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ALGEBRA – LESSON 73
Factoring the Difference of Two Squares,
Probability without Replacement
Be ready to grade the homework!
Warm up!
9 = 3
25 = 5
9 = 7
x2 = x
y2 = y
x2y2 = xy
9x2 = 3x
169x2 = 13x 36a2b2 = 6ab
Factoring a Difference of 2 Squares (not in your notes)
We have learned that factoring means to break into parts
that can be multiplied to result in the original equation.
We have learned a couple of strategies for factoring:
1 – We can “unfoil”.
2 – We can look for something that every term has in common.
Today we will learn another strategy that involves very
little math. It is simply recognizing the type of equation
and then using a formula for plugging in the numbers.
Factoring a Difference of 2 Squares (not in your notes)
We need to be on the lookout for a special kind of
factoring problems. If the equation has 2 terms which
are being subtracted and both terms are perfect squares.
25x2 - 9
• 2 terms
•Both terms formed from perfect squares.
(numbers AND variables)
•The terms are being subtracted.
•Hint 1: you may have to rearrange it for it to be subtraction.
•Hint 2: you may have to pull out a common factor first.
Factoring a Difference of 2 Squares
If the equation has 2 terms and both terms are perfect squares and 1 is
positive and 1 is negative, we call them the difference of 2 squares.
Steps
25x2 - 9
1. Find the square root of each term.
5x and 3
2. The answer will be (1st – 2nd)(1st + 2nd)
(5x – 3)(5x + 3)
* ALWAYS look for a common term to factor first!
Factoring a Difference of 2 Squares
Remember: this is a process problem. Recognize it then follow the steps.
16x2 – 9y2
1. Find the square root of each term.
4x and 3y
2. The answer will be (1st – 2nd)(1st + 2nd)
(4x – 3y)(4x + 3y)
* ALWAYS look for a common term to factor first!
#1
Factoring a Difference of 2 Squares
-49 + 16y2
1. Rearrange to make this subtraction.
16y2 - 49
2. Find the square root of each term.
4y and 7
3. The answer will be (1st – 2nd)(1st + 2nd)
(4y – 7)(4y + 7)
#2
Factoring a Difference of 2 Squares
64a2x2y2 – 9a2z2
1. Factor out a common term – even if it’s a perfect square.
a2(64x2y2 – 9z2)
2. Find the square root of each term.
8xy and 3z
3. The answer will be (1st – 2nd)(1st + 2nd).
Don’t forget the originally factored term.
a2(8xy – 3z)(8xy + 3z)
#3
Probability without replacement
We have calculated probability of independent events. But this process
will change as soon as we begin to see problems in which objects ARE
NOT replaced. For example – a marble is drawn and then NOT put
back in the bag.
We will calculate these probabilities by: multiplying each
probability taking into account that there are fewer total
items the 2nd time.
Probability without replacement
A bag contains 3 white marbles and 4 yellow marbles. One
marble is drawn at random and not replaced. Then a
second marble is drawn. What is the probability that the
first marble is yellow and the second one is white?
What is the probability for the yellow marble?
4
7
What is the probability for the white marble? (Remember, the 1st
marble was not put back in the bag.)
3
6
Multiply
4 x 3 =
7
6
2
7
#4
Probability without replacement
Problems will now be asking for you to calculate the
probability with replacement and without. Think about why
you are choosing the fractions you are choosing.
Also be aware that sometimes they’ll switch the order that
they ask the questions. Notice #5 and #6 on your notes.
Probability without replacement
Martha has a bag that contains 7 yellow gum drops and 6 black gum drops.
She randomly draws 2 gum drops, one after the other. What is the probability
that the first gum drop is yellow and the second gum drop is black?
a. With replacement:
b. Without replacement:
What is the probability for
the yellow gum drop?
7
13
What is the probability for
the yellow gum drop?
7
13
What is the probability for the
black gum drop? (Remember,
the 1st gum drop WAS put
back in the bag.)
6
13
What is the probability for the
black gum drop? (Remember,
the 1st gum drop WAS NOT
put back in the bag.)
6
12
Multiply
Multiply
7 x 6 = 42
13 13 169
7 x 6 = 7
13 12 26
#5
Probability without replacement
Larry has a urn that contains 7 orange marbles and 11 blue marbles. He
randomly draws 2 marbles, one after the other. What is the probability that
both marbles are orange?
a. Without replacement:
b. With replacement:
What is the probability for
the 1st orange?
7
18
What is the probability for
the 1st orange marble?
7
18
What is the probability for the
2nd orange marble?
(Remember, the 1st marble
WAS NOT put back in the bag.)
6
17
What is the probability for the
2nd orange marble?
(Remember, the 1st gum drop
WAS put back in the bag.)
7
18
Multiply
Multiply
7 x 6 = 7
18 17 51
7 x 7 = 49
18 18 324
#6
Homework: PS 73