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Mathway Packet Review Exercises to prepare for Math 1101/0099 Intensive Young Harris College Math Department Spring 2014 Adding Integers (signed numbers) For integers with the same sign: • The sum of two positive integers is positive • The sum of two negative integers is negative For integers with different signs, subtract their absolute values. The sum is: • Positive if the positive integer has the greater absolute value • Negative is the negative integer has the greater absolute value Example 1: Find 4 + (-­‐6) •
•
One number is positive and one is negative so we subtract their absolute values. |-­‐6| -­‐ |4| = 6-­‐4 = 2 The answer is negative because the absolute value of -­‐6 is greater than the absolute value of 4. So the answer is -­‐2. Subtracting Integers (signed numbers) •
•
To subtract an integer, add its opposite. Since you have now changed it to addition, follow the rules above for adding integers. Example 2: Find 6-­‐9 •
•
6-­‐9 = 6+ (-­‐9) = -­‐3 Example 3: Find -­‐10 -­‐ ( -­‐12) •
•
-­‐10-­‐(-­‐12) = -­‐10 + 12 =2 Mathway Packet
Adding and Subtracting Signed Numbers
Add or Subtract as indicated.
1) -22 + (-18)
2) -14 - (-7)
3) -2 + 21
4) 14 - 15
5) -3 - (-10)
6) 4 - (-1)
7) 4 + (-12)
8) -21 + 8
9) ⁻10 - 4 - 15
10) 11 - 4 - 19 - 11
11) 16 - ⁻13 - 2 - ⁻19
12) 2 + -10 + 7
13) ⁻4 + 15 + -12
14) 15 + -16 + -7
1
Answer Key
Testname: MATHWAY_1
1) -40
2) -7
3) 19
4) -1
5) 7
6) 5
7) -8
8) -13
9) ⁻29
10) ⁻23
11) 46
12) ⁻1
13) ⁻1
14) ⁻8
2
Simplifying Fractions •
•
•
Factor numerator factor denominator cancel common factors !"
Example: Simplify !" •
!"
!∙!
!∙!
!
= !∙! = !∙! = ! !"
Multiplying Fractions Multiply the numerators and multiply the denominators. Factor the numerators and denominators Cancel common factors Multiply any remaining factors back out ! !
Example: Multiply ! ∙ ! 6 10
6 ∙ 10 2 ∙ 3 ∙ 5 ∙ 2 2 ∙ 3 ∙ 5 ∙ 2 2 ∙ 2 4
∙
= =
=
=
= = 4 5 3
5∙3
5∙3
5∙3
1
1
•
•
•
•
Dividing Fractions To divide by a fraction, multiply by its reciprocal. Another way to say this is change the ÷ to Ÿ and to then ’flip’ the second fraction !"
!
Example: ! ÷ ! 10 2 10 9 10 ∙ 9 2 ∙ 5 ∙ 3 ∙ 3 2 ∙ 5 ∙ 3 ∙ 3 5 ∙ 3 15
÷ =
∙ =
=
=
=
=
= 15 3 9
3 2
3∙2
3∙2
3∙2
1
1
•
•
Mathway Packet
Fractions: simplifying, multiplying, and dividing
Write the fraction in lowest terms.
6
1)
8
2)
30
35
3)
15
47
4)
100
140
Multiply. Write the answer in lowest terms.
1 2
5) ∙
4 3
6)
2 ∙4
5 9
7)
12 ∙ 40 ∙ 15
25 66 32
8)
2 ∙4 ∙5
3 7 6
Divide. Write the answer in lowest terms and as a whole
or mixed number where possible.
3 4
9) ÷
4 5
10)
3 3
÷
8 4
11)
4 ÷8
7 5
1
Answer Key
Testname: MATHWAY_2
1)
3
4
2)
6
7
3)
15
47
4)
5
7
5)
1
6
6)
8
45
7)
3
22
8)
20
63
9)
15
16
10)
1
2
11)
5
14
2
Adding and Subtracting Fractions •
•
Like fractions are fractions that have the same denominator. To add or subtract like fractions, add or subtract the numerators and write the result over the denominator. Simplify if necessary. To add or subtract unlike fractions, rename the fractions with a least common denominator. Then add or subtract as with like fractions. !
!
Example: Subtract ! − ! !
!
Example: Add !" + !" 3 1
3−1 2 1∙2 1∙2 1
− = = =
=
= 4 4
4
4 2∙2 2∙2 2
The least common denominator of 15 and 12 is 60. !
!
!∙!
!
• Rename !" using the LCD: !" = !"∙! = !" !
!
!∙!
!
• Rename !" using the LCD: !" = !"∙! = !" !
!
!!!
!"
Now add the renamed fractions: !" + !" = !" = !" Mathway Packet
Adding and subtracting fractions
Add or subtract as indicated.
1)
2 + 3 + 1
15 15 15
2)
6
1
9
+
+
11 11 11
3)
6 - 4
21 21
4)
723 - 717
278 278
5)
3 +1
5 9
6)
5
1 1
+ +
18 3 6
7)
3 - 7
4 16
14)
1
8) - 1 + 30
6
9) - 2 - 1
3 6
10) - 1 + 3
16 4
11)
1 - 1
9 11
12)
4 - 3
5 20
13)
8 - 1
15 20
1
3 + 7
4 12
Answer Key
Testname: MATHWAY_3
1)
2
5
5
11
2) 1
3)
2
21
4)
3
139
5)
32
45
6)
7
9
7)
5
16
8) -
1
5
9) -
5
6
10)
11
16
11)
2
99
12)
13
20
13)
29
60
14)
4
3
2
Order of Operations -­‐ PEMDAS Operations "Operations" means things like add, subtract, multiply, divide, squaring, etc. If it isn't a number it is probably an operation. But, when you see something like ... 7 + (6 × 52 + 3) ... what part should you calculate first? Start at the left and go to the right? Or go from right to left? Warning: Calculate them in the wrong order, and you will get a wrong answer ! Order or Operations To eliminate this confusion, we have some rules of precedence, established at least as far back as the 1500s, called the "order of operations". The "operations" are addition, subtraction, multiplication, division, exponentiation, and grouping; the "order" of these operations states which operations take precedence (are taken care of) before which other operations. A common technique for remembering the order of operations is the abbreviation "PEMDAS", which is turned into the phrase "Please Excuse My Dear Aunt Sally". It stands for "Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction". This tells you the ranks of the operations: Parentheses outrank exponents, which outrank multiplication and division (but multiplication and division are at the same rank), and these two outrank addition and subtraction (which are together on the bottom rank). When you have a many operations of the same rank, you just operate from left to right. For instance, 15 ÷ 3 × 4 is not 15 ÷ 12, but is rather 5 × 4, because, going from left to right, you get to the division first. Mathway Packet
Order of Operations
Follow the order of operations to work the problem.
1) -20 - 1 + 14
2) 16 + (-8) - (-19) + 5
3) -8 + 17 - 11 - (-8) + 8
4) -17 - 1 + (-6)
5)
2 2 3 1
∙ + ∙
7 3 4 4
6) 1 +
1
4
∙
3
5
7)
7
7 7
÷
+
9
5 20
8)
1 - 7 + -5
3
5 15
9)
4
1 +1
∙
∙9
9
6 3
10)
3 -5 + - 7
10
5 2
1
Answer Key
Testname: MATHWAY_4
1) -7
2) 32
3) 14
4) -24
127
5)
336
1
15
6) 1
7)
4
9
8) -
29
15
9) 2
10) -
13
5
2
Using the FOIL method to multiply two binomials •
•
What if you have something like this: (4x + 6)(x + 2) to multiply out? That's where we use the FOIL method. FOIL means Firsts, Outers, Inners, Lasts. You use FOIL to multiply the terms inside the parenthesis in a specific order: first, outside, inside, last. Example: Multiply (4x + 6)(x + 2) Firsts -­‐ multiply the first term in each set of parenthesis: 4x * x = 4x2 Outers -­‐ multiply the two terms on the outside: 4x * 2 = 8x Inners -­‐ multiply both of the inside terms: 6 * x = 6x Lasts -­‐ multiply the last term in each set of parenthesis: 6 * 2 = 12 Now just add everything together to get 4x2 + 14x + 12. This method only works easily with two binomials. Mathway Packet
Multiplying Polynomials
Multiply the polynomials using the FOIL method.
Express the answer as a single polynomial in standard
form.
1) (x - 11)(x + 2)
2) (5x - 10)(-3x + 6)
3) (5x + 7y)(5x + 12y)
4) (x - 2y)(2x - 4y)
5) (x - 8y)(x - 2y)
6) (-3x + 7)(x - 8)
1
Answer Key
Testname: MATHWAY_5
1) x2 - 9x - 22
2) -15x2 + 60x - 60
3) 25x2 + 95xy + 84y 2
4) 2x2 - 8xy + 8y 2
5) x2 - 10xy + 16y 2
6) -3x2 + 31x - 56
2
Factoring out the Greatest Common Factor from a Polynomial Factoring polynomial expressions is not quite the same as factoring numbers, but the concept is very similar. When factoring numbers or factoring polynomials, you are finding numbers or polynomials that divide out evenly from the original numbers or polynomials. But in the case of polynomials, you are dividing numbers and variables out of expressions, not just dividing numbers out of numbers. Previously, you may have simplified expressions by distributing through parentheses, such as: 2(x + 3) = 2(x) + 2(3) = 2x + 6 Simple factoring in the context of polynomial expressions is backwards from distributing. That is, instead of multiplying something through a parentheses, you will be seeing what you can take back out and put in front of a parentheses, such as: 2x + 6 = 2(x) + 2(3) = 2(x + 3) The trick is to see what can be factored out of every term in the expression. Warning: Don't make the mistake of thinking that "factoring" means "dividing something off and making it magically disappear". Remember that "factoring" means "dividing out and putting in front of the parentheses". Nothing "disappears" when you factor; things merely get rearranged. Example: Factor 3x – 12. The only thing common between the two terms (that is, the only thing that can be divided out of each term and then moved up front) is a "3". So factor this number out to the front: 3x – 12 = 3( ) When we divided the "3" out of the "3x", we were left with only the "x" remaining. We put that "x" as my first term inside the parentheses: 3x – 12 = 3(x ) When we divided the "3" out of the "–12", I left a "–4" behind, so we put that in the parentheses, too: 3x – 12 = 3(x – 4) This is the final answer: 3(x – 4) Warning: Be careful not to drop "minus" signs when you factor. Mathway Packet
Factoring out Greatest Common Factors
Factor out the greatest common factor. Simplify the
factors, if possible.
1) 5t2 - 10t - 25
2) 5x3 + 20x
3) 6x2 - 12x
4) 5t2 - 10t - 25
5) 12wx - 9wy - 15wz
6) 48x8y 9 - 36x3y 6 - 42x6y 4
7) 3x + 27
8) 2x2 - 12x
9) 21x4 - 6x3 + 15x2
10) 8wx - 6wy - 10wz
11) 4x3 + 20x
12) 4x4 - 32x2
1
Answer Key
Testname: MATHWAY_6
1) 5(t2 - 2t - 5)
2) 5x(x2 + 4)
3) 6x(x - 2)
4) 5(t2 - 2t - 5)
5) 3w(4x - 3y - 5z)
6) 6x3y 4(8x5y 5 - 6y 2 - 7x3)
7) 3(x + 9)
8) 2x(x - 6)
9) 3x2(7x2 - 2x + 5)
10) 2w(4x - 3y - 5z)
11) 4x(x2 + 5)
12) 4x2(x2 - 8)
2
Factoring a trinomial where the leading coefficient is 1 A "quadratic" is a polynomial that looks like "ax2 + bx + c", where "a", "b", and "c" are just numbers. For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term "c", but also add up to equal "b", the coefficient on the x-­‐
term. For instance: Example: Factor x2 + 5x + 6. We need to find factors of 6 that add up to 5. Since 6 can be written as the product of 2 and 3, and since 2 + 3 = 5, then we’ll use 2 and 3. We know from multiplying polynomials that this quadratic is formed from multiplying two factors of the form "(x + m)(x + n)", for some numbers m and n. So we’ll draw our parentheses, with an "x" in the firsts of each: (x )(x ) Then we’ll write in the two numbers that we found above as our lasts: (x + 2)(x + 3) This is the answer: x2 + 5x + 6 = (x + 2)(x + 3) This is how all of the "easy" quadratics will work: we will find factors of the constant term that add up to the middle term, and use these factors to fill in our parentheses. Note that we can always check our work by multiplying back to get the original answer. Factoring a trinomial where the leading coefficient is not 1 In the Algebra lesson on factoring trinomials where a = 1, we learned that factoring may require that we put our investigative skills to work. Those skills will really be put to the test when our trinomial starts with an a-­‐value other than one. As with all factoring, there are several different methods that may be used. Let's see what is involved in the "Set-­‐up, Guess and Check" Method. When the leading coefficient is a number other than one, the number of possible answers increases ..... making our investigative efforts harder. 1. First, check to see if all of the terms share a common factor which may be removed. If each term can be factored before you begin, your work will be easier. The terms in this problem do not have a common factor. 2. Consider all of the possible factors of the leading coefficient, 2x². In this problem we only have one choice, 2x and x, making our work easy. So we can start with: (2x ) (x ) If the leading coefficient has several factors, make a list of the possible combinations, so as to be sure to include all possibilities. 3. Consider all of the possible factors of the last term, -­‐6. The possible factors are: +6 and -­‐1 -­‐6 and +1 +3 and -­‐2 -­‐3 and +2 You need to consider all of the possible ways of obtaining the number -­‐6. It appears that we have a multitude of "possible" answers: (2x + 6)(x -­‐ 1) (2x -­‐ 6)(x + 1) (2x + 3)(x -­‐ 2) (2x -­‐ 3)(x + 2) (x + 6)(2x -­‐ 1) (x -­‐ 6)(2x + 1) (x + 3)(2x -­‐ 2) (x -­‐ 3)(2x + 2) 4. We need to find the one "true" answer that will produce the desired middle term of -­‐x. Test each of the possible answers to see which will yield the correct middle term. (2x + 6)(x -­‐ 1) gives middle term 4x. (2x -­‐ 6)(x + 1) gives middle term -­‐4x. (2x + 3)(x -­‐ 2) gives middle term -­‐x. YEA!!!!! (2x -­‐ 3)(x + 2) gives middle term +x. (x + 6)(2x -­‐ 1) gives middle term 11x. (x -­‐ 6)(2x + 1) gives middle term -­‐11x. (x + 3)(2x -­‐ 2) gives middle term 4x. (x -­‐ 3)(2x + 2) gives middle term -­‐4x. 5. Answer: (2x + 3)(x -­‐ 2) Notice: While we initially had several options for answers, we really had only one true answer. The more options that a problem creates, the more detective work needed to find the true answer. Mathway Packet
Factoring Trinomials
Factor the polynomial. If there is a common factor - be
sure to factor that out FIRST.
1) x2 - x - 42
2) x2 + 6x + 8
3) x2 + 6x - 16
4) x2 - x - 72
5) 6x2 - 6x - 36
6) 3x2 - 21x + 36
7) 2x3 + 2x2 - 24x
8) 20z2 + 7z - 6
9) 20z2 - 3z - 9
10) 12x2 - 52x - 40
11) 12y 2 + 54y - 30
1
Answer Key
Testname: MATHWAY_7
1) (x + 6)(x - 7)
2) (x + 4)(x + 2)
3) (x + 8)(x - 2)
4) (x + 8)(x - 9)
5) 6(x + 2)(x - 3)
6) 3(x - 4)(x - 3)
7) 2x(x - 3)(x + 4)
8) (4z + 3)(5z - 2)
9) (4z - 3)(5z + 3)
10) 4(3x + 2)(x - 5)
11) 6(2y - 1)(y + 5)
2
Factor a polynomial of four or more terms by Grouping Below are the steps to accomplish factoring by grouping. It is assumed that there is no greatest common factor (GCF). • Put parentheses around the first two terms and around the last two terms. When putting in these parentheses, one must be careful to make sure negative signs go inside the parentheses. So it may be necessary to rewrite a subtraction as an addition. • Factor the GCF from the first set of parentheses and from the second set of parenthesis. These will be different GCFs. However, if the original four term polynomial is factorable, then what is left inside both sets of parentheses will be exactly the same. • Factor out the common binomial. Example 1: Factor x3 +3x2 + 5x + 15. Step one says put parentheses around the first two terms and around the last two terms. Since there are no negative signs, that is not a concern. (x3 +3x2) + (5x + 15) Step two says factor the GCF from each set of parentheses. In the first set of parentheses, x2 is the GCF; in the second, the GCF is 5. x2 (x + 3) + 5 (x + 3) Now what is in the parentheses should be exactly the same. It they are not the same, then the polynomial is not factorable or there has been a mistake made. Step three now says factor out the common binomial. (x + 3) ( x2 + 5) This is the answer. This answer can be checked by multiplying the binomials back out to get the original problem. Mathway Packet
Factor a polynomial with 4 or more terms by grouping
Factor by grouping.
1) x3 - 9x + 4x2 - 36
2) 24x3 + 18x2y - 20xy2 - 15y 3
3) 12a3 + 16a2b - 15ab 2 - 20b 3
4) x3 - 2x2 + 9x - 18
5) x3 + 8x2 + 7x + 56
6) t3 + 6t2 - 8t - 48
7) x3 - 5x2 - 3x + 15
8) p3 - 9p2 - 14p + 126
9) 5a3 + 10a2 + 8a + 16
10) 9a3 - 9a2 + 45a - 45
11) 9m3 + 18m2 - 3m - 6
12) 20x4 + 12x2 - 25x2 - 15
1
Answer Key
Testname: MATHWAY_8
1) (x + 3)(x - 3)(x + 4)
2) (6x2 - 5y 2)(4x + 3y)
3) (4a2 - 5b 2)(3a + 4b)
4) (x2 + 9)(x - 2)
5) (x + 8)(x2 + 7)
6) (t + 6)(t2 - 8)
7) (x - 5)(x2 - 3)
8) (p - 9)(p2 - 14)
9) (5a2 + 8)(a + 2)
10) 9(a - 1)(a2 + 5)
11) 3(3m2 - 1)(m + 2)
12) (4x2 - 5)(5x2 + 3)
2
Factor a polynomial with two terms by using Special Cases Special Case Difference of Two Squares If you see something of the form a2 -­‐ b2, you should remember the formula 𝑎! − 𝑏 ! = (𝑎 − 𝑏)(𝑎 + 𝑏) Example: x2 – 4 = (x – 2)(x + 2) This only holds for a difference of two squares. There is no way to factor a sum of two squares such as a2 + b2 into factors with real numbers. Special Case Difference or Sum of Two Cubes The other two special factoring formulas are two sides of the same coin: the sum and difference of cubes. These are the formulas: a3 + b3 = (a + b)(a2 – ab + b2) a3 – b3 = (a – b)(a2 + ab + b2) First, notice that the terms in each factorization are the same; then notice that each formula has only one "minus" sign. For the difference of cubes, the "minus" sign goes with the linear factor, a – b; for the sum of cubes, the "minus" sign goes in the quadratic factor, a2 – ab + b2. Some people use the mnemonic "SOAP" for the signs; the letters stand for "same" as the sign in the middle of the original expression, "opposite" sign, and "always positive". a3 ± b3 = (a [same sign] b)(a2 [opposite sign] ab [always positive] b2) Whatever method helps you best keep these formulas straight, do it, because you should not assume that you'll be given these formulas on the test. You really should know them. Note: The quadratic part of each cube formula does not factor, so don't attempt it. Example: Factor x3 – 8 This is x3 – 23, so I get: x3 – 8 = x3 – 23 = (x – 2)(x2 + 2x + 22) = (x – 2)(x2 + 2x + 4) Mathway Packet
Factor Special Cases when the polynomial has 2 terms
Factor the difference of two squares.
1) x2 - 9
2) 49 - x2
3) 25x2 - 64
4) 3x2 - 147
Factor the sum or difference of two cubes.
5) x3 - 27
6) 125 - 8x3
7) x3 + 27
Factor the assorted polynomials.
8) 36x2 - 25
9) 100x2 - 9y 2
10) x3 - 8
11) t3 + 8
12) 64mn4 - 64mp2
13) x2 - 25
14) z3 - 1
15) 25a2 - 25b 2
1
Answer Key
Testname: MATHWAY_9
1) (x + 3)(x - 3)
2) (7 - x)(7 + x)
3) (5x + 8)(5x - 8)
4) 3(x + 7)(x - 7)
5) (x - 3)(x2 + 3x + 9)
6) (5 - 2x)(25 + 10x + 4x2)
7) (x + 3)(x2 - 3x + 9)
8) (6x + 5)(6x - 5)
9) (10x + 3y)(10x - 3y)
10) (x - 2)(x2 + 2x + 4)
11) (t + 2)(t2 - 2t + 4)
12) 64m(n2 + p)(n2 - p)
13) (x - 5)(x + 5)
14) (z - 1)(z2 + z + 1)
15) 25(a + b)(a - b)
2