Transcript Factoring By Grouping

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Factor 𝑎𝑥 + 𝑏𝑥 + c
Factoring a Binomial
There are two possibilities when you are given a binomial.
• It is a difference of squares
• There is a monomial to factor out.
Common Monomial Factor
15𝑥 2 + 3𝑥
= 3𝑥(5𝑥 + 1)
8𝑥 2 + 12𝑥
= 4𝑥(2𝑥 + 3)
24𝑥 2 + 7𝑥
= 𝑥(24𝑥 + 7)
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Special Cases
The Difference of Squares :
𝑎2 − 𝑏2 = 𝑎 + 𝑏 𝑎 − 𝑏
For example :
Perfect Squares
Square root of
the first number
4𝑥 2 − 9 = (2𝑥 + 3)(2𝑥 − 3)
Square root of the
second number
Must be
subtraction
Addition and then subtraction
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Special Cases
The Difference of Squares :
𝑎2 − 𝑏2 = 𝑎 + 𝑏 𝑎 − 𝑏
Factor each expression.
a) 𝑥 2 − 16
= 𝑥 + 4 (𝑥 − 4)
b) 16𝑥 2 − 121
= 4𝑥 + 11 (4𝑥 − 11)
c) 81𝑑 4 − 16
= 9𝑑 2 + 4 (9𝑑 2 − 4)
d) 𝑥 2 𝑦 2 − 𝑧 2
= 𝑥𝑦 + 𝑧 (𝑥𝑦 − 𝑧)
= 9𝑑 2 + 4 (3𝑑 + 2)(3𝑥 − 2)
Factoring Quadratic Trinomials When a≠1
*Always look for a common factor amongst all of the terms first,
ESPECIALLY look to see if you can divide each term by the coefficient of 𝑥 2
to make the problem an a=1 problem.
1) −𝑥 2 − 4𝑥 + 32
= −1(𝑥 2 + 4𝑥 − 32)
= −1(𝑥 − 4)(𝑥 + 8)
-32  4
-1,32
-2,16
-4,8
2) 3𝑥 2 + 15𝑥 + 18
= 3(𝑥 2 + 5𝑥 + 6)
= 3(𝑥 + 2)(𝑥 + 3)
6  5
1,6
2,3
Factoring By Grouping
4𝑥 2 + 2𝑥 + 14𝑥 + 7
1) Group the first two terms and the second two terms
(4𝑥 2 + 2𝑥) + (14𝑥 + 7)
2) Factor the GCF out of each pair
2𝑥(2𝑥 + 1) + 7(2𝑥 + 1)
3) Factor out the common binomial factor.
(2𝑥 + 1)(2𝑥 + 7)
Factoring Quadratic Trinomials When a≠1
**We can not approach this problem the same way we approach problems
where a=1. We have to factor by grouping, so we need to re-write this
quadratic expressions with 4 terms.
6𝑥 2 + 11𝑥 + 3
6𝑥 2 + 2𝑥 + 9𝑥 + 3
2𝑥(3𝑥 + 1) + 3(3𝑥 + 1)
(3𝑥 + 1)(2𝑥 + 3)
1) Find factors of ac that add up to b.
ac = 6 ∙ 3 = 18
1×18=18
2×9=18
b=11
1+18=19
2+9=11
2) Replace the middle term using these factors
(11x = 2x+9x)
3) Factor by grouping
Factoring Practice
Factor each quadratic expression.
1) 2𝑥 2 − 7𝑥 + 6
ac=12
1x12
-1(-12)
-2(-6)
-3(-4)
b=-7
1+12=13
-1+(-12)=-13 **
-2+(-6) = -8
-3+(-4) = -7
2) −2𝑥 2 + 5𝑥 − 3
−2𝑥 2 + 2𝑥 + 3𝑥 − 3
−2𝑥 𝑥 − 1 + 3(𝑥 − 1)
ac = 6 c = 5
1x6 1+6=6
2x3 2+3=5
𝑥 − 1 (−2𝑥 + 3)
2𝑥 2 − 3𝑥 − 4𝑥 + 6
𝑥(2𝑥 − 3) − 2(2𝑥 −3)
(2𝑥 − 3)(𝑥 − 2)
**Think ahead. You know your second binomial
needs to match the first. Anticipate when you need
to factor out a negative.
Factoring Practice
3) −𝑥 2 + 5𝑥 + 14
= −(x − 2)(x + 7)
4) 3𝑥 2 − 2𝑥 − 5
= 3𝑥 2 − 5𝑥 + 3𝑥 − 5 = (3𝑥 − 5)(𝑥 + 1)
5) 2𝑛2 + 3𝑛 − 9
=2𝑛2 − 3𝑛 + 6𝑛 − 9 = (𝑛 − 3)(𝑛 + 3)
6) 3𝑦 2 − 8𝑦 + 4
=3𝑦 2 − 2𝑦 − 6𝑦 + 4 = (3𝑦 − 2)(𝑦 − 2)
7) 5𝑥 2 + 19𝑥 + 12
= 5𝑥 2 + 4𝑥 + 15𝑥 + 12 = (5𝑥 + 4)(𝑥 + 3)
8) 2𝑣 2 + 11𝑣 + 5
=2𝑣 2 + 1𝑣 + 10𝑣 + 5 = (2𝑣 + 1)(𝑣 + 5)
9) 2𝑥 2 + 5𝑥 + 2
= 2𝑥 2 + 1𝑥 + 4𝑥 + 2 = (2𝑥 + 1)(𝑥 + 2)