Exponents and Radicals

Exponents and Radicals

Integer Exponents: An integer exponent, such as the 2 in 42, tell us how many times to multiply the base number by itself. 42 tells us to multiply 4 by itself twice, that is 4 * 4 = 16, while 43 tells us to multiply it three times, that is 4 * 4 * 4 = 64. In this case 4 is the base, and 2 and 3 are exponents.

Working with Exponents:

1) a0 = 1 (when a 0)

2) a-n =

3) am an =am+n x4 x3 = x4+3 =x7

4) and

5) (am)n = amn (x3)5 = x15

6) (ab)n = anbn (3x)2 = 9x2

7)

8) x m y –n =

Working with nth Roots:

• Roots are the opposite of powers.
• Roots are the numbers, which, when multiplied, give the power.
• 2 to the fifth power is 32, the fifth root of 32 is 2.
• 7 to the third power is 343, the third (or cube) root of 343 is 7.
• If a is any positive integer, then the principal nth root of a is defined as follows:

means bn = a. If n is even, we must have a greater than zero, and b greater than zero.

Example:

Try:

Properties of nth roots:

1.

2.

3.

4. if n is odd,

5. if n is even

Simplifying Expressions Involving nth Roots:

Examples:

a.)

b.)

c.) We cannot add these are they are, because they are different radicals. Let’s simplify them and see what we get. so

Now we have two expressions with the same number under the radical, so we can add them.

d.)If b>0, then

Definition of Rational Exponents:

• For any rational exponent m/n in lowest terms, where m and n are integers and n>0, then . If n is even, then a must be equal to or greater than 0.

• By the same definition, . All the rules of exponents hold for rational exponents.

Examples:

or alternately

Practice Problems

1. Basic exponential definitions using integers:

54 = -73 =210 =

50 =-60 =2,7560 =

5 -4 =-9-2 =6-3 =

Laws of Exponents: (Be sure to express all answers as positive exponents.

x4x7 = (x4)7 = x –4 x 7 = (x-4)7 =

= =

(2a3b5)(3ab2)2 = (2s3t –1)(1/4 x5)(16t-2)=

==

Roots:

Simplify the expressions: Assume the letters denote any real numbers.

==

==

==

Rational Exponents:

Simplify: 1251/3==x -5/6=

Simplify the expression and eliminate any negative exponent(s). Assume that all letters denote positive numbers.

(2x3/2)(4x) –1/2=(-2a3/4)(5a3/2)=

(a2/5) –3/4=(x –4y 3 z 2/5) –3/5 =

==

Algebraic Expressions

Definitions:

Variable is a letter that can represent any number from a given set of numbers.

Algebraic expression is some combination of variables, such as 2x2 – 3x + 4

Monomial is an expression of the form axk, it has only one term.

Binomial is the sum of two monomials, such as 3x + 9

Trinomial is the sum of three monomials, such as 7x2 + 4x – 3

Polynomial is a general term for anything above a monomial.

Degree of a polynomial is the highest exponent in the expression. 7x2+ 4x is degree 2, or second degree, 7x3 +4x2 is of degree 3.

Combining Algebraic Expressions:

The basic rule is that you can add or subtract like terms, that is, terms with the same variable in the same power.

Example: 5x7 + 3x7 = 10x7 5x7 + 3x6 = 5x7 + 3x6 Since the powers are different, they cannot be combined by addition or subtraction.

Example: (x3 – 6x2 + 2x + 4) + (x3 + 5x2 – 7x) = (x3 + x3) + (-6x2 + 5x2) + (2x – 7x) + (4) =

2x3 –x2 –5x + 4

Example: (x3 + 6x2 + 2x + 4) – (x3 + 5x2 – 7x) = x3 + 6x2 + 2x + 4 - x3 - 5x2 + 7x

= (x3 – x3) + (6x2 –5x2) + (2x + 7x) + 4

= -11x2 + 9x + 4

Multiplication of Algebraic Expressions:

To multiply by a monomial, use the distributive property.

Example 3(4x + 7) = 12x + 21 3x( 2x – 6) = 6x2 – 18x

To multiplytwo binomials, use the FOIL method: First, Outside, Inside, Last

Example: (2x + 1)(3x – 5)

First --2x * 3x = 6x2

Outside -- 2x * -5 = -10x

Inside -- 1 * 3x = 3x

Last -- +1 * -5 = -5

Then put it together: 6x2 –10x +3x –5 = 6x2 –7x –5

Example: =

= =

Special Product Formulas:

If you memorize these, you will find that you will be using them often.

1. (A + B)(A – B) = A2 – B2 Sum and difference of same terms

2. (A + B)2 = A2 +2AB + B2Square of a sum

3. (A - B)2 = A2 - 2AB + B2Square of a difference

4. (A + B)3 = A3 + 3A2B + 3 AB2 + B3Cube of a sum

5. (A - B)3 = A3 - 3A2B + 3 AB2 - B3 Cube of a difference

Examples:

(6x + 3)(6x – 3) If A = 6x and B = -3 we have (6x)2 – (3)2 or 36x2 - 9

(3x + 5)2 If A = 3x and B = 5, we have (3x)2 + 2(3x)(5) + (5)2 or 9x2 + 30x + 25

(x – 2)2 If A = x and B = -2, we have (x)2 –2(x)(-2) + (-2)2 or x2 +4x +4

FACTORING:

Factoring is the opposite of multiplication. The factors are the numbers, that when multiply, give us the product. The factors of 21 are 7 and 3. The factors of 45 are 9 and 5. The factors of 12 are 2 and 6 or 3 and 4. When we have several sets of factors, we often find the prime factors, so that the factors of 12 are 2, 2, and 3, all of which are prime numbers.

Algebraic factoring follows the same pattern.

Common factors:

The easiest way to factor an algebraic expression is to find something common to all the terms, and divide by that factor. In fact, that is what we should look for first.

Factor: 3x2 – 6x 3 is a factor of both numbers, and x is a factor of both variables. So we will “take out” 3x. This gives us 3x(x – 2)

Factor: 8x4y2 + 6x3y3 – 2 xy4 Let’s look, we have numbers, xs and ys. 2 is the highest number that goes into all the numbers. The lowest power of x is 1, and the lowest power of y is 2

2xy2(4x3 + 3x2 – y2)

Factor: (2x + 4)(x – 3) – 5(x – 3): Here we have two terms separated by a minus sign. The first term has two factors, and the second term has two factors. Both terms have a common factor of (x – 3) so we will “take out” (x – 3). This gives us (x - 3)(2x + 4 –5) or (x – 3)(2x – 1)

Factoring trinomials with no common factors:

Factor: x2 + 7x + 12 Most of the time, trinomials are products of two binomials, as we did above. We used the FOIL method. Now we will try to do it backwards. We need to find two x terms that when multiplied gives us x2. That one is easy, x and x. We also need to find two numbers that multiply to 12. We could have +1 and +12, -1 and –12, +2 and +6, -2 and –6, +3 and +4, or -3 and -4. But the middle term is 7x. Since the coefficient of x2 is one, we do not need to worry about it here. We need two numbers that multiply to 12 and add to 7. That would be +3 and +4. So the factors are (x + 3)(x + 4). You can check this by multiplying them together.

Factor: 6x2 + 7x - 5: This one isn’t quite as easy, since 6x2 has a number of factors, such as 6x and x, 2x and 3x, as well as –6x and –x, and –2x and –3x. -5 could be +1 and –5 or – 1 and +5. We need to find the correct combination of these to give us +7x. This is why this method is called trial and error. After some guessing and combining the correct answer is (3x + 5)(2x – 1). If we were to FOIL this, the two middle terms would be –3x and 10x which gives us the 7x we were looking for.

Special Factoring Formulas:

A2 - B2 = (A – B)(A + B)Difference of squares

A2 + 2AB + B2 = (A + B)2Perfect Square

A2 - 2AB + B2 = (A - B)2Perfect Square

A3 – B3 = (A – B)(A2 + AB + B2)Difference of Cubes

A3 + B3 = (A + B)(A2 - AB + B2)Sum of Cubes

These you will need to memorize, and try to be able to recognize.

Examples:

4x2 – 25 4, x2, and 25 are all perfect squares, so we have the difference between two squares.

(2x)2 – 52 = (2x – 5)(2x + 5)

(x + y)2 – z2 = (x + y – z)(x + y + z)

27x3 – 1 27, x3, and 1 are all perfect cubes, so we have the difference of cubes.

(3x)3 – (1)3 = (3x – 1)[(3x)2 + (3x)(1) + 12] = (3x – 1)(9x2 + 9x + 1)

x6 + 8 x6 = (x2)3 8 = (2)3 so this is the sum of cubes

(x2 + 2)(x4 – 2x + 4)

x2 + 6x + 9 x2 and 9 are both perfect squares, and 6 is twice the square root of 9

(x + 3)2

4x2 – 4xy + y2 4, x2, and y2 are all perfect squares, and 4xy is twice the product of their square roots. So (2x – y)(2x + y)

Factoring by Grouping: Sometimes there is not a common factor for all terms, and it is not a trinomial, nor a special form. Then we need to check to see if by grouping terms in a certain way, we can find a common factor, usually a binomial factor.

Example: Factor x3 + x2 + 4x + 4 The 4’s sort of look like they belong together, and the first two terms seem similar. So let’s group them. (x3 + x2) + (4x + 4). Each of these has a common factor, giving x2(x + 1) +4(x + 1). Since the (x + 1) is not common to both groups, we factor it out giving us (x + 1)(x2 + 4).

Example: Factor x3 – 2x2 – 3x + 6 = (x3 – 2x2) – (3x – 6) -- Watch signs:

= x2(x – 2) –3(x – 2) = (x – 2) (x2 – 3)

Factoring Completely: Sometimes we need to check our factors to be sure they can’t be factored more. This is similar to saying that 12 = 3 * 4, but since 4 = 2 * 2, the prime factors of 12 are 3 * 2 * 2.

Example: 2x4 –8x2 First we factor out the common factor, 2x2. This gives us 2x2(x2 – 4). The second factor is the difference of two squares, so the final factorization is 2x2(x – 2)(x + 2)

Example: x5y2 – xy6 = xy2(x4 – y4) = xy2(x2 + y2)(x2 – y2) = xy2(x2 + y2)(x + y)(x – y)

Factoring Expressions with Fractional Exponents: Since fractional exponents follow the same rules as integral exponents, the factoring is really the same, although it looks more complicated.

Example: Factor 3x3/2 – 9x1/2 + 6x –1/2 There are three terms with fractional exponents of x. The smallest of these is –1/2. So we will factor this first (along with the common factor 3) to get 3x –1/2 (x2 –3x + 2) -- OK let’s check this out. x3/2 divided by x –1/2 = x 3/2 –(-1/2) = x4/2 = x2

So far we have 3x –1/2 (x2 –3x + 2). The second term factors further to 3x –1/2(x – 2)(x – 1).

Example: Factor: (2 + x) –2/3 x + ( 2 + x)1/3

(2 + x) is a common base, so factor out the smallest exponent (-2/3)

(2 + x) –2/3(x + (2 + x)1) = (2 + x) –2/3(2 + 2x) = (2 + x) –2/3(2)(1 + x) or in better form

2((2 + x) –2/3(1 + x)

Summary:

1. Always look for common factors first.
2. If there are two terms, are they the difference of two squares, or the sum or difference of two cubes?
3. If there are three terms, can they be factored using FOIL? or are they prefect squares?
4. If there are four terms, can they be grouped?

Practice Problems

Perform the indicated operations and simplify:

1.(5 – 3x) + (2x – 8)

2.(3x2 + x + 1) – ( 2x2 – 3x – 5)

3.8(2x + 5) –7(x – 9)

4.4(x2 – 3x + 5) – 3(x2 – 2x + 1)

5.5(3t – 4) – (t2 + 2) – 2t(t – 3)

Multiplication of binomials

6.(4x – 1)(3x + 7)

7.(3x + 4)(3x + 4)

8.

9.

Special Products

10.(x + 2)(x – 2)

11.(2x + 6)2

12. (5x – 4)2

13. (x + 6)3

14. (2x – 3)3

Common factors: Factor:

15.2x4 + 4x3 – 14x2

16.(z + 2)2 +4(z + 2)

17.–7x4y2 + 14xy3 + 21 xy4

Factor these trinomials:

18.x2 – 6x + 5

19.6y2 + 11y – 21

20.a2 +10a + 16

21.6x2 + x - 12

Special Factors: Factor the following:

1. x2 – 36
2. 49 – 4y2
3. 27a3 – b3
4. 8x3 + 125
5. (x + 3)2 – 4
6. 16x2 – 24z + 9
7. x2 + 4x + 4

Factor by grouping:

1. 3x3 –x2 + 6x – 2
2. –9x3 – 3x2 + 3x + 1

Factor completely:

1. 5ab – 8abc
2. y2 – 8y + 15
3. x6 - 64
4. y4(y + 2)3 + y5(y + 2)4
5. x –3/2 + 2x –1/2 + x1/2
6. x4 + 3x2 - 4

Inequalities:

An inequality uses one of the inequality signs, (greater than), (less than), ≥ (greater than or equal to)

or ≤ (less than or equal to). Inequalities look a lot like equations, except that one of the inequality signs replaces the = sign.

Inequalities do not have one solution, but a range of solutions. x < 3 includes all the real numbers less than (but not including) 3, while x ≤ includes all the real number less than 3, including 3.

There are several ways of showing the solution to an inequality:

Interval notation: (-∞, 3) and (-∞, 3]

Set notation: {x | x < 3} and {x | x ≤ 3}

Graph or number line: The open circle says that the number is not included, a closed circle would indicate that the number is included.

-1012345

Solving Inequalities: The rules for solving inequalities are the same as the rules for solving equations, that is, you can do the same thing to both sides, such as add the same number, subtract the same number, etc.

However, there is an IMPORTANT exception. Multiplying or dividing by a negative number reverses thedirection of the inequality. That is, if the inequality was < it becomes >.

Example -x > -6; dividing by -1 gives x < 6

Example -3x < 12; dividing by -3 gives x > -4

Linear Inequalities: Just like a linear equation, a linear inequality must have no higher power than 1.

Example: Solve the inequality 3x < 9x + 4 and sketch the solution set.

3x – 9x < 9x – 9x + 4; -6x < 4 dividing by -6 gives x > -4/6 or x > -2/3.

The solution consists of all numbers greater than -2/3. In interval notation (-2/3, ∞). The graph is:

-2/30

Example: Solve the inequality 4 ≤ 3x – 2 < 13. Here we have two inequality signs, so it is really two inequalities abbreviated to one. However the rules remain the same. First add 2 to all sides of the inequality:

4 + 2 ≤ 3x – 2 + 2 < 13 + 2 This gives 6 ≤ 3x < 15. Next divide all sides by positive 3 to get

2 ≤ x < 5. So the solution set is [2, 5) and is graphed:

01234567

Non Linear Inequalities: To solve a non linear inequality:

1. Move all terms to one side

2. Factor the non zero side

3. Find the intervals by determining the values for which each factor equals zero.

4. Make a number line, and mark the values found in # 3

5. Test each section of the number line to determine which solve the inequalities.

Quadratic Inequalities: Follow the steps above.

Example: Solve the inequality x2 – 5x + 6 ≤ 0.

Step 1: This is already in the correct form

Step 2: (x – 3) (x – 2) ≤ 0

Step 3: 3 and 2 would make a factor equal 0.

Section ABC

Step 4:

01234

Step 5: Take a value out of each of the sections to see if it makes the inequality true or false;

Section A Choose 1 x2 – 5x + 6 ≤ 0. (1)2 -5(1) + 6 ≤ 0? 1 -5 + 6 = 2 which is not less than or equal to 0, so section A returns a FALSE.

Section B Choose 2.5 (2.5)2 – 5(2.5) + 6 ≤ 0? 6.25 – 12.5 + 6 = -0.25 which is less than 0, so section B returns a TRUE

Section C Choose 4 (4)2 – 5(4) + 6 ≤ 0? 16 – 20 + 6 = 2 which is not less than 0, so section C returns a FALSE.

The only section that returned a TRUE was section B, so this is the solution to the inequality. [2, 3]

An Inequality Involving a Quotient:

Example: Solve:

Step 1: Move to left side This needs to be simplified:

So we have

Step2: Neither the numerator nor the denominator can be factored

Step 3: 0 and 1 are the values that would make the numerator or the denominator equal to zero

Step 4:

Section A BC

-2-10123

From Section A -2 ≥1? No. Section A is FALSE

From Section B ½ 3≥ 1? Yes Section B is TRUE

From Section C 2 -3 ≥1? No Section C is FALSE

So the solution is [0, 1) -- Notice that the 1 is not included, because 1 would make the denominator equal to 0, and division by zero is never allowed.

Inequalities with Three Factors: These follow all of the steps we have been following, but we will have more divisions to the number line, and more sections to check.

Example: x <

Step 1: Move all to one side: x - <0 Simplify:

Step 2: Factor

Step 3: -1, 2, and 1 are the values that would make the numerator factors or the denominator equal to zero.

Step 4:

Section ASection B Section CSection D

-2-1012 3

Step 5:

From Section A -2 -2 is less than -2/3 so Section A is TRUE

From Section B 0 ? 0<-1? No Section B is FALSE

From Section C 1½ (or 1.5) 1.5 < 4? Yes Section C is TRUE

From Section D 3 3<1? No Section D is FALSE

The solution is (-∞,-1) U (1, 2)

Absolute Value Inequalities:

The properties for absolute value inequalities are similar to those for absolute value equations.

Example: Solve |x - 5| < 2.

We will get two inequalities for this. They are:

+(x – 5) < 2 and –(x – 5) < 2.

x – 5 < 2; x < 7 -x + 5 < 2; -x < -3; x > 3

So the solution is x < 7 and x > 3, which gives us (3, 7) as the solution (x is between 3 and 7)

Example: Solve: |3x + 2| ≥ 4

The two inequalities are:

+(3x + 2) ≥ 4 and -(3x + 2) ≥ 4

3x + 2 ≥ 4 -3x – 2 ≥ 4

3x ≥ 2 -3x ≥ 6

x ≥ 2/3 x ≤ -2

So the solution is x ≥ 2/3 and x ≤ -2. In interval notation this is (-∞, -2] U [2/3, ∞)

Modeling with Inequalities:

Example: A carnival has two plans for tickets:

Plan A: $5 entrance fee and 25 cents each ride

Plan B: $2 entrance fee and 50 cents each ride.

How many rides would you have to take for plan A to be less expensive than plan B?

Let x = number of rides;

Plan A < Plan B

5 + 0.25x < 2 + 0.50x; 3 < 0.25x; 12 < x which can be rewritten as x > 12.

If you take more than 12 rides, plan A is less expensive.

Example: A group of students decide to attend a concert. The cost of chartering a bus to take them to the concert is $450, which is to be shared equally among the students. The concert promoters offer discounts to groups arriving by bus. Tickets normally cost $50 each but are reduced by 10 cents per ticket for each person in the group (up to the maximum capacity of the bus). How many students must be in the grou for the total cost per student to be less than $54?

Let x = number of persons in group.

Bus cost + ticket cost < 54

+ (50 – 0.10x) < 54 Remember with inequalities we cannot just multiply both sides by the denominator, (since we don’t know the sign of x, so we don’t know if we have to change the direction of the inequality). So we have to first gather everything on one side.

+ 50 – 0.10x – 54 < 0; - 0.10x - 4 < 0; Now we multiply everything by x

450 – 0.10x2 – 4x < 0. Multiplying by 10 to get rid of the decimal gives

4500 – x2 – 40x < 0 Rearranging

-x2 – 40x + 4500 < 0 Factoring gives

(-x + 50)(x + 90) < 0 so the values that make a factor 0 are 50 and -90.

There is no need to check below -90, since we can’t have fewer than 0 people.

Less than 50, say 40 gives (-40+50)(40+90) or 10(130) = 1300 This is more than 0, and we want less than 0, so that region is FALSE

More than 50, say 60, gives (-60+50)(60+90) or (-10)(150) = -1500 This is less than 0, so that is the region that is TRUE.

If there are more than 50 people in the group, the total cost per person will be less than $54.

Practice Problems:

Linear Inequalities:

Solve each of the following inequalities, then give the solution set in interval notation, and graph it.

1.) 3x + 11 < 5

2.) 6 – x ≥ 2x + 9

3.) 5 ≤ 3x – 4 ≤ 14

4.) -3 ≤ 3x + 7 < ½

Quadratic Inequalities: Solve. Give the final answer in interval notation:

5.) x2 – x ≥ 20

6.) x2 +2x < 3

7.) 5x2 + 3x ≥ 3x2 + 2

Fractional Inequalities:

8.)

9.)

10.)

Inequalities with three factors:

11.)

Absolute Value Inequalities:

12.) |3x| < 15

13.) |x + 1| ≥ 1

14.) 3 - |2x + 4| ≤ 1

Modeling with Inequalities:

15.) A telephone company offers two long distance plans.

Plan A: $25 per month and 5 cents per minute

Plan B: $5 per month and 12 cents per minute.

For how many minutes of long-distance calls would plan B be financially advantageous?

16.) A riverboat theater offers bus tours to groups on the following basis: Hiring the bus costs the group $300, to be shared equally by the group members. Theater tickets, normally $40 each, are discounted by 25 cents times the number of people in the group. How many members must be in the group so that the cost of the theater tour (bus fare plus theater ticket) is less than 50 dollars a person?

17..) A determined gardener has 120 feet of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area enclosed to be at least 800 square feet. What range of values is possible for the length of her garden?

GRAPHING ANSWER SETS

AND

USING INTERVAL AND SET-BUILDER NOTATION TO REPRESENT ANSWER SETS

Graphing on Real Number Line / Representing Same Set in Interval Notation / Representing Same Set in Set-Builder Notation
----|----|----|----|----|----|----|----|----|----|---- / [ 3 , 5 ) / {x | 3 x < 5 }
----|----|----|----|----|----|----|----|----|----|---- / ( , 4 ] / { x | x 4 }
----|----|----|----|----|----|----|----|----|----|---- / ( 20 , ) / { x | x > 20 }
----|----|----|----|----|----|----|----|----|----|---- / ( - , 3 ] [ 7 , 10 ) / { x | x 3 or 7 x < 10 }
----|----|----|----|----|----|----|----|----|----|---- / (-, -2 ) ( -2 , 1 ) ( 1 , 3 ) ( 3 ,) / { x | x -2, 1, 3 }

UNION VS. INTERSECTION

Union “”

“all assets combined”

“every item in the first set along with every item in the second set”

Intersection “”

“what is common to both sets”

“only if it shows up in the first set and it also shows up in the second set”

Example:

Let sets A and B be as follows.

A = {2, 4, 6, 8, 10}

B = {1, 2, 3, 4, 5, 6}

Then we would have the following.

AB = {1, 2, 3, 4, 5, 6, 8, 10}

AB = {2, 4, 6}