8.3 Factoring trinomials
in the form
2
Ax + Bx
+ Bx + C
+C
Where A does not equal 1
Where A does not equal 1
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Look at the coefficients of the outer terms: A and C in standard form
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3x2 + 14x + 8
Multiply them together
Multiply them together
3∙8=24
Now find factors of 24
1∙24,
1
24, 2
2 ∙12,
12, 3 3 ∙8,
8, 4 4 ∙6
6
• Again, find the pair that adds up to B, the coefficient of the middle term 14: 2
ffi i t f th
iddl t
14 2 and 12
d 12
3x2 + 14x
+ 14x + 8
+8
Multiply outer coefficients together (A∙C)
Find the pair that adds up to 14: (B)
Find the pair that adds up to 14: (B)
2 and 12
Rewrite the expression using 2x and 12x
instead of 14x
• 3x2 + 12x + 2x + 8
• Break into two groups of sums
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3x2 + 12x
+ 12x + 2x
+ 2x + 8
+8
• (3x2 + 12x) + (2x + 8)
• Notice it is two sums that are ADDED!
Notice it is two sums that are ADDED!
not multiplied!!
• Factor each group separately
h
l
• 3x(x+4)+2(x+4)
( ) ( )
• Notice each group has the same binomial in ( )
3x(x+4)+2(x+4)
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Notice each group has the same binomial in ( )
Pretend (x+4) is ☺ 3x☺+2☺
Pretend (x+4) is ☺
Can you factor out the ‘☺’? Of course: ☺(3x+2)
Now put the (x+4) in for ☺
Now put the (x+4) in for ☺
(x+4)(3x+2)
Check with calculator to 3x2 + 14x + 8
2x2 —3x
3x — 9
Multiply A∙C: 2∙(—9)=—18
Factor pairs:
Factor pairs: —1∙18, —2∙9, —3∙6, —6∙3 , —9∙2, , —18∙1
Which pair sums to —3?
Rewrite original with this sum for the middle
Rewrite original with this sum for the middle term
• 2x2 —6x + 3x — 9
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2x2 —6x
6x + 3x
+ 3x — 9
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Split into groups
(2x2 — 6x) + (3x
6x) + (3x — 9)
Factor each group
2x (x — 3) + 3(x — 3)
Factor out the matching binomial of each group
Factor out the matching binomial of each group
(x — 3)(2x + 3)
check
15x4 — 39x3 +18x2
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Always, ALWAYS!! check for common factors
3x2 (5x2 — 13x + 6)
+ 6)
Now use the method on the ( ) part
A ∙ C is 5 ∙ 6 = 30
Factors of 30: 1∙30
Factors of 30: 1
30, 2
2∙15
15, 3
3∙10
10, 5
5∙6
6
Which adds to B (—13)?
3 and 10, but they both are negative
3x2 (5x2 — 13x + 6)
+ 6)
• —3 and —10 are the factor pair of A∙C with sum of B
• Rewrite expression with those as coefficients of x
of x
• 3x2 [5x2 — 10x — 3x + 6] • Break into groups 3x2 [(5x2—10x)+(—3x+6)]
• Factor each group
F t
h
3 2 [5x(x—2)+(—3)(x—2)]
3x
[5 ( 2)+( 3)( 2)]
3x2 [5x(x—2)+(—3)(x—2)]
[5x(x 2)+( 3)(x 2)]
• Bring out the binomial factor (x—2)
• 3x2 (x
(x—2)(5x—3)
2)(5x 3)
• Check against 15x4 — 39x3 +18x2 with table
6x2—19x+8
19x+8
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6∙8=48
1∙48,
1
48, 2
2∙24,
24, 3
3∙16,
16, 4
4∙12,
12, 6
6∙8
8
3 and 16 sum to 19, and since 19 is negative
6x2—3x—16x+8
(6x2 — 3x) + (
3x) + (—16x
16x + 8)
+ 8)
3x(x—1)+(—8)(x—1)
(x—1)(3x—8) check it with table on calculator
10x2+19xy+6y2
• Two variables: not much harder. Just remember to include y with last terms
y
• Common factors?
• A∙C=10∙6=60
A C 10 6 60
• 1∙60, 2∙30, 3∙20, 4∙15, 5∙12, 6∙10
,
,
,
,
,
• 4+15=19
• 10x2+15xy+4xy+6y2
10x2+15xy+4xy+6y2
• (10x2+15xy)+(4xy+6y2)
• Since there is that pesky y, remember to Since there is that pesky y, remember to
factor it out…or leave it in, as needed
• 5x(2x+3y)+2y(2x+3y)
(2 3 ) 2 (2 3 )
• (2x+3y)(5x+2y)
(
y)(
y)
• The second variable should not be an issue!
6x3y2+26x2y3+24xy4
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If it says ‘Factor Completely’, look carefully
Common Factors: 2xy2
Common Factors: 2xy
2xy2(3x2+13xy+12y2)
Done? Not yet…
3 12=36
3∙12=36
Factors: 1∙36, 2∙18, 3∙12, 4∙9 4+9=13, rewrite ( ) with 4 and 9
2xy2(3x2+13xy+12y2)
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2xy2(3x
2
(3 2+9xy+4xy+12y
9
2 2)
Break into groups and factor
2xy2[(3x2+9xy)+(4xy+12y2)]
2xy2[3x(x+3y)+4y(x+3y)]
Factor out matching binomial
2 2 (x+3y)(3x+4y)
2xy
( +3 )(3 +4 )
calculator check is not definitive for 2 variables
check by multiplying: – first by FOIL, (x+3y)(3x+4y) then distribute 2xy2
6x2—7x+2
7x+2
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A∙C
A
C = 6∙2 = 12
6 2 12
Factors: 1∙12, 6∙2 , 3∙4
3+4=7, and since —7 is in expression
Use —3 and —4
6x2—3x—4x+2
Group: (6x2—3x)+(—4x+2)
Factor each group 3x(2x—1)+(—2)(2x—1)
Factor out (2x—1)(3x—2)
(
)(
)
Check with table in calculator
20x4 — 40x3 — 25x2
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Common factor first: 5x2
5x2 (4x2—8x—5)
8x 5) Now: A∙C = —20, find factor pairs —1∙20, —2∙10, —4∙5, —5∙4, —10∙2, —20∙1 Which pair do we want? —10
Which pair do we want? 10 + 2 = + 2 = —8
8
Rewrite: 5x2 (4x2 + 2x — 10x — 5)
5x2 (4x2 + 2x
+ 2x — 10x — 5)
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Group: 5x2 [(4x2 + 2x) + (—10x — 5)]
Factor groups: 5x2 [2x(2x + 1)
Factor groups: 5x
+ 1) + (
+ (—5)(2x
5)(2x + 1)]
+ 1)]
Factor out binomial: 5x2 (2x + 1)(2x — 5)
Check with table in calculator