Transcript Review

Chapters 5.1 – 5.6
REVIEW OF FACTORING
Factors
 Factors are numbers or variables that are multiplied in a multiplication
problem.
 Factor an expression means to write the expression as a product of its factors
 a · b = c, then a and b are said to be factors of c
 Examples:

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

3 · 5 = 15, then 3 and 5 are factors of the product 15
x3 · x4 = x7, then x3 and x4 are factors of x7
x(x + 2) = x2 + 2x, then x and x + 2 are factors of x2 + 2x
(x – 1) (x + 3) = x2 + 2x – 3, then x – 1 and x + 3 are factors of x2 + 2x – 3
Prime Numbers and Composite Numbers
 Prime Number is an integer greater than 1 that has exactly two
factors, itself and one.
 The first 15 prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
 Composite is a positive integer (other than 1) that is not a
prime number.
 The number 1 is neither prime nor composite, it is called a
unit.
Greatest Common Factor
 Greatest Common Factor (GCF) of two or more numbers is the greatest
number that divides into all the numbers.
 GCF (two or more numbers) – Write each number as a product of prime
factors. Determine the prime factors common to all numbers. Multiply the
common factors.
 Example: Find the GCF of 40 and 140
40 = 2 · 20 = 2 · 2 · 10 = 2 · 2 · 2 · 5 = 23 · 5
140 = 2 · 70 = 2 · 2 · 35 = 2 · 2 · 5 · 7 = 22 · 5 · 7
GCF = 22 · 5 = 4 · 5 = 20
Greatest Common Factor
 GCF (two or more terms) – Take each factor the largest
number of times that it appears in all of the terms.
 Example: Find the GCF of 18y2, 15 y3, and 27y5
18y2 = 2 · 9 · y · y
= 2·3·3·y·y
15 y3 =
=
3·5·y·y·y
27y5 = 3 · 9 · y · y · y · y · y = 3 · 3 · 3 · y · y · y · y · y
GCF = 3 · y · y = 3 y2
Factor a Monomial from a Polynomial
1.
2.
3.
Determine the greatest common factor (GCF) of all the terms in
the polynomial.
Write each term as the product of the GCF and its other factor.
Use the distributive property to factor out the GCF.
a (b + c) = ab + ac

Example #9 pg 348:
7x – 35
7(x – 5)

GCF = 7
FOIL = First, Outer, Inner, Last
First = (a)(c)
Outer = (a)(d)
(a + b) (c + d)
Inner = (b)(c)
Last = (b)(d)
Factor a Monomial from a Polynomial
1.
2.
3.
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Determine the greatest common factor (GCF) of all the terms in
the polynomial.
Write each term as the product of the GCF and its other factor.
Use the distributive property to factor out the GCF.
Distributive property says the if a, b and c are all real numbers
then
a (b + c) = ab + ac
Monomial has one term: Exp: 5 , 4x, -6x2
Binomial has two terms: Exp: x + 4 , x2 – 6 , 2x2 – 5x
Trinomial has three terms: Exp: x2 – 2x + 3 , 5x2 – 6x +7
Polynomial has infinite number of terms: Exp: 2x4 – 4x2 – 6x + 3
Factor a Monomial from a Polynomial
 Example:
6a4 + 27a3 - 18a2
3a2(2a2 + 9a – 6)
GCF = 3a2
x(5x-2) + 7(5x-2)
(x+7)(5x-2)
(5x-2) is a factor
 Example:
Factor a Four-Term Polynomial by Grouping
Determine if there is a common factor, if yes then factor out.
2. Arrange the four terms so that the first two terms and the last
two terms have a common factor.
3. Use distributive property to factor each group of terms.
(first two, last two)
4. Factor GCF from the results.
1.

Example #27, pg 348
x2 + 3x – 2xy – 6y
x(x + 3) – 2y(x + 3)
(x + 3) (x – 2y)
Example #22, pg 348:
x2 – 5x + 4x – 20
x(x – 5) + 4(x – 5)
(x – 5) (x + 4)
Signs
 If the 3rd term is positive the factor of last two terms will be positive.
 If the 3rd term is negative the factor of the last two terms will be negative.
 2 positives – Factor positive
x2 + b + c
( _ + _ ) ( _+ _ )
 b = Negative, c = Positive – Factor Negative
x2 – b + c
( _ - _ ) ( _- _ )
 b = Positive, c = Negative – Factor Positive Negative
x2 + b – c
( _ + _ ) ( _- _ )
 b = Negative, c = Negative – Factor Positive Negative
x2 – b – c
( _ + _ ) ( _- _ )
Factoring Trinomials, a = 1
 In the form of ax2 + bx + c, where a = 1
1.
Find two numbers whose product equals the constant, c, and whose
sum equals the coefficient of the x-term, b.
2.
Use the two numbers found, including their signs, to write the
trinomial in factored form . (x + one number)(x + second number)
3.
Check using the FOIL method.

Example # 37, pg 348
x2 + 11x + 18
(x + 9) (x+ 2)
Factors of c that add to b
8 = (2)(9)
2 + 9 = 11
Factoring Trinomials
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Example # 40, pg 348
x2 – 15x + 56
(x - 7) (x - 8)
Factors of c (56)that add to b (-15)
56 = (-7)(-8)
-7 + -8 = -15
 A prime polynomial is a polynomial that cannot be factored using
only integer coefficients.
 Example #36, pg 348
x2 + 4x – 15
PRIME
Factors of c (-15) that add to b (+4)
-15 = (-3)(5)
-15 = (3)(-5)
-3 + 5 ≠ 4
5 + -3 ≠ 4
Factoring Trinomials, a ≠ 1
 In the form of ax2 + bx + c, where a ≠1, by Trial and Error
1.
2.
3.
4.
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Factor out any common factors to all three terms
Write all pairs of factors of the coefficient of the squared term, a
Write all pairs of factors of the constant term, c
Try various combinations of these factors until the correct middle term,
bx, is found.
Example:
3x2 + 20x + 12
(3x + 2) (x + 6)
Factors Possible
Sum
of 12
Factors
Inner/Outer
(1)(12) (3x+1)(x+12) 36x + x = 37x
(2)(6) (3x+2)(x+6) 18x + 2x = 20x
(3)(4) (3x+3)(x+4)
12x + 3x = 15x
Factoring Trinomials, a ≠ 1

Example:
2x2 + 2x – 12
2(x2 + x – 6)
2x(x + 3) (x – 2)
1. Common Factor = 2
2. Factors of -6 that add to 1
(-1)(6) add -1 + 6 = 5
(-2)(3) add -2 + 3 = 1
Remember you can check with FOIL
2(x + 3) (x – 2)
2(x2 – 2x + 3x – 6)
2(x2 + x – 6)
2x2 + 2x – 12
Factoring Trinomials
 In the form of ax2 + bx + c, where a ≠1, by Grouping
1.
Factor out any common factors to all three terms
2.
Fine two numbers whose product is equal to the product of a times c, and whose sum
is equal to b
3.
Rewrite the middle term, bx, as the sum or difference of two terms using the numbers
found
4.
Factor by grouping.
5.
HINT: If no factors of (a)(c) add up to (b) then cannot factor.
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Example:
3x2 + 20x + 12
3x2 + 18x + 2x + 12
3x(x + 6) + 2(x + 6)
(3x + 2) (x + 6)
No common factors
a = 3 b = 20 c = 12
(a)(c) = (3)(12) = 36
factors of 36 that add to 20
(1)(36)
1 + 36 = 37
(2)(18)
2 + 18 = 20
(3)(12)
3 + 12 = 15
Difference of Two Squares
 a2 – b2 = (a + b) (a – b)
 Example #64, pg 349
x2 – 36
x2 – 62
(x + 6) (x – 6)
Example # 70, page 349
64x6 – 49y6
(8x 3) 2 – (7y3) 2
(8x 3 + 7y3 )(8x 3 – 7y3 )
Sum of Two Cubes
 a3 + b3 = (a + b) (a2 – ab + b2)
 Example #74, pg 349
x3 + 8
x 3 + 23
a = x, b = 2
(x + 2) (x 2 – 2x + 22)
(x + 2) (x 2 – 2x + 4)
Example # 77, page 349
125a3 + b3
5a3 + b3
a = 5a, b = b
(5a + b) ((5a) 2 – 5ab + b2)
(5a + b) (25a 2 – 5ab + b2)
Difference of Two Cubes
 a3 – b3 = (a - b) (a2 + ab + b2)
 Example #76, pg 349
b3 – 64
b3 – 43
a=b, b=4
(b – 4) (b2 + 4b + 42)
(b – 4)(b2 + 4b + 16)
Example # 73, page 349
x3 – 1
x3 – 13
a=x,b=1
(x – 1)(x2 + 1x + 12)
(x – 1)(x2 + x + 1)
General Procedure for Factoring a Polynomial
1. Factor any GCF of all terms
2. If a two term polynomial determine if it is a special
factor. If so factor using the formula
3. If three term polynomial, factor according to methods
discussed for a = 1 or a ≠ 1.
4. If more than three terms try factoring by grouping
5. Determine if there are any common factors, and factor
them out.
Quadratic Equation
 Standard Form:
a + bx + c = 0
a, b and c are real numbers
 Zero factor Property – if ab = 0, then a = 0 or b = 0
 Solve the quadratic Equation by factoring
1.
Write the equation in standard form
2.
Factor the side of the equation that is not 0
3.
Set each factor equal to 0 and solve
4.
Check each solution.
Quadratic Equation
Example
x2 – 3x = -2
x2 – 3x + 2 = 0
Factors of 2 that sup to -3
(-1)(-2) = 2
(-1) + (-2) = -3
(x – 1)(x – 2)
x – 1 = 0 and x – 2 = 0
x=1
x=2
Check (x = 1)
x2 – 3x = -2
12 – 3(1) = -2
1 – 3 = -2
-2 = -2
True
Check (x = 2)
x2 – 3x = -2
22 – 3(2) = -2
4 – 6 = -2
-2 = -2
True
Homework – Review Factoring
 Page 348 – 349:
#11, 13, 15, 35, 39, 41, 55, 63, 69, 85