§5.4 Factor TriNomials Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§5.4 Factor

TriNomials

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot College Mathematics

1

Bruce Mayer, PE

BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt

Review § 5.3

MTH 55

 Any QUESTIONS About

• §5.3 → Factoring by GCF and/or

Grouping

 Any QUESTIONS About HomeWork

• §5.3 → HW-13


2

Bruce Mayer, PE


Factor (+1) ·x 2 + bx + c

 Recall the FOIL method of multiplying two binomials:

F O I L

( x + 2)( x + 5) = x 2 + 5 x + 2 x + 10

= x 2 + 7 x + 10


3

Bruce Mayer, PE


Factor (+1) ·x 2 + bx + c

 To factor x 2 + 7 x + 10, think of FOIL:

The first term, x 2 , is the product of the

First terms of two binomial factors, so the first term in each binomial must be x .

 The challenge is to find two numbers p and q such that p • q = c , and p + q = b x 2 + 7 x + 10 = ( x + p )( x + q )


4

= x 2 + qx + px + pq

Bruce Mayer, PE


Factor (+1) ·x 2 + bx + c

 Need to find two numbers p & q such that x 2 + 7 x + 10 = ( x + p )( x + q ) = x 2 + qx + px + pq

 Thus the numbers p and q must be selected so that their

• PRODUCT is 10

• SUM is 7

 The Factor Pairs for 10 [and their sums]

• 1·10 [11]; (−1)·(−10) [−11]; 2·5 [7];

(−2)·(−5) [−7];


5

Bruce Mayer, PE


Factor (+1) ·x 2 + bx + c

 In this case, Examination of the “c” term factor-pairs revealed the desired numbers p = 2 & q = 5

 Thus the factorization for x 2 + 7 x + 10

( x + 2)( x + 5) or ( x + 5)( x + 2).

 Note that the FACTORING Process converted an ADDITION-chain into a

6 pure MULTIPLICATION-Chain

Chabot College Mathematics Bruce Mayer, PE


Example



FOIL Factoring

 Multiplying binomials uses the FOIL method,

Factoring uses the FOIL method backwards

Product of x and x is x

2

.

F

L Product of 5 and –7 is –35.

Sum of the product of outer and inner terms


7

O I

Bruce Mayer, PE


Factor x 2 + bx + c for Positive c

 When the constant term of a trinomial is

positive

, look for two numbers with the

same sign

. The sign is that of the middle term: x 2 – 7 x + 10 = ( x – 2)( x – 5); x 2 + 7 x + 10 = ( x + 2)( x + 5);


8

Bruce Mayer, PE


Example



Factor x 2 + 7x + 12

 SOLUTION: Think of FOIL in reverse:

( x + )( x + )

 We need a constant term that has a product of 12 and a

SUM of 7.

 We list some pairs of numbers that multiply to 12

Pairs of

Factors of 12

1, 12

2, 6

Sums of

Factors

13

8

3, 4 7



1,



12



13



2,



6



3,



4





8

7


9

Bruce Mayer, PE


Example



Factor x 2 + 7x + 12

 Since 3



4 = 12 and 3 + 4 = 7 , the factorization of x 2 + 7 x + 12 is

( x + 3 )( x + 4 ).

 To check we simply multiply the two binomials.

 CHECK by FOIL:

( x + 3)( x + 4) = x 2 + 4 x + 3 x + 12

= x 2 + 7 x + 12




10

Bruce Mayer, PE


Example  Factor y 2 – 8y + 15

 SOLUTION: Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 15 in which both factors are negative. Their SUM must be −8.

Pairs of

Factors of 15

–1, –15

–3, –5

Sums of

Factors

–16

–8 y 2

Sum of −8

− 8 y + 15 =

( y − 3)( y – 5)


11

Bruce Mayer, PE


Factor x 2 + bx + c for Negative c

 When the constant term of a trinomial is negative , look for two numbers whose product is negative. One must be positive and the other negative : x 2 – 4 x – 21 = ( x + 3)( x – 7); x 2 + 4 x – 21 = ( x – 3)( x + 7).

12

 Select the two numbers so that the number with the

LARGER absolute value has the SAME SIGN as b , the coefficient of the middle term

Chabot College Mathematics Bruce Mayer, PE


Example  Factor x 2 – 5x – 24

 SOLUTION: The constant term must be expressed as the product of negative & positive numbers.


13

 Since the sum of the two numbers must be negative, the negative number must have the greater absolute value.

Pairs of

Factors of 24

1,



24

2,



12

3,



8

4,



6

6,



4

8,



3

Sums of

Factors



23



10



5



2

2

5 x 2 − 5 x − 24 =

( x + 3)( x – 8)

Bruce Mayer, PE


Example  Factor t 2 – 32 + 4t

 SOLUTION:

 Rewrite the trinomial t 2 + 4 t − 32.

 We need one positive and one negative factor. The sum must be 4, so the positive factor must have the larger absolute value

Pairs of

Factors of 32



1, 32



2, 16



4, 8

Sums of

Factors

31

14

4 t 2 + 4 t − 32 =

( t + 8)( t − 4)


14

Bruce Mayer, PE


Example



Two Variables

 Factor: a 2 + ab − 30 b 2

 SOLUTION

 We need the factors of a 2 & 30 b 2 that when added equal ab.

• Those factors are a , and −5 b & 6 b .

a 2 + ab − 30 b 2 = ( a − 5 b )( a + 6 b )


15

Bruce Mayer, PE


Prime Polynomials

 A polynomial that canNOT be factored is considered prime .

• Example: x 2 − x + 7

 Often factoring requires two or more steps. Remember, when told to factor, we should factor completely . This means the final factorization should contain only prime polynomials.


16

Bruce Mayer, PE


Example



Factor 2x 3 −24x 2 +72x

 SOLUTION

Always look first for a common factor.

In this case factor out 2 x :

2 x ( x 2 − 12 x + 36)

 Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of

36 in which both factors are negative.

• Their SUM must be −12.


17

Bruce Mayer, PE


Example



Factor 2x 3 –24x 2 +72x

 The factorization of

(

( x x

2 or (

– 12

– 6)( x x x

– 6) 2

+ 36) is

– 6)


2 x 3 – 24 x 2 + 72 x is

2 x ( x – 6) 2 or 2 x ( x – 6)( x – 6)

Pairs of

Factors of 36

Sums of

Factors



1,



36



37



2,



18



20



3,



12



15



4,



9



13



6,



6



12

2x 3 – 24 2 x – 72 x = 2 x ( x – 6 )( x – 6)


18

Bruce Mayer, PE


To Factor (+1) ·x 2 + bx + c

1. Distribute out Common Factors

2. Find a pair of factors that have c as their product and b as their sum.

a) If c is positive, its factors will have the same sign as b.

b) If c is negative, one factor will be positive and the other will be negative. Select the factors such that the factor with the larger absolute value has the same sign as b.

3. CHECK by MULTIPLYING


19

Bruce Mayer, PE


Factoring When: LeadCoeff ≠ 1

 Factoring Trinomials of the

Type a x 2 + bx + c

• Factoring with FOIL

• The

Grouping

Method


20

Bruce Mayer, PE


Factor ax 2 +bx+c by FOIL


21

Bruce Mayer, PE


Example



Factor 3x 2 – 14x – 5

1. First, check for a common factor, or GCF for all Terms. There is none other than 1 or −1.

2. Find the F irst terms whose product is 3 x 2 .

The only possibilities are 3 x and x :

(3 x + )( x + )

3. Find the L ast terms whose product is −5.

Possibilities are ( −5)(1) & (5)(−1)

 Important!: Since the First terms are not identical, we must also consider the above factors in reverse order: (1)( −5), & (−1)(5).


22

Bruce Mayer, PE


Example



Factor 3x 2 – 14x – 5

4. Knowing that the F irst and L ast products will check, inspect the O uter and I nner products resulting from steps (2) and (3)

Look for the combination in which the sum of the products is the middle term.

(3 x – 5)( x + 1) = 3 x 2 + 3 x – 5 x – 5

= 3 x 2 – 2 x – 5

Wrong middle term

(3 x – 1)( x + 5) = 3 x 2 + 15 x – x – 5

= 3 x 2 + 14 x – 5

Wrong middle term


23 close

Bruce Mayer, PE


Example



Factor 3x 2 – 14x – 5

4. Keep Trying the factors of −5.

(3 x + 5)( x – 1) = 3 x 2 – 3 x + 5 x – 5

= 3 x 2 + 2 x – 5

Wrong middle term

(3 x + 1)( x – 5) = 3 x 2 – 15 x + x – 5

= 3 x 2 – 14 x – 5

CORRECT middle term!

 Thus

3 x 2

–

14 x

–

5 = (3 x + 1)( x

–

5)


24

Bruce Mayer, PE


LdCoeff ≠ 1 Factorization Notes

 Reversing the signs in the binomials reverses the sign of the middle term

 Organize your work so that you can keep track of which possibilities you have checked.

 Remember to include the largest common factor - if there is one - in the final factorization.

 ALWAYS CHECK by multiplying


25

Bruce Mayer, PE


Example



Factor 14x + 5 – 3x 2

 SOLUTION:

 It is an important problem-solving strategy to find a way to make problems look like problems we already know how to solve. Rewrite the equation in descending order.

14 x + 5 – 3 x 2 = – 3 x 2 + 14 x + 5


26

Bruce Mayer, PE


Example



Factor 14x + 5 − 3x 2

 Starting with −3 x 2 + 14 x + 5

 Factor out the –1:

−3 x 2 + 14 x + 5 = −1 (3 x 2 − 14 x − 5)

= −1 (3 x + 1)( x − 5)


14 x + 5 − 3 x 2 is −1(3 x + 1)( x − 5).

or ( −3 x − 1)( x − 5) or (3 x + 1)( − x + 5)


27

Bruce Mayer, PE


Example



2Vars: 6x 2 − xy − 12y 2

 SOLUTION: No common factors exist, we examine the first term, 6 x 2 . There are two possibilities:

(2 x + )(3 x + ) or (6 x + )( x + ).

 The last term −12 y 2 , has pairs of factors:

12 y , − y 6 y , −2 y 4 y , −3 y and −12 y , y −6 y , 2 y −4 y , 3 y as well as each pairing reversed .


28

Bruce Mayer, PE


Example



2Vars: 6x 2 − xy − 12y 2

 SOLUTION:

 Some trials such as (2 x – 6 y )(3 x + 2 y ) and (6 x + 4 y )( x – 3 y ), cannot be correct because (2 x – 6 y ) and (6 x + 4 y ) contain a common factor, 2.

 Trial  Product

(2 x + 3 y )(3 x − 4 y ) 6 x 2 − 8 xy + 9 xy − 12

= 6 x 2 + xy − 12 y 2 y 2


29

Bruce Mayer, PE


Example



2Vars: 6x 2 − xy − 12y 2

 SOLUTION: the Trial (2 x + 3 y )(3 x − 4 y ) incorrect, but only because of the sign of the middle term.

 To correctly factor, simply change the signs in the binomials.

 Trial  Product

(2 x − 3 y )(3 x + 4 y ) 6 x 2 + 8 xy − 9 xy − 12 y 2

= 6 x 2 − xy − 12 y 2

30

The factorization: (2 x − 3 y )(3 x + 4 y )

Bruce Mayer, PE Chabot College Mathematics


Example



18m 2 – 19mn – 12n 2

 SOLUTION

 There are no common factors.

 Factor the first term, 18 m 2 and get the following possibilities: 18 m

 m , 9 m



2 m , and 6 m



3 m .

 Factor the last term, −12 n 2 , which is negative. The possibilities are:

( −12 n )( n ), ( − n )(12 n ), ( −2 n )(6 n ),

(6 n )( −2 n ), ( −4 n )(3 n ) or ( −3 n )(4 n )


31

Bruce Mayer, PE


Example



18m 2 – 19mn – 12n 2

 Look for combinations of factors such that the sum of the outside and the inside products is the middle term, ( −19 mn ).

• (9 m + n )(2 m − 12 n ) = 18 m 2 − 106 mn − 12 n 2

• (9 m − 12 n )(2 m + n ) = 18 m 2 − 15 mn − 12 n 2

• (9 m − 3 n )(2 m + 4 n ) = 18 m 2 + 30 mn − 12 n 2

• (9 m + 4 n )(2 m – 3 n ) = 18 m 2 − 19 mn − 12 n 2

 Thus ANS →

(9 m + 4 n )(2 m − 3 n )


32 correct middle term

Bruce Mayer, PE


Factoring by Substitution

 Some times substituting a single varaible for a (complicated) expression reveals an easily factored PolyNomial

 Example

 factor p 2 q 2 + 7 pq + 6

 SOLUTION

 Rewrite using Product-to-Power

Exponent rule → ( pq ) 2 + 7( pq ) + 6

 Now engage a substitution


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Bruce Mayer, PE



 factor p 2 q 2 + 7 pq + 6 = ( pq ) 2 + 7( pq ) + 6

 Now to engage a substitution LET

u = pq

 Replace in the Expression pq with u u 2 + 7 u + 6

 The Expression in u is easily

FOIL-factored u 2 + 7 u + 6 = ( u + 6)( u + 1)


34

Bruce Mayer, PE



 Now BACK Substitute u = pq

( u ) 2 + 7 ( u ) + 6 = ( u + 6)( u + 1)



( pq ) 2 + 7( pq ) + 6 = ( pq + 6)( pq + 1)

 p 2 q 2 + 7pq + 6 = (pq + 6)(pq + 1)


35

Bruce Mayer, PE


WhiteBoard Work

 Problems From §5.4 Exercise Set

• 30, 44, 66, 82, 92

 Shaded Area

Equals



2 x



4 x











2







1

2

 x

2







  x

2







A



8 x

2 

2

 x

2 

2 x

2



4

  


36

Bruce Mayer, PE


All Done for Today

F.O.I.L.

Factoring


37

Bruce Mayer, PE


Chabot Mathematics

Appendix

r

2  s

2 

 r

 s

 r

 s



Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

–


38

Bruce Mayer, PE


Factoring ax 2 +bx+c by Grouping

1. Factor out the largest common factor, if one exists.

2. Multiply the leading coefficient a and the constant c ; i.e., form the a • c product

3. Find a pair of factors of a • c whose sum is b.

4. Rewrite the middle term, bx , as a sum or difference using the factors found in step (3).

5. Factor by grouping.

6. Include any common factor from step (1) and check by multiplying.


39

Bruce Mayer, PE


Example



Factor 4x 2 – 5x – 6

 SOLUTION

1. First, we note that there is no common factor (other than 1 or −1).

2.

We multiply the leading coefficient, 4 and the constant, −6: (4)(−6) = −24.

3. We next look for the factorization of

−24 in which the sum of the factors is the coefficient of the middle term, −5.


40

Bruce Mayer, PE


Example



Factor 4x 2 – 5x – 6

3.

Pairs of

Factors of -24

Sums of

Factors

1, –24

–1, 24

2, –12

–2, 12

–23

23

–10

10

–5


41

3, –8

–3, 8

4, –6

–4, 6

5

–2

2

We would normally stop listing pairs of factors once we have found the one we need

Bruce Mayer, PE


Example



Factor 4x 2 – 5x – 6

4. Next, we express the middle term as a sum or difference using the factors found in step (3):

−5 x = −8 x + 3 x.

5. We now factor by grouping as follows:

4 x 2 − 5 x − 6 = [4 x 2 − 8 x ] + [ 3 x − 6]

= 4 x ( x − 2 ) + 3( x − 2 )

= ( x − 2 )(4 x + 3)


42

Bruce Mayer, PE


Example



Factor 4x 2 – 5x – 6

6. CHECK by FOIL:

( x − 2)(4 x + 3) = 4 x 2 + 3 x − 8 x − 6

= 4 x 2 − 5 x − 6 

 The factorization of 4 x 2 − 5 x − 6 is

( x − 2)(4 x + 3).


43

Bruce Mayer, PE


Example: Factor 8x 3 + 10x 2 – 12x

 SOLUTION

1. Factor out the Greatest Common

Factor (GCF), 2 x :

8 x 3 + 10 x 2 − 12 x = 2 x (4 x 2 + 5 x − 6)

2. To factor 4 x 2 + 5 x − 6 by grouping, we multiply the leading coefficient, 4 and the constant term ( −6): 4(−6) = −24.


44

Bruce Mayer, PE



3. next look for pairs of factors of −24 whose sum is 5.

Pairs of Factors of −24 Sums of Factors

3, −8

−3, 8

−5

5

4. We then rewrite the 5 x in 4 x 2 + 5 x − 6 using:

5 x = −3 x + 8 x


45

Bruce Mayer, PE



5. Next, factor by grouping:

4 x 2 + 5 x − 6 = [4 x 2 − 3 x ] + [8 x − 6]

= x (4 x − 3) + 2 (4 x − 3)

= ( x + 2) (4 x − 3)

6. The factorization of the original trinomial 8 x 3 + 10 x 2 − 12 x is

2 x ( x + 2)(4 x − 3)


46

Bruce Mayer, PE


Graph y = |x|

6

 Make T-table

y = |x | x

5

6

3

4

-1

0

1

2

-6

-5

-4

-3

-2

5

6

3

4

1

2

1

0

4

3

6

5

2

5

4

3

2

1

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

0

-1

0

-2

-3

-4

-5

-6 file =XY_Plot_0211.xls


47 y

1 2 3

Bruce Mayer, PE


4 5 x

6

4

5 y

-3 -2

3

2

1

-1

0

0

-1

-2

M55_§JBerland_Graphs_0806.xls

-3

1 2 3 4 5 x


48

Bruce Mayer, PE


§5.4 Factor TriNomials Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§5.4 Factor

TriNomials

positive

same sign

All Done for Today

Chabot Mathematics

Appendix

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§5.4 Factor TriNomials Chabot Mathematics Bruce Mayer, PE

Chabot Mathematics

§5.4 Factor

TriNomials

positive

same sign

All Done for Today

Chabot Mathematics

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