Math083 day 1 chapter 6 2013 fall

Math 083 College Algebra
•Fall 2013
•Instructor Mr. Bianco

Math 083 Bianco
•Today :
–Syllabus
–Multiplying
–Factoring

Mr. Bianco
•15 years of teaching High
School Math- Algebra to
Calculus
•5 Years at Essex

What you need to know
• BE ON TIME!!!!
• Try all homework
• Check website(s) if you miss
class- But make sure you inform
me first
• Don’t wait to get help!
• No electronic devices on during
class

Which number doesn’t belong and be
ready to explain why
9 16 25 43

6
POLYNOMIALSPOLYNOMIALS
MULTIPLY:
1. (7) (2x - 5)
(7) (2x) - (7) (5)
14x - 35

7
POLYNOMIALSPOLYNOMIALS
MULTIPLY:
2. (2x2
)(5x4
+ 7x)
(2x2
) (5x4
)+ (2x2
) (7x)
10x6
+ 14x3

8
Multiplying by FOILMultiplying by FOIL
(3x + 2) (5x + 4)
F O I L

9
(3x + 2) (5x + 4)
F O I L
15x2
F ---> First Terms

10
(3x + 2) (5x + 4)
F O I L
O ---> Outer Terms
15x2
+ 12x

11
(3x + 2) (5x + 4)
F O I L
I ---> Inner Terms
15x2
+ 12x + 10x

12
(3x + 2) (5x + 4)
F O I L
L ---> Last Terms
15x2
+ 12x + 10x + 8

13
(3x + 2) (5x + 4)
F O I L
Combine Like Terms
15x2
+ 22x + 8
15x2
+ 12x + 10x + 8

14
(3x + 2) (5x + 4)
F O I L
15x2
+ 22x + 8

15
(5x + 1) (5x + 1)
F O I L

16
(5x + 1) (5x + 1)
F O I L
25x2
F ---> First Terms

17
(5x + 1) (5x + 1)
F O I L
O ---> Outer Terms
25x2
+ 5x

18
(5x + 1) (5x + 1)
F O I L
I ---> Inner Terms
25x2
+ 5x + 5x

19
(5x + 1) (5x + 1)
F O I L
L ---> Last Terms
25x2
+ 5x + 5x + 1

20
(5x + 1) (5x + 1)
F O I L
Combine Like Terms
25x2
+ 10x + 1
25x2
+ 5x + 5x + 1

21
(5x + 1) (5x + 1)
F O I L
25x2
+ 10x + 1

22
(5x + 1)2
(5x + 1) (5x + 1)
25x2
+ 10x + 1

In the process of using the FOIL method on products
of certain types of binomials, we see specific
patterns that lead to special products.
Square of a Binomial
(a + b)2
= a2
+ 2ab + b2
(a – b)2
= a2
– 2ab + b2
Product of the Sum and Difference of Two Terms
(a + b)(a – b) = a2
– b2
Special Products

Although you will arrive at the same results
for the special products by using the
techniques of this section or last section,
memorizing these products can save you
some time in multiplying polynomials.
Special Products

Factors
Factors (either numbers or polynomials)
When an integer is written as a product of
integers, each of the integers in the product is a
factor of the original number.
When a polynomial is written as a product of
polynomials, each of the polynomials in the
product is a factor of the original polynomial.
Factoring – writing a polynomial as a product of
polynomials.

Greatest common factor – largest quantity that is a
factor of all the integers or polynomials involved.
Finding the GCF of a List of Monomials
1) Find the GCF of the numerical coefficients.
2) Find the GCF of the variable factors.
3) The product of the factors found in Step 1 and 2 is
the GCF of the monomials.
Greatest Common Factor

Find the GCF of each list of numbers.
1) 12 and 8
12 = 2 · 2 · 3
8 = 2 · 2 · 2
So the GCF is 2 · 2 = 4.
1) 7 and 20
7 = 1 · 7
20 = 2 · 2 · 5
There are no common prime factors so the
GCF is 1.
Example:

Find the GCF of each list of numbers.
1) 6, 8 and 46
6 = 2 · 3
8 = 2 · 2 · 2
46 = 2 · 23
So the GCF is 2.
1) 144, 256 and 300
144 = 2 · 2 · 2 · 3 · 3
256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
300 = 2 · 2 · 3 · 5 · 5
So the GCF is 2 · 2 = 4.
Example:

1) x3
and x7
x3
= x · x · x
x7
= x · x · x · x · x · x · x
So the GCF is x · x · x = x3
• 6x5
and 4x3
6x5
= 2 · 3 · x · x · x
4x3
= 2 · 2 · x · x · x
So the GCF is 2 · x · x · x = 2x3
Find the GCF of each list of terms.
Example:

Find the GCF of the following list of terms.
a3
b2
, a2
b5
and a4
b7
a3
b2
= a · a · a · b · b
a2
b5
= a · a · b · b · b · b · b
a4
b7
= a · a · a · a · b · b · b · b · b · b · b
So the GCF is a · a · b · b = a2
b2
Notice that the GCF of terms containing variables will use the smallest exponent
found amongst the individual terms for each variable.
Example:

The first step in factoring a polynomial is to
find the GCF of all its terms.
Then we write the polynomial as a product by
factoring out the GCF from all the terms.
The remaining factors in each term will form a
polynomial.
Factoring Polynomials

Factor out the GCF in each of the following
polynomials.
1) 6x3
– 9x2
+ 12x =
3 · x · 2 · x2
– 3 · x · 3 · x + 3 · x · 4 =
3x(2x2
– 3x + 4)
2) 14x3
y + 7x2
y – 7xy =
7 · x · y · 2 · x2
+ 7 · x · y · x – 7 · x · y · 1 =
7xy(2x2
+ x – 1)
Factoring out the GCF
Example:

Factor out the GCF in each of the following
polynomials.
1) 6(x + 2) – y(x + 2) =
6 · (x + 2) – y · (x + 2) =
(x + 2)(6 – y)
2) xy(y + 1) – (y + 1) =
xy · (y + 1) – 1 · (y + 1) =
(y + 1)(xy – 1)
Factoring out the GCF
Example:

Factoring polynomials often involves additional
techniques after initially factoring out the GCF.
One technique is factoring by grouping.
Factor xy + y + 2x + 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have a
GCF of y and the last 2 terms have a GCF of 2.
xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =
y(x + 1) + 2(x + 1) = (x + 1)(y + 2)
Factoring by Grouping
Example:

Factoring a Four-Term Polynomial by Grouping
1) Arrange the terms so that the first two terms have a
common factor and the last two terms have a common
factor.
2) For each pair of terms, use the distributive property to
factor out the pair’s greatest common factor.
3) If there is now a common binomial factor, factor it out.
4) If there is no common binomial factor in step 3, begin
again, rearranging the terms differently.
• If no rearrangement leads to a common binomial factor, the
polynomial cannot be factored.

1) x3
+ 4x + x2
+ 4 = x · x2
+ x · 4 + 1 · x2
+ 1 · 4 =
x(x2
+ 4) + 1(x2
+ 4) =
(x2
+ 4)(x + 1)
2) 2x3
– x2
– 10x + 5 = x2
· 2x – x2
· 1 – 5 · 2x – 5 · (– 1) =
x2
(2x – 1) – 5(2x – 1) =
(2x – 1)(x2
– 5)
Factor each of the following polynomials by grouping.
Example:

Factor 2x – 9y + 18 – xy by grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y =
2(x + 9) – y(9 + x) =
2(x + 9) – y(x + 9) = (make sure the factors are identical)
(x + 9)(2 – y)

Remember that factoring out the GCF from the terms of
a polynomial should always be the first step in factoring
a polynomial.
This will usually be followed by additional steps in the
process.
Factor 90 + 15y2
– 18x – 3xy2
.
90 + 15y2
– 18x – 3xy2
= 3(30 + 5y2
– 6x – xy2
) =
3(5 · 6 + 5 · y2
– 6 · x – x · y2
) =
3(5(6 + y2
) – x (6 + y2
)) =
3(6 + y2
)(5 – x)
Factoring
Example:

Factoring polynomials often involves additional
techniques after initially factoring out the GCF.
One technique is factoring by grouping.
Factor xy + y + 2x + 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have
a GCF of y and the last 2 terms have a GCF of 2.
xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =
y(x + 1) + 2(x + 1) = (x + 1)(y + 2)
Example:

Factoring a Four-Term Polynomial by Grouping
1) Arrange the terms so that the first two terms have a
common factor and the last two terms have a common
factor.
2) For each pair of terms, use the distributive property to
factor out the pair’s greatest common factor.
3) If there is now a common binomial factor, factor it out.
4) If there is no common binomial factor in step 3, begin again,
rearranging the terms differently.
• If no rearrangement leads to a common binomial factor,
the polynomial cannot be factored.

Factor the polynomial x2
+ 13x + 30.
Since our two numbers must have a product of 30 and
a sum of 13, the two numbers must both be positive.
Positive factors of 30 Sum of Factors
1, 30 31
2, 15 17
3, 10 13
Note, there are other factors, but once we find a
pair that works, we do not have to continue
searching.
So x2
+ 13x + 30 = (x + 3)(x + 10).
Factoring Trinomials of the Form x2
+ bx + c
Example:

– 11x + 24.
Since our two numbers must have a product of 24 and
a sum of –11, the two numbers must both be negative.
Negative factors of 24 Sum of Factors
– 1, – 24 – 25
– 2, – 12 – 14
– 3, – 8 – 11
So x2
– 11x + 24 = (x – 3)(x – 8).
Factoring Trinomials of the Form x2
+ bx + c
Example:

– 6x + 10.
Since our two numbers must have a product of 10 and a
sum of – 6, the two numbers will have to both be
negative.
Negative factors of 10 Sum of Factors
– 1, – 10 – 11
– 2, – 5 – 7
Since there is not a factor pair whose sum is – 6,
x2
– 6x +10 is not factorable and we call it a prime
polynomial.
Prime Polynomials
Example:

You should always check your factoring
results by multiplying the factored
polynomial to verify that it is equal to the
original polynomial.
Many times you can detect computational
errors or errors in the signs of your numbers
by checking your results.
Check Your Result!

Factoring Trinomials of the Form ax2
+ bx + c
Returning to the FOIL method,
F O I L
(3x + 2)(x + 4) = 3x2
+ 12x + 2x + 8
= 3x2
+ 14x + 8
To factor ax2
+ bx + c into (#1·x + #2)(#3·x + #4), note
that a is the product of the two first coefficients, c is
the product of the two last coefficients and b is the
sum of the products of the outside coefficients and
inside coefficients.
Note that b is the sum of 2 products, not just 2
numbers, as in the last section.

Factor the polynomial 2x2
+ 11x + 5.
Factoring Trinomials of the
Form ax2
+ bx + c
Example:

+ 11x + 5.
Form ax2
+ bx + c
Example:
My Fat Dog Sam Stinks

+ 11x + 5.
Form ax2
+ bx + c
Example:
Multiply

+ 11x + 5.
Form ax2
+ bx + c
Example:
Factor

+ 11x + 5.
Form ax2
+ bx + c
Example:
Divide

+ 11x + 5.
Form ax2
+ bx + c
Example:
Simplify

+ 11x + 5.
Form ax2
+ bx + c
Example:
Slide

Factor the polynomial 2x2 + 11x + 5.
Possible factors of 25x2
are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.
We need to methodically try each pair of factors until we
find a combination that works, or exhaust all of our possible
pairs of factors.
Keep in mind that, because some of our pairs are not
identical factors, we may have to exchange some pairs of
factors and make 2 attempts before we can definitely decide
a particular pair of factors will not work.
Form ax2
+ bx + c
Continued.
Example:

We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to
20x.
{x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x
(x + 4)(25x + 1) x 100x 101x
{x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x
Factors of
25x2
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Factors of
4
{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x
+ bx + c
Continued.
Example continued:

Check the resulting factorization using the FOIL method.
(5x + 2)(5x + 2) =
= 25x2
+ 10x + 10x + 4
5x(5x)
F
+ 5x(2)
O
+ 2(5x)
I
+ 2(2)
L
= 2x2 + 11x + 5
So our final answer when asked to factor 2x2 + 11x + 5
will be (5x + 2)(5x + 2) or (5x + 2)2
.
+ bx + c
Example continued:

– 41x + 10.
Possible factors of 21x2
are {x, 21x} or {3x, 7x}.
Since the middle term is negative, possible factors of
10 must both be negative: {-1, -10} or {-2, -5}.
We need to methodically try each pair of factors until
we find a combination that works, or exhaust all of our
possible pairs of factors.
+ bx + c
Continued.
Example:

We will be looking for a combination that gives the sum of
the products of the outside terms and the inside terms
equal to −41x.
Factors of
21x2
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Factors of
10
{x, 21x}{1, 10}(x – 1)(21x – 10) –10x −21x – 31x
(x – 10)(21x – 1) –x −210x – 211x
{x, 21x} {2, 5} (x – 2)(21x – 5) –5x −42x – 47x
(x – 5)(21x – 2) –2x −105x – 107x
+ bx + c
Continued.
Example continued:

Factors of
21x2
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Factors of
10
(3x – 5)(7x – 2) −6x −35x −41x
{3x, 7x}{1, 10}(3x – 1)(7x – 10) −30x −7x −37x
(3x – 10)(7x – 1) −3x −70x −73x
{3x, 7x} {2, 5} (3x – 2)(7x – 5) −15x −14x −29x
+ bx + c
Continued.
Example continued:

Check the resulting factorization using the FOIL method.
(3x – 5)(7x – 2)=
= 21x2
– 6x – 35x + 10
3x(7x)
F
+ 3x(-2)
O
- 5(7x)
I
- 5(-2)
L
= 21x2
– 41x + 10
So our final answer when asked to factor 21x2
– 41x + 10
will be (3x – 5)(7x – 2).
+ bx + c
Example continued:

Math083 day 1 chapter 6 2013 fall

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