[email protected] mth55_lec-22_sec_5-3_gcf-n-grouping.ppt 1 bruce mayer, pe chabot college...

[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§5.3 GCF§5.3 GCFGroupingGrouping



Review §Review §

Any QUESTIONS About• §5.2 → PolyNomial Multiplication

Any QUESTIONS About HomeWork• §5.2 → HW-17

5.2 MTH 55



PolyNomial Factoring DefinedPolyNomial Factoring Defined

To factor a polynomial is to find an equivalent expression that is a product. An equivalent expression of this type is called a factorization of the polynomial• Factoring Breaks an algebraic

expression into its simplest pieces–“Simplest” Smallest Powers



Example Example Factoring Monomials Factoring Monomials

Find three factorizations of 24x3. SOLUTION

a) 24x3 = (6 4)(x x2)

= 6x 4x2

b) 24x3 = (6 4)(x2 x)

= 6x2 4x

c) 24x3 = ((−6)(−4))x3

= (−6)(−4x3)



Greatest Common Factor (GCF)Greatest Common Factor (GCF)

Find the prime factorization of 105 & 60• Use Factor-Tree

105

5 21

3 7

60

2 30

2 15

3 5



Example Example GCF GCF

Thus

753105

532260

Recognize the Factors that both numbers have in COMMON

The GREATEST Common Factor is the PRODUCT of all the COMMON Factors

In This Case the GCF:

1553GCF 715105

41560



Examples Examples GCF GCF

Find the GCF for Monomials:14p4q and 35pq3

The Prime Factorizations• 14p4q = 2 7 p p p p q

• 35pq3 = 5 7 p q q q

Thus the GCF = 7 p q = 7pq



Examples Examples GCF GCF

Find the GCF for Three Monomials:15x2 30xy2 57x3y

The Prime Factorizations• 15x2 = 3 5 x x

• 30xy2 = 2 3 5 x y y

• 57x3y = 3 19 x x x y

Thus the GCF = 3 x = 3x

ID the Commo

n Factors



Factoring When Terms Have a Factoring When Terms Have a Common FactorCommon Factor To factor a polynomial with two or more terms of the form ab + ac, we use the distributive law with the sides of the equation switched:

ab + ac = a(b + c). Multiply Factor

4x(x2 + 3x − 4) 4x3 + 12x2 − 16x = 4xx2 + 4x3x − 4x4 = 4xx2 + 4x3x − 4x4 = 4x3 + 12x2 − 16x = 4x(x2 + 3x − 4)



Example Example Factor by Distributive Factor by Distributive

Factor: 9a − 21 SOLUTION The prime factorization of 9a is 33a The prime factorization of 21 is 37 The largest common factor is 3. 9a − 21 = 33a − 37 (UNdist the 3)

= 3(3a − 7) Chk: 3(3a − 7) = 3 3a − 3 7 = 9a −

21



Example Example Factor by Distributive Factor by Distributive Factor: 28x6 + 32x3. SOLUTION The prime factorization of 28x6 is 2 2 7 x x x x x x The prime factorization of 32x3 is 2 2 2 2 2 x x x The largest common factor is

2 2 x x x or 4x3. 28x6 + 32x3 = (4x3 7x ) + (4x3 8)

= 4x3(7x3 + 8)



Factor Factor 1212xx55 −− 21 21xx44 + 24 + 24xx33

The prime factorization of 12x5 is

2 2 3 x x x x x The prime factorization of 21x4 is

3 7 x x x x The prime factorization of 24x3 is

2 2 2 3 x x x The largest common factor is 3 x x x or 3x3. 12x5 – 21x4 + 24x3 = 3x3 4x2 – 3x3 7x + 3x3 8

= 3 x x x 2 2 x x

= 3 x x x 7 x

= 3 x x x 2 2 2

= 3x3(4x2 – 7x + 8)



Example Example Distributive factoring Distributive factoring

Factor: 9a3b4 + 18a2b3

SOLUTION The Prime

Factorizations:

2 3

2 3

3 4

2 3

3 3

3

9

1 2 38

a b a b

a

a b

b a b

The Greatest Common Factor is 9a2b3

Distributing OUT the GCF Produces the factorization:

9a3b4 + 18a2b3 = 9a2b3(ab + 2)



Example Example Distributive factoring Distributive factoring

Factor: −4xy + 8xw − 12x SOLUTION The Expanded Factorizations

• −4xy = −4x y

• +8xw = − 2 −4x w

• − 12x = 3 −4x

Thus the Factored expression:

−4xy + 8xw − 12x = −4x(y − 2w + 3)



Factoring Out a Negative GCFFactoring Out a Negative GCF When the coefficient of the term of greatest

degree is negative, it is sometimes preferable to factor out the −1 that is understood along with the GCF• e.g. Factor Out the GCF for

Factor out only the 3 .

Or factor out the –3

Both areCorrect

533

53333159333

3434

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wwww



PolyNomial Factoring TipsPolyNomial Factoring Tips

Factor out the Greatest Common Factor (GCF), if one exists.

The GCF multiplies a polynomial with the same number of terms as the original polynomial.

Factoring can always be checked by multiplying. • Multiplication should yield the

original polynomial.



Factoring by GROUPINGFactoring by GROUPING

Sometimes algebraic expressions contain a common factor with two or more terms.

Example: Factor x2(x + 2) + 3(x + 2) SOLUTION: The binomial (x + 2) is a

factor of BOTH x2(x + 2) & 3(x + 2). Thus, (x + 2) is a common factor; so

x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3

= (x + 2)(x2 + 3)



Grouping Game PlanGrouping Game Plan

If a polynomial can be split into groups of terms and the groups share a common factor, then the original polynomial can be factored.

This method, known as factoring by grouping, can be tried on any polynomial with four or more terms



Examples Examples Grouping Grouping

Factor by grouping.

a) 3x3 + 9x2 + x + 3

b) 9x4 + 6x − 27x3 − 18 Solution

a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3)

= 3x2(x + 3) + 1(x + 3)

= (x + 3)(3x2 + 1)

Don’t Forget the “1”



Examples Examples Grouping Grouping

Factor by grouping.

a) 3x3 + 9x2 + x + 3

b) 9x4 + 6x − 27x3 − 18 Solution

b) 9x4 + 6x − 27x3 − 18

= (9x4 + 6x) + (−27x3 − 18)

= 3x(3x3 + 2) + (−9)(3x3 + 2)

= (3x3 + 2)(3x − 9)



Example Example Grouping Grouping

Factor: y5 + 5y3 + 3y2 + 15 SOLUTION

y5 + 5y3 + 3y2 + 15

= (y5 + 5y 3) + (3y

2 + 15)

= y 3 (y

2 + 5) + 3(y 2 + 5)

= (y 2 + 5) (y

3 + 3)

Grouping

Factoring each binomial

Factoring out the common factor(a BiNomial)



Factor Factor 4ab + 2ac + 8xb + 4xc

Try grouping terms which have something in common. Often, this can be done in more than one way.

For example

4 2 8 4ab ac xb xc

( ) ( )4 2 8 4ab ac xb xc ( ) ( )4 8 2 4ab xb ac xc orGrp-1 Grp-2

a’s & x’s Grouping b’s & c’s Grouping




Next, find the greatest common factor for the polynomial in each set of parentheses.

( ) ( )4 2 8 4ab ac xb xc

The GCF for (4ab + 2ac) is 2a

The GCF for (8xb + 4xc) is 4x

The GCF for (4ab + 8xb) is 4b

The GCF for (2ac + 4xc) is 2c

( ) ( )4 8 2 4ab xb ac xc Grouping Set-1 Grouping Set-2




Write each of the polynomials in parentheses as the product of the GCF and the remaining polynomial

( ) ( )4 2 8 4ab ac xb xc ( ) ( )4 8 2 4ab xb ac xc

2 2 4 2a b c x b c( ) ( ) 4 2 2 2b a x c a x( ) ( )

Apply the distributive property to any common factors

( )( )2 4 2a x b c ( )( )4 2 2b c a x




Examine the Factorizations ( ) ( )4 2 8 4ab ac xb xc ( ) ( )4 8 2 4ab xb ac xc

Notice that it did not matter how the terms were originally grouped, the factored forms of the polynomials are IDENTICAL

2 2 2( )( )a x b c 2 2 2( )( )b c a x



WhiteBoard WorkWhiteBoard Work

Problems From §5.3 Exercise Set• 22, 32, 52, 56, 68, 84

Factor byGrouping



All Done for TodayAll Done for Today

Factoring4-Term

Polynomials



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



533

53333159333

3434

ww

wwww



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls



x

y

-3

-2

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3 4 5

M55_§JBerland_Graphs_0806.xls




Divide each polynomial in parentheses by the GCF

( ) ( )4 2 8 4ab ac xb xc ( ) ( )4 8 2 4ab xb ac xc

4 2

2

4

2

2

2

ab ac

a

ab

a

ac

a

2b c

8 4

4

8

4

4

4

xb xc

x

xb

x

xc

x

2b c

4 8

4

4

4

8

4

ab xb

b

ab

b

xb

b

a x2

2 4

2

2

2

4

2

ac xc

c

ac

c

xc

c

a x2

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