INTRODUCTION TO FACTORING POLYNOMIALS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice...

INTRODUCTION TO FACTORING

POLYNOMIALS

MSJC ~ San Jacinto CampusMath Center Workshop

SeriesJanice Levasseur

Definitions

• Recall: Factors of a number are the numbers that divide the original number evenly.

• Writing a number as a product of factors is called a factorization of the number.

• The prime factorization of a number is the factorization of that number written as a product of prime numbers.

• Common factors are factors that two or more numbers have in common.

• The Greatest Common Factor (GCF) is the largest common factor.

Ex: Find the GCF(24, 40).

Prime factor each number:

24

40

2 122 62 3 24 = 2*2*2*3 = 23*3

2 202 102 5 40 = 2*2*2*5 = 23*5

GCF(24,40) = 23 = 8

The Greatest Common Factor of terms of a polynomial is the largest factor that the original terms share

• Ex: What is the GCF(7x2, 3x)

7x2 = 7 * x * x

3x = 3 * x

The terms share a factor of x

GCF(7x2, 3x) = x

Ex: Find the GCF(6a5,3a3,2a2)

• 6a5 = 2*3*a*a*a*a*a• 3a3 = 3*a*a*a• 2a2 = 2*a*a

The terms share two factors of a

GCF(6a5,3a3,2a2)= a2

Note: The exponent of the variable in the GCF is the smallest exponent of that variable the terms

Definitions

• To factor an expression means to write an equivalent expression that is a product

• To factor a polynomial means to write the polynomial as a product of other polynomials

• A factor that cannot be factored further is said to be a prime factor (prime polynomial)

• A polynomial is factored completely if it is written as a product of prime polynomials

To factor a polynomial completely, ask

• Do the terms have a common factor (GCF)?• Does the polynomial have four terms? • Is the polynomial a special one?

– Is the polynomial a difference of squares?• a2 – b2

– Is the polynomial a sum/difference of cubes?• a3 + b3 or a3 – b3

– Is the trinomial a perfect-square trinomial?• a2 + 2ab + b2 or a2 – 2ab + b2

• Is the trinomial a product of two binomials?• Factored completely?

Ex: Factor 7x2 + 3x

Think of the Distributive Law: a(b+c) = ab + ac reverse it ab + ac = a(b + c)

Do the terms share a common factor? What is the GCF(7x2, 3x)?

Recall: GCF(7x2, 3x) = x

7 x2 + 3 x = x( + ) What’s left?

x x

Factor out

7x2 + 3x = x(7x + 3)

Ex: Factor 6a5 – 3a3 – 2a2

Recall: GCF(6a5,3a3,2a2)= a2

6a5 – 3a3 – 2a2 = a2( - - )

a2 a2 a2

3 1

6a3 3a 2

6a5 – 3a3 – 2a2 = a2(6a3 – 3a – 2)

Your Turn to Try a Few

Ex: Factor x(a + b) – 2(a + b)

Always ask first if there is common factor the terms share . . .

x(a + b) – 2(a + b) Each term has factor (a + b)

x(a + b) – 2(a + b) = (a + b)( – )

(a + b) (a + b)

x 2

x(a + b) – 2(a + b) = (a + b)(x – 2)

Ex: Factor a(x – 2) + 2(2 – x)

As with the previous example, is there a common factor among the terms?

Well, kind of . . . x – 2 is close to 2 - x . . . Hum . . .

Recall: (-1)(x – 2) = - x + 2 = 2 – x

a(x – 2) + 2(2 – x) =

a(x – 2) + 2((-1)(x – 2))

= a(x – 2) + (– 2)(x – 2)

= a(x – 2) – 2(x – 2)

a(x – 2) – 2(x – 2) = (x – 2)( – )

(x – 2) (x – 2)

a 2

Ex: Factor b(a – 7) – 3(7 – a)

Common factor among the terms?

Well, kind of . . . a – 7 is close to 7 - a

Recall: (-1)(a – 7) =

- a + 7 = 7 – a

b(a – 7) – 3(7 – a) =

b(a – 7) – 3((-1)(a – 7))

= b(a – 7) + 3(a – 7)

= b(a – 7) +3(a – 7)

b(a – 7) + 3(a – 7) = (a – 7)( + )

(a – 7) (a – 7)

b 3

Factor by Grouping

• If the polynomial has four terms, consider factor by grouping

1. Factor out the GCF from the first two terms2. Factor out the GCF from the second two terms

(take the negative sign if minus separates the first and second groups)

3. If factor by grouping is the correct approach, there should be a common factor among the groups

4. Factor out that GCF5. Check by multiplying using FOIL

Ex: Factor 6a3 + 3a2 +4a + 2

Notice 4 terms . . . think two groups: 1st two and 2nd two

Common factor among the 1st two terms?

6a3 + 3a2 = 3a2( + )

GCF(6a3, 3a2) = 3a2

3a2 3a2

2a 12a 1

Common factor among the 2nd two terms?

GCF(4a, 2) = 2

4a + 2 = 2( + )

2 2

2 12a 1

Now put it all together . . .

6a3 + 3a2 +4a + 2 = 3a2(2a + 1) + 2(2a + 1)

Four terms two terms. Is there a common factor?

Each term has factor (2a + 1)

3a2(2a + 1) + 2(2a + 1) = (2a + 1)( + )

(2a + 1) (2a + 1)

3a2 2

6a3 + 3a2 +4a + 2 = (2a + 1)(3a2 + 2)

Ex: Factor 4x2 + 3xy – 12y – 16x



4x2 + 3xy = x( + )

GCF(4x2, 3xy) = x

x x

4x 3y4x 3y


GCF(-12y, - 16x) = -4

-12y – 16x = - 4( )

-4 -4

3y4x

3y + 4x


4x2 + 3xy – 12y – 16x = x(4x + 3y) – 4(4x + 3y)


Each term has factor (4x + 3y)

x(4x + 3y) – 4(4x + 3y) = (4x + 3y)( )

(4x + 3y) (4x + 3y)

x – 4

4x2 + 3xy – 12y – 16x = (4x + 3y)(x – 4)

Ex: Factor 2ra + a2 – 2r – a



2ra + a2 = a( + )

GCF(2ra, a2) = a

a a2r a


GCF(-2r, - a) = -1

-2r – a = - 1( )

-1 -1

2r + a


2ra + a2 –2r – a = a(2r + a) – 1(2r + a)


Each term has factor (2r + a)

a(2r + a) – 1(2r + a) = (2r + a)( )

(2r + a) (2r + a)

a – 1

2ra + a2 – 2r – a = (2r + a)(a – 1)


• Do the terms have a common factor (GCF)?

• Does the polynomial have four terms? • Is the polynomial a special one?




Special PolynomialsIs the polynomial a difference of squares?

• a2 – b2 = (a – b)(a + b)

Is the trinomial a perfect-square trinomial?• a2 + 2ab + b2 = (a + b)2

• a2 – 2ab + b2 = (a – b)2

Ex: Factor x2 – 4

Notice the terms are both perfect squares

x2 = (x)2 4 = (2)2

x2 – 4 = (x)2 – (2)2

a2 – b2

and we have a difference

= (x – 2)(x + 2)

difference of squares

= (a – b)(a + b)factors as

Ex: Factor 9p2 – 16


9p2 = (3p)2 16 = (4)2

9a2 – 16 = (3p)2 – (4)2

a2 – b2


= (3p – 4)(3p + 4)



Ex: Factor y6 – 25


y6 = (y3)2 25 = (5)2

y6 – 25 = (y3)2 – (5)2

a2 – b2


= (y3 – 5)(y3 + 5)



Ex: Factor 81 – x2y2


81 = (9)2 x2y2 = (xy)2

81 – x2y2 = (9)2 – (xy)2

a2 – b2


= (9 – xy)(9 + xy)



FOIL Method of Factoring

• Recall FOIL – (3x + 4)(4x + 5) = 12x2 + 15x + 16x + 20 = 12x2 + 31x + 20

The product of the two binomials is a trinomialThe constant term is the product of the L termsThe coefficient of x, b, is the sum of the O & I

productsThe coefficient of x2, a, is the product of the F

terms

FOIL Method of Factoring

1. Factor out the GCF, if any2. For the remaining trinomial, find the F

terms (__ x + )(__ x + ) = ax2

3. Find the L terms ( x + __ )( x + __ ) = c

4. Look for the outer and inner products to sum to bx

5. Check the factorization by using FOIL to multiply

Ex: Factor b2 + 6b + 5

1. there is no GCF

2. the lead coefficient is 1 (1b )(1b )

3. Look for factors of 5 1, 5 & 5, 1

(b + 1)(b + 5) or (b + 5)(b + 1)

4. outer-inner product?

(b + 1)(b + 5) 5b + b = 6b

or (b + 5)(b + 1) b + 5b = 6b Either one works b2 + 6b + 5 = (b + 1)(b + 5)

5. check: (b + 1)(b + 5) = b2 + 5b + b + 5= b2 + 6b + 5

Ex: Factor y2 + 6y – 55

1. there is no GCF

2. the lead coefficient is 1 (1y )(1y )

3. Look for factors of – 55

1, -55 & 5, - 11 & 11, - 5 & 55, - 1

(y + 1)(y – 55) or (y + 5)(y - 11) or ( y + 11)(y – 5) or (y + 55)(y – 1) 4. outer-inner product?

(y + 1)(y - 55) -55y + y = - 54y (y + 55)(y - 1) -y + 55y = 54y y2 + 6y - 55 = (y + 11)(y – 5)

5. check: (y + 11)(y – 5) = y2 – 5y + 11y - 55= y2 + 6y – 55

(y + 5)(y - 11) -11y + 5y = -6y (y + 11)(y - 5) -5y + 11y = 6y

Factor completely – 3 Terms• Always look for a common factor

– immediately take it out to the front of the expression all common factors

– show what’s left inside ONE set of parenthesis• Identify the number of terms. • If there are three terms, and the leading coefficient is positive:

– find all the factors of the first term, find all the factors of the last term– Within 2 sets of parentheses,

• place the factors from the first term in the front of the parentheses• place the factors from the last term in the back of the parentheses

– NEVER put common factors together in one parenthesis.– check the last sign,

• if the sign is plus: use the SAME signs, the sign of the 2nd term• if the sign is minus: use different signs, one plus and one minus

– “smile” to make sure you get the middle term• multiply the inner most terms together then multiply the outer most

terms together, and add the two products together.

Factor completely: 2x2 – 5x – 7

• Factors of the first term: 1x & 2x

• Factors of the last term: -1 & 7 or 1 & -7

• (2x – 7)(x + 1)

Factor Completely.

4x2 + 83x + 60

• Nothing common• Factors of the first term: 1 & 4 or 2 & 2• Factors of the last term: 1,6 2,30 3,20 4,15 5,12

6,10• Since each pair of factors of the last has an even number,

we can not use 2 & 2 from the first term

• (4x + 3)(1x + 20 )

Sign Pattern for the Binomials

Trinomial Sign Pattern Binomial Sign Pattern

+ + ( + )( + )

- + ( - )( - )

- - 1 plus and 1 minus

+ - 1 plus and 1 minus

But as you can tell from the previous example, the FOIL method of factoring requires a lot of trial and error (and hence luck!) . . . Better way?

ac Method for factoring ax2 + bx + c

1. Factor out the GCF, if any

2. For the remaining trinomial, multiply ac

3. Look for factors of ac that sum to b

4. Rewrite the bx term as a sum using the factors found in step 3

5. Factor by grouping

6. Check by multiplying using FOIL

Ex: Factor 3x2 – 4x – 15

1. Is there a GCF? No

2. Multiply ac a = 3

3

and c = – 15

– 15

3(-15) = - 45

3. Factors of -45 that sum to

– 4

– 4

4. Rewrite the middle term

3x2 – 4x – 15 = 3x2 – 9x + 5x – 15

1 – 45 – 44 3 – 15 – 12 5 – 9 – 4

Note: although there are more factors of – 45, we don’t have to check them since we found what we were looking for!

Four-term polynomial . . . How should we proceed to factor?

Factor by grouping . . . 3x2 – 9x + 5x – 15


3 x 2 – 9x = 3x( – )

3x

3x 3x

3

x 3

Common factor among the 2nd two terms? 5

5 x – 15 = 5( – )5 5

3

x 3

3x2 – 9x + 5x – 15 = 3x(x – 3) + 5(x – 3)

= (x – 3)( ) 3x + 5

Ex: Factor 2t2 + 5t – 12



2

and c = – 12

– 12

2(-12) = - 24

3. Factors of -24 that sum to

+ 5

5


2t2 + 5t – 12 = 2t2 – 3t + 8t – 12

1 – 24 – 23 2 – 12 – 10 3 – 8 – 5

Four-term polynomial . . . Factor by grouping . . .

Close but wrong sign so reverse it

- 3 8 5

2t2 – 3t + 8t – 12


2 t 2 – 3t = t( – )

t

t t

3

2t 3


8 t – 12 = 4( – )4 4

3

2t 3

2t2 – 3t + 8t – 12 = t(2t – 3) + 4(2t – 3)

= (2t – 3)( ) t + 4

2

Ex: Factor 9x4 + 18x2 + 8



9

and c = 8

8

9(8) = 72

3. Factors of 72 that sum to

18

18


9x4 + 18x2 + 8 = 9x4 + 6x2 + 12x2 + 8

1 72 73 3 24 27 6 12 18


Bit big think bigger factors

9x4 + 6x2 + 12x2 + 8


9x4 + 6x2 = 3x2( + )

3x2

3x2 3x2

2

3x2 2


12x2 + 8 = 4( + )4 4

3

3x2 2

9x4 + 6x2 + 12x2 + 8 = 3x2(3x2 + 2) + 4(3x2 + 2)

= (3x2 + 2)( + ) 3x2 4

3

3x2

Ex: Factor 12x2 – 17xy + 6y2

1. Is there a GCF? No, but notice two variables

2. Multiply ac a = 12x2

12

and c = 6y2

6y2

12x2(6y2) = 72y2

3. Factors of 72x2y2 that sum to

– 17 y

- 17xy


12x2 – 17xy + 6y2 = 12x2 – 8xy – 9xy + 6y2

-1xy -72xy -73xy -6xy -12xy -18xy -8xy -9xy -17xy


Each factor need a y, both need to be negative

Too big, think bigger factors

Pick one to be the variable

12x2 – 8xy – 9xy + 6y2


12x2 – 8xy = 4x( – )

4x

4x 4x

2y

3x 2y

Common factor among the 2nd two terms? - 3y

– 9xy + 6y2 = - 3y( )-3y -3y

-2y

3x – 2y

12x2 – 8xy – 9xy + 6y2 = 4x(3x – 2y) – 3y(3x – 2y)

= (3x – 2)( ) 4x – 3y

3

3x

Ex: Factor x3 + 3x2 – 4x – 12


2. Notice four terms grouping

Common factor among the 1st two terms? x2

x3 + 3x2 = x2( + ) x2 x2

x x 3

Common factor among the 2nd two terms? - 4

– 4x – 12 = – 4( ) - 4 - 4

3 x + 3

x3 + 3x2 - 4x – 12 = x2(x + 3) – 4(x + 3) = (x + 3)( ) x2 – 4

Cont: we have (x + 3)(x2 – 4)

But are we done? No. We have to make sure we factor completely.

Is (x + 3) prime? can x + 3 be factored further? No . . . It is prime

What about (x2 – 4)? Recognize it?

Difference of Squares

x2 = (x)2 4 = (2)2

x2 – 4 = (x)2 – (2)2 = (x – 2)(x + 2)

Therefore x3 + 3x2 – 4x – 12 = (x + 3)(x2 – 4)

= (x + 3)(x – 2)(x + 2)

Special PolynomialsIs the polynomial a sum/difference of cubes?

• a3 + b3 = (a + b)(a2 - ab + b2)• a3 – b3 = (a - b)(a2 + ab + b2)

Ex: Factor 8p3 – q3

Notice the terms are both perfect cubes

8p3 = (2p)3 q3 = (q)3

8p3 – q3 = (2p)3 – (q)3

a3 – b3


= (2p – q)((2p)2 + (2p)(q) + (q)2)

difference of cubes

= (a – b)(a2 + ab + b2)factors as

= (2p – q)(4p2 + 2pq + q2)

Ex: Factor x3 + 27y9

Notice the terms are both perfect cubes

x3 = (x)3 27y9 = (3y3)3

x3 + 27y9 = (x)3 + (3y3)3

a3 + b3

and we have a sum

= (x + 3y3)((x)2 - (x)(3y3) + (3y3)2)

sum of cubes

= (a + b)(a2 - ab + b2)factors as

= (x + 3y3)(x2 – 3xy3 + 9y6)

INTRODUCTION TO FACTORING POLYNOMIALS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice...

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Transcript of INTRODUCTION TO FACTORING POLYNOMIALS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice...