Order of Operations:ParenthesisExponents (including roots)Multiplication & DivisionAddition & Subtraction

Always Work Left to Right

Review:

Properties of Arithmetic

3, 2, 1…Lets work some

problems!

9 + 6 ÷ (2 - 8)

Answer:

8

6 · 6 - (7 + 5)

Answer:

24

2 - (2 + 3 - 8)

Answer:

5

-3 - (-4)(-5) - (-6)

Answer:

-17

Reciprocal & Multiplicative Inverse Property

If one fraction is the inverted (upside down) form of another fraction, each of the fractions is said to be the RECIPROCAL of the other fraction.

For example:

3/2 is the reciprocal of 2/3

-4/11 is the reciprocal of -11/4

1/4 is the reciprocal of 4

-5 is the reciprocal of -1/5

1/0 is MEANINGLESS and therefore ZERO is the ONLY real number which DOES NOT have a RECIPROCAL.

If a number is multiplied by its reciprocal the PRODUCT is the number 1.

Definition:

For any nonzero real number a, the RECIPROCAL, or MULTIPLICATIVE INVERSE,

of the number is 1/a.

Evaluate:

4 ÷ ¼

Answer:

16

Challenging Problems

8 · (5 – 7)⁴ + 2 · 3² ÷ 2 - 33

Answer:

110

2 (10² + 3 · 19) ÷ (-5² ÷ ¼)

Answer:

-3.14

(39 + 2)(83 - 4)

Answer:

3239

Distributive Property

Vocabulary

• The Distributive Property is a mathematical property that allows you to multiply a number on the outside of the parentheses by each number inside the parentheses.

What does it look like?• 5(2 + 4) = 5(2) + 5(4) Multiply the 5 by EVERYTHING in the parentheses!!

5(6) = 10 + 20 30 = 30

• (2 + 3)6 = (2)(6) + (3)(6) (5)(6) = (12) + (18)

• a(b + c) = a(b) + a(c) a(b + c) = ac + ab

• (a + b)c = (a)(c) + (b)(c)(a +b)c = ac + bc

It works with subtraction in the Parentheses too!

• 5(6 – 3) = 5(6) – 5(3) Multiply the 5 everything in the parentheses!

5(3) = 30 – 1515 = 15

• (12 – 3)(-4) = (12)(-4) – (3)(-4)(9)(-4) = -48 - (-12) -36 = -36

• a(b – c) = a(b) – a(c)• (a – b)c = a(c) – b(c)

Use the Distributive Property to Simplify

• Find 20(102) 20(102) = 20(100 + 2) = 20(100) + 20(2) = 2,000 + 40 = 2,040

• Find 53(40) 53(40) = (50 + 3)(40) = (50)(40) + 3(40) = 2000 + 120 = 2,120

• Find 9(199) 9(199) = 9(200 – 1)

= 9(200) – 9(1) = 1800 – 9 = 1,791

Let’s try some more

problems…

Simplify:

7(b + 2) 7b + 7(2) 7b + 14 4(x + 1) 4x + 4(1) 4x + 4 (-2)(3 + x) (-2)(3) + (-2)x -6 + (-2x) -6 – 2x

2(x – 4) 2x – (2)(4) 2x - 8 (-3)(4 – y) (-3)(4) – (-3)(y) -12 – (-3y) -12 + 3y (4n-6)5

(4n)(5) – 6(5) 20n - 30

More Problems… 12(a + 3) 12a + 12(3) 12a + 36

(c-4)(-2) c(-2) – 4(-2) -2c – (-8) -2c + 8

5(x + y) 5x + 5y

4(x + y + z) 4x + 4y + 4z

-(x +2)(-1)(x + 2)(-1)x + (-1)(2) -x - 2

Let’s put our thinking cap on…

Recall the distributive property of multiplication over addition . . .

symbolically:

a × (b + c) = a × b + a × cand pictorially (rectangular array area model):

a × b a × ca

b c

An example: 6 x 13 using your mental math skills . . .

symbolically:

6 × (10 + 3) = 6 × 10 + 6 × 3and pictorially (rectangular array area model):

6 × 10 6 × 36

10 3

What about 12 x 23? Mental math skills?

(10+2)(20+3) = 10×20 + 10×3 + 2×20 + 2×3

10 × 20 10 × 310

20 3

2 × 32 2 × 20

200

30

40

+ 6

276

And now for multiplying binomials

(a+b)×(c+d) = a×(c+d) + b×(c+d) = a×c + a×d + b×c + b×d

a × c a × da

c d

b × db b × c

We note that the product of the two binomials has four terms – each of these is a partial product. We multiply each term of the first binomial by each term of the second binomial to get the four partial products.

Product of the FIRST terms of the binomials

Product of the OUTSIDE terms of the binomials

Product of the INSIDE terms of the binomials

Product of the LAST terms of the binomials

F + O + I + L

( a + b )( c + d ) = ac + ad + bc + bd

Because this product is composed of the First, Outside, Inside, and Last terms, this pattern is often referred to as FOIL method of multiplying two binomials. Note that each of these four partial products represents the area of one of the four rectangles making up the large rectangle.

Are the two expressions equal? (y/n) give answer to

both expressions

(5+6)²

5² + 6²

Answer:

NO!121 ≠ 61

How to expand a sum: “FOIL”(x+1)² =

(x+1)·(x+1)

x·(x+1) + 1·(x+1) x·x +(x)·(1) +(1)·(x) + 1·1

x² + x + x + 1² x² + 2x + 1

Challenging Problems

(101)²

Answer:

10,201

(7+5)² -7² - 5²

Answer:

70

(99)(101)

Answer:

9999

Expand the following algebraic expression:

(x+3)²

Answer:

x² + 6x + 9