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Factoring Polynomials
Copyright © 2011 Pearson Education, Inc.
Method 1: Factoring Trinomials by Trial and Error
Factoring by Grouping
Example
Factor 3x  14 x  8.
2
Solution
• Factor 3x2 + 14x + 8, the result will be of the form
(3x +?)(x +?)
• Product of the last terms must be 8, so the last terms
must be 1 and 8 or 2 and 4
• Rule out negative last terms in the factors, because
2
the middle term of 3x +14x +8 has the positive
coefficient 14
Copyright © 2011 Pearson Education, Inc.
Slide 2
Method 1: Factoring Trinomials by Trial and Error
Factoring by Grouping
Solution Continued
• Decide between the two pairs of possible last terms
by multiplying:
Copyright © 2011 Pearson Education, Inc.
Slide 3
Method 1: Factoring Trinomials by Trial and Error
F a c t o rin g b y Tr i a l a n d E r r o r
Example
Factor 2 x  5 x  25.
2
Solution
2
• Factor 2x – 5x – 25, the result will be of the form
(2x +?)(x +?)
• Product of the last terms must be –25, so the last
terms must be 1 and –25, 5 and –5, or –1 and 25
• Decide amongst the three pairs of possible last terms
by multiplying:
Copyright © 2011 Pearson Education, Inc.
Slide 4
Method 1: Factoring Trinomials by Trial and Error
F a c t o rin g b y Tr i a l a n d E r r o r
Solution Continued
Copyright © 2011 Pearson Education, Inc.
Slide 5
Method 1: Factoring Trinomials by Trial and Error
F a c t o rin g b y Tr i a l a n d E r r o r
Property
To factor a trinomial of the form ax2+bx +c by trial
and error: If the trinomial can be factored as a
product of two binomials, then the product of the
coefficients of the first terms of the binomials is equal
to a and the product of the last terms of the binomials
is equal to c. For example, to find the
correct factored expression, multiply the
possible products and
identify those for which
the coefficient of x is b.
Copyright © 2011 Pearson Education, Inc.
Slide 6
Ruling Out Possibilities While Factoring By Trial and Error
Ruling Out Possibilities
Example
Factor 10 x 2  19 x  6.
Solution
•Result will be in one of these two forms:
(10x +?)(x +?)
(5x +?)(2x +?)
•The product of the last terms must be 6
•Last terms must be −1 and −6, or −2 and −3, where we
can write each pair in either order
•Rule out positive last terms because the middle term of
2
10x − 19x + 6 has a negative coefficient, −19
Copyright © 2011 Pearson Education, Inc.
Slide 7
Ruling Out Possibilities While Factoring By Trial and Error
Ruling Out Possibilities
Solution Continued
2
• Terms of 10x −19x +6 do not have a common factor
of 2,
• Also rule out products that have a factor of 2
• For example, we can rule out (10x −6)(x −1),
because it has a factor of 2:
(10x − 6)(x − 1) = 2(5x − 3)(x − 1)
• Decide among the remaining possible last terms by
multiplying:
Copyright © 2011 Pearson Education, Inc.
Slide 8
Ruling Out Possibilities While Factoring By Trial and Error
Ruling Out Possibilities
Solution Continued
2
(10x − 1)(x − 6) = 10x − 61x + 6
Contains factor of 2, rule out: (10x − 2)(x − 3)
2
(10x − 3)(x − 2) = 10x − 23x + 6
Contains factor of 2, rule out: (5x − 1)(2x − 6)
2
(5x − 6)(2x − 1) = 10x − 17x + 6
2
(5x − 2)(2x − 3) = 10x − 19x + 6 ← Success!
Contains factor of 2, rule out: (5x − 3)(2x − 2)
2
So, 10x − 19x + 6 = (5x − 2)(2x − 3).
Copyright © 2011 Pearson Education, Inc.
Slide 9
Factoring Out the GCF, Then Factoring by Trial and Error
Completely Factoring a Polynomial
Example
Factor 6 x3 y 2  26 x 2 y 3  24 xy 4 .
Solution
2
• First factor out the GCF, 2xy
2
2
2
2xy (3x + 13xy + 12y )
• Factor further, the result will be in the form
2
2xy (3x +?)(x +?)
2
• Product of the last terms must be 12y , so the last
terms must be y and 12y, 2y and 6y, or 3y and 4y,
where we can write each pair in either order
Copyright © 2011 Pearson Education, Inc.
Slide 10
Factoring Out the GCF, Then Factoring by Trial and Error
Completely Factoring a Polynomial
Solution
Example
Continued
• Decide by multiplying and rule out any choice that
contains a factor the original equation does not:
2
(3x + y)(x + 12y) = 3x2 + 37xy + 12y
Contains factor of 3, rule out: (3x + 12y)(x + y)
2
(3x + 2y)(x + 6y) = 3x2 + 20xy + 12y
Contains factor of 3, rule out: (3x + 6y)(x + 2y)
Contains factor of 3, rule out: (3x + 3y)(x + 4y)
2
2
(3x + 4y)(x + 3y) = 3x + 13xy + 12y ← Success!
Copyright © 2011 Pearson Education, Inc.
Slide 11
Factoring Out the GCF, Then Factoring by Trial and Error
Completely Factoring a Polynomial
Solution
Example
Continued
3 2
2 3
4
2
2
2
6x y +26x y +24xy = 2xy (3x +13xy+12y )
2
= 2xy (3x +4y)(x +3y)
Copyright © 2011 Pearson Education, Inc.
Slide 12
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Process
2
To factor a trinomial of the form ax + bx + c by
grouping (if it can be done),
1.
Find pairs of numbers whose product is ac.
2.
Determine which of the pairs of numbers from step
1 has the sum b. Call this pair of numbers m and n.
3.
Write the bx term as mx + nx:
ax2 + bx + c = ax2 + mx + nx + c
4.
Factor ax2 + mx + nx + c by grouping.
Another name for this technique is the ac-method.
Copyright © 2011 Pearson Education, Inc.
Slide 13
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Example
Factor 3x  14 x  8 by grouping.
2
Solution
Here, a = 3, b = 14, and c = 8.
Step 1. Find the product ac: ac = 3(8) = 24.
Step 2. We want to find two numbers m and n that
have the product ac = 24 and the sum b = 14:
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Slide 14
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Solution
Example
Continued
Product = 24 Sum = 14?
1(24) = 24 1 + 24 = 25
2(12) = 24 2 + 12 = 14 ← Success!
3(8) = 24
3 + 8 = 11
4(6) = 24
4 + 6 = 10
Since 2(12) = 24 and 2 + 12 = 14, we conclude that
the two numbers m and n are 2 and 12.
Copyright © 2011 Pearson Education, Inc.
Slide 15
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Solution
Example
Continued
Step 3. We write the bx term, 14x, as the sum
mx + nx:
2
Step 4. We factor 3x + 2x + 12x + 8 by grouping:
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Slide 16
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Example
Factor 6 x  7 x  2 by grouping.
2
Solution
Here, a = 6, b = −7, and c = 2.
Step 1. Find the product ac: ac = 6(2) = 12.
Step 2. We want to find two numbers m and n that
have the product ac = 12 and the sum b = −7:
Copyright © 2011 Pearson Education, Inc.
Slide 17
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Solution
Example
Continued
Product = 12
Sum = −7?
−1(−12) = 12 −1 + (−12) = −13
−2(−6) = 12
−2 + (−6) = −8
−3(−4) = 12
−3 + (−4) = −7 ← Success!
Since −3(−4) = 12 and −3 + (−4) = −7, we conclude
that the two numbers m and n are −3 and −4.
Step 3. We write
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Slide 18
Method 2: Factoring Trinomials by Grouping
Factoring by Grouping
Solution
Example
Continued
2
Step 4. We factor 6x − 3x − 4x + 2 by grouping:
Example
Factor 20 x  40 x  25 x .
4
3
2
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Slide 19
Method 2: Factoring Trinomials by Grouping
F a c t o r i n g O u t t h e G C F, T h e n F a c t o r i n g b y G r o u p i n g
Solution
2
First, we factor out the GCF, 5x :
4
3
2
2
2
20x − 40x − 25x = 5x (4x − 8x − 5)
2
Next, we use grouping to try to factor 4x −8x−5,
where a = 4, b = −8, and c = −5.
Step 1. Find the product ac: ac = 4(−5) = −20.
Step 2. We want to find two numbers m and n that
have the product ac = −20 and the sum b = −8:
Copyright © 2011 Pearson Education, Inc.
Slide 20
Method 2: Factoring Trinomials by Grouping
F a c t o r i n g O u t t h e G C F, T h e n F a c t o r i n g b y G r o u p i n g
Solution Continued
Product = −20
Sum = −8?
1(−20) = −20 1 + (−20) = −19
2(−10) = −20 2 + (−10) = −8 ← Success!
4(−5) = −20 4 + (−5) = −1
5(−4) = −20 5 + (−4) = 1
10(−2) = −20 10 + (−2) = 8
20(−1) = −20 20 + (−1) = 19
2
(We have temporarily put aside the GCF, 5x .)
Copyright © 2011 Pearson Education, Inc.
Slide 21
Method 2: Factoring Trinomials by Grouping
F a c t o r i n g O u t t h e G C F, T h e n F a c t o r i n g b y G r o u p i n g
Solution Continued
Since 2(−10) = −20 and 2 + (−10) = −8, we conclude
that the two numbers m and n are 2 and −10.
Step 3. We write
2
Step 4. We factor 4x + 2x − 10x − 5 by grouping:
4
3
2
2
2
So, 20x − 40x − 25x = 5x (4x − 8x − 5)
2
= 5x (2x − 5)(2x + 1).
Copyright © 2011 Pearson Education, Inc.
Slide 22
Method 2: Factoring Trinomials by Grouping
F a c t o r i n g O u t t h e G C F, T h e n F a c t o r i n g b y G r o u p i n g
Solution Continued
We use a graphing calculator table to verify our work:
Copyright © 2011 Pearson Education, Inc.
Slide 23