solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

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Bell Ringer Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

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Page 1: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Bell Ringer

Solve for x.

30°

2x + 10

2x + 10 = 60 – 10 – 10 2x = 50

2 2 x = 25

Page 2: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Final Exam Review

Page 3: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Exam

The exam will be 40 multiple choice questions with 2 extra credit questions.

You will have 1 hour to complete the exam.

No extra time will be given. You may bring ONE sheet of notes to

use on the exam.

Page 4: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Topics

The final exam will cover:1. Inequalities2. Probability3. Area & Perimeter of Polygons & Circles4. Angles & Lines5. Exponents6. Radicals7. Polynomials

Page 5: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Inequalities

Page 6: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

1. Graphing Linear Inequalities

< or > = Open Circle< or > = Closed Circle< or < = Shade to the left.> or > = Shade to the right.

.

.

.

Greater Than Less Than

Page 7: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

1. Graphing Linear Inequalities

x < 3Open or Closed?

Right or Left?

x > -4Open or Closed?

Right or Left?

.

Page 8: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

-5 + x < -1 + 5 +5

x < 4

Now graph it!

2. One-Step Linear Inequality

Page 9: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

3. Two-Step Linear Inequality

4x + 1 > -11 – 1 – 1 4x > -12

4 4 x > -3

Page 10: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

4. Reversing the Sign!

Whenever we multiply or divide by a negative number, we must REVERSE the inequality sign.

-2x < 6 -2 -2 x > -3

We have to divide

by -2.So we have to reverse the sign.

Page 11: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

4. Reverse the Sign

-3x + 1 < 10 – 1 – 1

-3x < 9 -3 -3

x > -3Reverse the Sign!

Page 12: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Key Words

“At least”means greater than or equal to (>)

“No more than”means less than or equal to (<)

“More than”means greater than (>)

“Less than”means less than (<)

Page 13: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

5. Word Problems

Chris has $200 in his bank account. He makes $10 an hour at his job. He wants to save at least $400 to buy some chickens. What is the minimum number of hours Chris will have to work?

200 + 10h > 400 – 200 – 200

10h > 200 10 10

h > 20 hours

Has already!

Wants more than

this amount!

Mo’ Money, mo’ money, mo’ money!

Page 14: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

5. Word Problems

Tom calls a cab which charges $2.50 plus

$0.50 a mile. If Tom has no more than $20.00 in his pocket, how far can he go?

$2.50 + $0.50m < $20 -2.50 -2.50 0.50m < 17.50 0.50 0.50

m < 35 miles

That’s a lie. I got big bank!

Page 15: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Probability

Page 16: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Probability

Event – This is the selected outcome. Ex. If event A is the probability of rolling a 5 or higher, the probability is 2/7, so P(A) = 2/7.

Complement – This is the probability of everything other than the event.

Ex. In the example above, the complement is rolling 4 or lower, so the complement of event A is 5/7, or

P(A) = 5/7.Probability of “A Bar”

Page 17: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Coin Toss

If you toss a coin twice, what are the possible outcomes?

HH, TT, HT, TH What is the probability of two heads?

HH, TT, HT, TH =

What is the probability of at least one head?

HH, TT, HT, TH =

It’s complement would be 3/4!

It’s complement would be 1/4!

1/4

3/4

Page 18: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Two Independent Events

To find the probability of two independent events occurring together, multiply their probabilities!

Ex. Find the probability of tossing a coin twice and having heads occur twice.

12

12

. 14

=Probability of

Toss #1 coming up

heads.Probability of Toss #2

coming up heads.

Probability of two heads!

Page 19: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Independent Events

Ex. A coin is tossed and a card is drawn from a standard deck.a. What is the probability of tossing heads and drawing an ace?

b. What is the probability of tossing tails and drawing a face card?

12

113

. 126

=

12

313

. 326

=

Page 20: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Working with Factorials

4! + 3! =

3! – 2! =

4! 2! =

=

(4•3•2•1) + (3•2•1) = 30

(3•2•1) – (2•1) = 4

(4•3•2•1) (2•1) = 48

6!4!

(6•5•4•3•2•1) (4•3•2•1)

= 30

Page 21: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Combinations (Formula)

(Order doesn’t matter! AB is the same as BA)

nCr =

Where:n = number of things you can choose

fromr = number you are choosing

n!r! (n – r)!

Page 22: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Combinations (Formula)

There are 6 pairs of shoes in the store. Your mother says you can buy any 2 pairs. How many combination of shoes can you choose?

So n = 6 and r = 2

6C2 = = 6!2! (6 – 2)!

6•5•4•3•2•12•1(4•3•2•1)

=30 2 = 15 combinations!

Page 23: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Permutations (Formula)

(Order does matter! AB is different from BA)

nPr =

Where:n = number of things you can choose

fromr = number you are choosing

n! (n – r)!

Page 24: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Permutations (Formula)

In a 7 horse race, how many different ways can 1st, 2nd, and 3rd place be awarded?

So n = 7 and r = 3

7P3 = = 7!(7 – 3)!

7•6•5•4•3•2•1 (4•3•2•1)

= 210 permutations!

Page 25: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Fundamental Counting Principle

You have a choice of 3 meats, 4 cheeses, and 2 breads. How many different types of sandwiches could you make?

Multiply the choices!

3 • 4 • 2 = 24 different sandwiches

Page 26: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area & Perimeter: Polygons and Circles

Page 27: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Key Vocabulary Perimeter – The distance around an

polygon.

Area – The amount of space inside a two dimensional shape.

Page 28: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perimeter Estimate or calculate the length of

a line segment based on other lengths given on a geometric figure.

x

17 in

8 in

8 + x = 17 - 8 - 8 x = 9 in Easy!

Page 29: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perimeter Compute the perimeter of polygons

when all side lengths are given

8 in

7 in

7 in

7 in

7 in8 in

Add all the sides:8 + 7 + 7 + 8 + 7 + 7 =

44 inEven I can do this!

Page 30: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area Compute the area of rectangles

when whole number dimensions are given.

25 in

6 in

Area of Rectangle = Length • Width

A = L • WA = 25 • 6 A = 150 in2

Page 31: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area Compute the area and perimeter of

triangles and rectangles in simple problems.

Area of Triangles = 1/2 • Base • Height

Base Base

Heig

ht

Heig

ht

Page 32: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area Kyla mows lawns for $1.20 per

square feet. How much did she charge to cut the lawn below?

23 ft

11

ft

A = L • WA = 23 • 11A = 253 ft 2

Price = 253•$1.20Price = $303.60

Page 33: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Distance of Sides Find the missing value, x.

x

28 inx + 2x + x =

284x = 28

4 4x = 7 inches

I get it. Add up the bottom sides to equal the top!

x

2x

Page 34: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perimeter Find the perimeter.

17 in

Find x:x + 5 =

17 – 5 – 5

x = 12 If x = 12, then x – 1

is 11!

x – 1

x + 5

Use x to find perimeter:17 + 17 + 11 + 11

Perimeter = 56 inches

Page 35: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

I can find the area by cutting it!

Area Find the area.

5 ft

15 ftArea of ‘A’A = 10 • 7A = 70 ft2

Area of ‘B’A = 5 • 8A = 40 ft2

A = 70 + 40A = 110 ft2

8 ft

7 ft

5 ft1

0 f

t

AB

Page 36: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area of Trapezoids Definition – Quadrilaterals with at least one

pair of parallel sides.

Area of a Trapezoid =

(Find the average of the bases and multiply by the height!)

b1

b2

h

b1 & b2 are the top and

bottom bases.

h is the height.

(b1+ b2) 2

• h

Page 37: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

= • 9

Area of Trapezoids Find the area of the trapezoid below.

12

9

(b1+ b2) 2

A = • h

16

(12 + 16) 2

= 126 ft2

Page 38: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Parts of a Circle Circumference – The distance around a

circle. (Perimeter of a circle.) Radius – The distance from the center of a

circle to any point on its circumference. Diameter – The distance from one side of a

circle, passing through the center, to the other side of the circle.

.

Radius

Circumference Diameter

Page 39: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Circumference of a Circle The diameter of a circle is equal to

twice the radius, or

d = 2r Circumference of a circle is equal to

the diameter multiplied by pi, or

C = 2πr or

C = πd

Page 40: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Circumference of a Circle

C = 2πrC = 2 π 5C = 10π

or 31.42

5

Page 41: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Circumference of a Circle

C = dπC = 25.5π

or 80.1125.5

Diameter is 25.5!

Page 42: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Area of a Circle The area of a circle is equal to:

A = πr2.

6A = π62 A = 36πA = 113.10

Page 43: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Word Problems Pizza World offers two types of pizzas:

rectangles and circles. If each pizza cost $12.50, which is the better buy?

12 in

12

in

12 in diameter

A = 12 • 12= 144 in2

A = π • 62 = 113.10 in2

This is the

better buy!

Page 44: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Angles & Lines

Page 45: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Parallel Lines

You can identify parallel lines by their equations!

y = 3x + 7y = 3x – 9

These two lines are parallel. Their slopes are the same!

(Notice that they have different y-intercepts!)

Page 46: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perpendicular Lines Lines that intersect at right angles (900)

are perpendicular. Perpendicular lines have slopes that are

negative reciprocals. The product of their slopes = -1.

These two lines are perpendicular. They

intersect at a right angle.

Page 47: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perpendicular Lines

Negative reciprocals

1. What is the reciprocal of ?

2. What is the reciprocal of 3?

23

32

So the negative reciprocal is – ! 32

13

So the negative reciprocal is – ! 13

Page 48: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Perpendicular Lines

These equations are perpendicular:

y = 2x + 8

y = - x – 5

y = - x – 7

y = x + 5

12

45

32

54

Page 49: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

YX

Complementary Angles

Complementary Angles – Two angles that add up to 90°

Angles X and Angle Y are

complementary and add up to 90

°.

Page 50: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try this…

Find the missing angle.

36°

90 – 36 = 54°

Page 51: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Supplementary Angles

Supplementary angles - Two angles that add up to 180°

Angles X and Angle Y are

supplementary and add up to

180 °.YX

Page 52: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try this…

Solve for x.

x 138°

180 – 138 = 42°

Page 53: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Vertical Angles

Vertical Angles - A pair of opposite angles formed by the intersection of two lines. Vertical angles are always equal.

A

B

Angle A and Angle B are vertical angles. They are equal!

Page 54: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Corresponding Angles

nLine n is a transversal.

A

B

Corresponding angles – Two congruent angles that lie on the same side of the transversal.

A and B are corresponding angles.

Page 55: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Alternate Interior Angles

n

Line n is a transversal.

A

C

B

D

A = CÐB = D

They are alternate interior angles.

(Interior = Inside!)

Page 56: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Alternate Exterior Angles

n

Line n is a transversal.

A

C

B

D

A = CÐB = D

They are alternate exterior angles.

(Exterior = Outside!)

Page 57: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try this…

Find the missing angles.

42° A

A = 138°B = 138°C = 42°

B C

FED

G

D = 42°E = 138°F = 138°G = 42°

Page 58: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try this…

d ° 65 °

70 ° 70 °

Hint: The 3 angles in a triangle sum to 180°.

Find the missing angles.

70 + 70 + b = 180140 + b = 180b = 40°

40 °

40 + 65 + d = 180105 + d = 180d = 75°

Page 59: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

The Pythagorean Theorem

In a RIGHT Triangle, if sides “a” and “b” are the legs and side “c” is the hypotenuse, then

a2 + b2 = c2

a

b

c

Page 60: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Finding the Hypotenuse

Find the length of the hypotenuse.

12

16

C

a2 + b2 = c2

122 + 162 = c2

144 + 256 = c2

400 = c2

√400 = c20 = c

Page 61: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Finding the Leg

Find the length of the missing leg.

29

20

a

a2 + b2 = c2

a2 + 212 = 292

a2 + 400 = 841a2 = 441a = √441

a = 21

Page 62: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Exponents

Page 63: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Laws of Exponents

Zero Exponent Property – Any number raised to the zero power is 1.

x0 = 1 30 = 1 120 = 1Negative Exponent Property – Any number raised to a negative exponent is the reciprocal of the number.

x-4 = 5-1 = 5x-3y5 = 1x4

5y5

x3

15

Page 64: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Laws of Exponents

Product of Powers – When multiplying numbers with the same bases, ADD the exponents.

Quotient of Powers – When dividing numbers with the same bases, SUBTRACT the exponents.

x2•x8 = x10 3x4•x-2 = 3x2

x6

x2= x4 10x4y3

2x7y5y2

x3=

Page 65: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Laws of Exponents

(95)3 = 915

Power of a Power - When you have an exponent raised to an exponent, multiply the exponents!

(x4y5)2 = x8y10

Power of a Product - Distribute the exponent on the outside of the parentheses to all of the terms inside of the parentheses.

(xy)3 = x3y3 (2bc)2 = 4b2c2

Page 66: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Laws of Exponents

Power of a Quotient – Distribute the exponent on the outside of the parentheses to the numerator and the denominator of the fraction.

2x

=( )3 23

x3( )=8x3

3x2y

=( )4 34

x8y4( )= 81x8y4

Page 67: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Radicals

Page 68: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Simplifying Square Roots

√8 = √4 ∙ √2 = 2√2

8 is not a perfect

square, so we will

simplify it!

8 is made up of 4 ∙ 2. Look! 4 is a

perfect square!

√4 = 2We can’t

simplify √2, so we leave him alone.

Page 69: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Simplifying Square Roots

√75 =

√25 ∙ √3 =

5√3

Find factors of 75!

25 and 3 are factors of 75!

Find the square root of 25!

√72 =√36 ∙ √2 = 6√2

Page 70: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try these…

√27

√32

√20

√75

= √9 ∙ √3

= √16 ∙ √2

= √4 ∙ √5

= √25 ∙ √3

= 3√3

= 4√2

= 2√5

= 5√3

Page 71: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Combining Square Roots

To combine square roots, combine the coefficients of like square roots.

4√3 + 5 √3= 9√3

7√5 – 4√2 =

Combine the coefficients, keep the radical!

You cannot combine unlike radicals!

7√5 – 4√2

Page 72: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Simplify and Combine√20 + √5 =√4 ∙ √5 + √5

=

√12 + √27 =

2√3

+√3 ∙ √43√3 =

√9 ∙ √3 =

5√3

2√5 + √5 =

3√5

+

Page 73: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Multiplying RadicalsWhen multiplying radicals, you can multiply the two numbers and put the answer under one radical. Simplify!

√3 ∙ √2 =

√6

√3 ∙ √3 =

√9 = 3

√3 ∙ √6 =

√18 = √9 ∙ √2

= 3√2

Page 74: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Multiplying Radicals

2√5 ∙ 3√5 1. Multiply the coefficients.

2 ∙ 3 = 62. Multiply the radicals.

√5 ∙ √5 = √253. Solve.

6√25 = 6 ∙ 5 = 30

Page 75: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Try This…

3√7 ∙ 2√5 =6√35

2√3 ∙ 5√3 =10√9 =10 ∙ 3

4√2 ∙ 3√8 =12√16= 12 ∙ 4

= 48

2√5 ∙ 3√2 =6√10

= 30

Page 76: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Polynomials

Page 77: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Add: (x2 + 3x + 1) + (4x2 +5)

Step 1: Identify like terms:

Step 2: Add the coefficients of like terms, do not change the powers of the variables:

Adding PolynomialsAdding Polynomials

(x2 + 3x + 1) + (4x2 +5)

Notice: ‘3x’ doesn’t have a like term.

(x2 + 4x2) + 3x + (1 + 5)

5x2 + 3x + 6

Page 78: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Subtract: (3x2 + 2x + 7) – (x2 + x + 4)

Subtracting PolynomialsSubtracting Polynomials

Step 1: Change subtraction to addition.

Step 2: Add like terms.

(3x2 + 2x + 7) + (- x2 + - x + - 4)

(3x2 + 2x + 7)+ (- x2 + - x + - 4)

2x2 + x + 3

Change signs of all terms after subtraction sign.

Page 79: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Multiplying Polynomials

Distributive Property – Review!!!

3(x + 5) = 3x + 15

We will need that same concept to multiply polynomials.

3(2x2 + 4x + 3)= 6x2

We will distribute the outside term to everything on the inside.

+ 12x

+ 9

Page 80: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Multiplying Polynomials

4(3x2 – 2x + 1) = 12x2 – 8x + 4

2x(2x2 + x + 5) = 4x3 + 2x2 + 10x

4 ∙ 3x2 4 ∙ –2x

4 ∙ 1

2x ∙ 2x2

2x ∙ x

2x ∙ 5

12x2 –8x +

4

4x3 +2x2

+10x

Page 81: Solve for x. 30° 2x + 10 2x + 10 = 60 – 10 – 10 2x = 50 2 2 x = 25

Multiplying Polynomials: Expansion Boxes VS. FOIL

(2x + 1)(4x – 3)

Combine:8x2 – 6x + 4x – 3

8x2 – 2x – 3

Mult. 4x  – 3

2x     

 + 1    

8x2 – 6x

+ 4x

– 3

First: +2x ∙ +4x = +8x2

Outer: +2x ∙ -3 = -6xInner: +1 ∙ +4x = +4xLast: +1 ∙ -3 = -3Combine:

8x2 – 6x + 4x – 3 8x2 – 2x – 3

Same Answer!Same Answer!