Lecture 03 special products and factoring

Special Products and Factoring

Multiply: a Monomial and a Polynomial
Multiply each polynomial term by that monomial:
 Positive numbers – law of distribution
 3x2(6xy + 3y2)
 3x2(6xy) + 3x2(3y2)
 18x3y + 9x2y2
 Negative coefficients – be careful!
 -2ab2(3bz – 2az + 4z3)
 -2ab2(3bz) – (-2ab2)(2az) + (-2ab2)(4z3)
 -6ab2z + 4a2b2z – 8ab2z3
 Number following polynomial – other distributive law
 (2x3 – 3x2 – 5)(3x)
 3x(2x3) – (3x)(3x2) – (3x)(5)
 6x4 – 9x3 – 15x

Multiply: Two Polynomials
Horizontal Method: (use the distributive property repeatedly)
(2a + b)(3a – 2b)
= (2a + b)(3a) – (2a + b)(2b)
= 6a2 + 3ab – (4ab + 2b2)
= 6a2 + 3ab – 4ab – 2b2
= 6a2 – ab – 2b2
Vertical Method: 3x2 + 2x – 5
4x + 2 Multiply top by rightmost
6x2 + 4x – 10 Then by term to the left
12x3 + 8x2 – 20x____ Lastly, add like terms
12x3 + 14x2 – 16x – 10

Concept: Similar Binomials
 Any pair of binomials with matching variable parts
 When multiplied, they produce a trinomial or
binomial
 Examples:
x and x
(3 5) ( 2)
 
x x x x
2     
(3 5)( 2) 3 10
a b and a b
    
(  ) ( 
2 )
a b a b a ab b
x y and x Not Similar
( 5 )( 7) 7 5 35
( 5 ) ( 7)
2 2 ( )( 2 ) 3 2
2
 
2      
x y x x x xy y
a b and b a
( ) ( 2 )
a b b a a a b b
x and x Not Similar
(2  3) (2 
3)
2 3 2
x x x x x
2 2
(2  3)(2  3)  4  3  6 
9
 
2 2 4 2 2
( )( 2 ) 2 3
    
x y and x y
     
(  ) (  
)
x y x y x y
2 2 ( )( )

The FOIL Method
Useful for Multiplying Two Similar Binomials
(x + 8)(x - 5) = x2 – 5x + 8x – 40
(4t2 + 5)(3t2 - 2) = 12t4 – 8t2 + 15t2 – 10
(y – 8)(y2 + 5) = y3 + 5y – 8y2 – 40

Practice
Using the FOIL Method in Your Head
If the polynomials are similar, combine the middle
terms
(x + 2)(x – 5) = x2 – 3x – 10
(2y + 3)(4y + 1) = 8y2 + 14y + 3
(m – 3n)(m – 2n) = m2 – 5mn + 6n2
(2x + 3y2)(x – 7y2) = 2x2 – 11xy2 – 21y4
(a + b)(c + d) = ac + ad + bc + bd

Multiplying 3 or more
Polynomials
 Use same technique as you use for numbers:
 Multiply any 2 together and simplify the temporary
product
 Multiply that temporary product times any
remaining polynomial and simplify
-2r(r – 2s)(5r – s) = (use foil on the binomials)
-2r(r2 – 11rs + s2) = (distribute the monomial)
-2r3 – 22r2s – 2rs2

The Product of Conjugates
(F + L)(F – L) = F2 – L2
 The middle term disappears ONLY when the binomials are
conjugates: identical except for different operations
 Multiplying these is easier than using FOIL!
 (x + 4)(x – 4) = x2 – 42 = x2 – 16
 (5 + 2w)(5 – 2w) = 25 – 4w2
 (3x2 – 7)(3x2 + 7) = 9x4 – 49
 (-4x – 10)(-4x + 10) = 16x2 – 100
 (6 + 4y)(6 – 4x) = use the foil method
= 36 – 24x + 24y – 16xy

Squaring a Binomial Sum
(F + L)(F + L) = F2 + 2FL + L2
Square the 1st term
Multiply 1st times 2nd,
double it, add it
Square the 2nd term
Try:
(2x + 3)2
(2x)2 + 2(6x) + 32
4x2 + 12x + 9
(½x + 5)2
(½x)2 + 2(5x/2) + 52
¼x2 + 5x + 25

Squaring a Binomial Difference
(F – L)(F – L) = F2 – 2FL + L2
Square the 1st term
Multiply 1st times 2nd,
double it, subtract it
Square the 2nd term and add it
Try:
(3x - 4)2 =
(3x)2 – 2(12x) + 42 =
9x2 – 24x + 16
(5a – 2b)2 =
(5a)2 – 2(10ab) + (2b)2 =
25a2 – 20ab + 4b2

Practice
Binomial Conjugates and Squares
(F + L)(F – L) = F2 – L2
(F + L)2 = F2 + 2FL + L2
(F – L)2 = F2 – 2FL + L2
 (x + 3)(x – 3) = x2 – 9
 (2y – 5)(2y – 5) = 4y2 – 20y + 25
 (m + 3n)2 = m2 + 6mn + 9n2
 (2y – 5)(2y + 5) = 4y2 – 25
 (a + b)(a + b) = a2 + 2ab + b2
 (3x – 7y)2 = 9x2 – 42xy + 49y2

 Find the product of the following:

Find the product
Use the FOIL method

Find the product
Use Dist. Prop. twice

Factoring by Grouping
 video

Factoring Trinomials
 Video1
 video2

Perfect Square Trinomial
 video

Perfect Square Factoring
 video

 FACTORING IS THE REVERSE of multiplying.
2x² + 9x − 5
(2x ?)(x ?)
(2x 5)(x 1)
or with x --
(2x 1)(x 5) ?
(2x − 1)(x + 5) = 2x² + 9x − 5.

Problem 1. Place the correct signs to give the middle
term.
a) 2x² + 7x − 15 = (2x − 3)(x + 5)
b) 2x² − 7x − 15 = (2x + 3)(x − 5)
c) 2x² − x − 15 = (2x + 5)(x − 3)
d) 2x² − 13x + 15 = (2x − 3)(x − 5)

Problem 2. Factor these trinomials.
a) 3x² + 8x + 5 = (3x + 5)(x + 1)
b) 3x² + 16x + 5 = (3x + 1)(x + 5)
c) 2x² + 9x + 7 = (2x + 7)(x + 1)
d) 2x² + 15x + 7 = (2x + 1)(x + 7)
e) 5x² + 8x + 3 = (5x + 3)(x + 1)
f) 5x² + 16x + 3 = (5x + 1)(x + 3)

Problem 3. Factor these trinomials.
a) 2x² − 7x + 5 = (2x − 5)(x − 1)
b) 2x² − 11x + 5 = (2x − 1)(x − 5)
c) 3x² + x − 10 = (3x − 5)(x + 2 )
d) 2x² − x − 3 = (2x − 3)(x + 1)
e) 5x² − 13x + 6 = (5x − 3)(x − 2)

Factor completely 6x8 + 30x7 + 36x6.
To factor completely means to first remove any common
factor.
6x8 + 30x7 + 36x6 = 6x6(x² + 5x + 6).
Now continue by factoring the trinomial:
= 6x6(x + 2)(x + 3).

Problem 4. Factor completely. First remove any common
factors.
a) x3 + 6x² + 5x = x(x2 + 6x + 5) = x(x + 5)(x + 1)
b) x5 + 4x4 + 3x3 = x3(x2 + 4x + 3) = x3(x + 1)(x + 3)
c) x4 + x3 − 6x² = x²(x² + x − 6) = x²(x + 3)(x − 2)
d) 4x² − 4x − 24 = 4(x² − x − 6) = 4(x + 2)(x − 3)
e) 2x3 − 14x² − 36x = 2x(x2 − 7x − 18) = 2x(x + 2)(x − 9)
f) 12x10 + 42x9 + 18x8 = 6x8(2x² + 7x + 3) = 6x8(2x + 1)(x + 3).

Quadratics in different arguments
Here is the form of a quadratic trinomial with argument x :
ax² + bx + c.
The argument is whatever is being squared. x is
being squared. x is called the argument. The
argument appears in the middle term.
a, b, c are called constants.
In this quadratic,
3x² + 2x − 1,
the constants are 3, 2, −1.

Now here is a quadratic whose argument is x3:
3x6 + 2x3 − 1.
x6 is the square of x3.

Now, since the quadratic with argument x can be factored
in this way:
3x² + 2x − 1 = (3x − 1)(x + 1),
then the quadratic with argument x3 is factored in the
same way:
3x6 + 2x3 − 1 = (3x3 − 1)(x3 + 1).
Whenever a quadratic has constants 3, 2, −1, then for any
argument, the factoring will be
(3 times the argument − 1)(argument + 1).

Problem 5. Multiply out each of the following, which have
the same constants, but different argument.
a) (z + 3)(z − 1) = z² + 2z − 3
b) (y + 3)(y − 1) = y² + 2y − 3
c) (y6 + 3)(y6 − 1) = y12 + 2y6 − 3
d) (x5 + 3)(x5 − 1) = x10 + 2x5 − 3

Problem 6. Factor each quadratic.
a) x² − 6x + 5 = (x − 1)(x − 5)
b) z² − 6z + 5 = (z − 1)(z − 5)
c) x8 − 6x4 + 5 = (x4 − 1)(x4 − 5)
d) x10 − 6x5 + 5 = (x5 − 1)(x5 − 5)
e) x6y6 − 6x3y3 + 5 = (x3y3 − 1)(x3y3 − 5)

Problem 7. Factor each quadratic.
a) x4 − x² − 2 = (x² − 2)(x² + 1)
b) y6 + 2y3 − 8 = (y3 + 4)(y3 − 2)
c) z8 + 4z4 + 3 = (z4 + 1)(z4 + 3)
d) 2x10 + 5x5 + 3 = (2x5 + 3)(x5 + 1)
e) x4y² − 3x²y − 10 = (x²y + 2)(x²y − 5)
f) cos²x − 5 cos x + 6 = (cos x − 3)(cos x − 2)

Reference
 http://cnx.org/content/m21901/latest/

 The following are some of the products which occur frequently in
Mathematics.
 
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    
    
    
  
I a c  d  ac 
ad
II a  b a  b  a 
b
2 2
2 2 2
III a  b a  b  a  b  a  ab 
b
2 2 2
IV a b a b a b a ab b
V x a x b x 2
a b x ab
VI ax b cx d acx 2
ad bc x bd
VII a b c d ac bc ad bd
.
.
. 2
. 2
.
.
.
      
     
     
     

Lecture 03 special products and factoring

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Lecture 03 special products and factoring