomials can be factored
Looking for common factors in 2 or more terms …
is the first step in factoring polynomials
Remember a(b + c) = ab + ac
(distributive law)
Consider that a is a common factor of ab + ac
so we can factor ab + ac into a(b + c)
For x2 + 3x the only common factor is x , so
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x2 + 3x = x (? + ?) = x(x + 3)
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Another example: 4y2 + 6y – 10
The common factor is 2
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4y2 + 6y – 10 = 2(? + ? – ?) = 2(2y2 + 3y – 5)
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Check by multiplying: 2(2y2) + 2(3y) – 2(5) = 4y2 + 6y – 10
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Find the Greatest Common Factor
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7a – 21 =
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19x3 + 3x =
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7(? – ?) =
7(a – 3)
x(? + ?) =
x(19x2 + 3)
18y3 – 12y2 + 6y =
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6y(? – ? + ?) =
6y(3y2 – 2y + 1)
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Introduction to Factoring by Grouping:
Factoring Out Binomials
 x2(x
+ 7) + 3(x + 7) =
(x + 7)(? + ?) =
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 y3(a
+ b) – 2(a + b) =
(a + b)(? – ?) =
 (a + b)(y3 – 2)
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You try: 2x2(x – 1) + 6x(x – 1) + 17(x – 1) =
(x – 1)(? + ? – ?)
 (x – 1)(2x2 + 6x + 17)
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Factoring by Grouping
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1.
2.
For polynomials with 4 terms:
Arrange the terms in the polynomial into 2 groups
such that each group has a common monomial
factor
Factor out the common monomials from each group
(the binomial factors produced will be either identical or opposites)
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Factor out the common binomial factor
Example: 2c – 2d + cd – d2
2(c – d) + d(c – d)
(c – d)(2 + d)
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Factor by Grouping
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8t3 + 2t2 – 12t – 3
2t2(4t + 1) – 3(4t + 1)
(4t + 1)(2t2 – 3)
4x3 – 6x2 – 6x + 9
2x2(2x – 3) – 3(2x – 3)
(2x – 3)(2x2 – 3)
y4 – 2y3 – 12y – 3
y3(y – 2) – 3(4y + 1)
Oops – not factorable via grouping
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What Next?
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Next time: Section 5.4 –
Factoring Trinomials
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