Download MATH 60 EXAM 3 REVIEW (Chapters 7 and 8)

MATH 60 EXAM 3 REVIEW (Chapters 7 and 8)
CHAPTER 7 REVIEW - Factoring
polynomial – is a term or sum of terms in which all variables have whole number exponents.
Example: 3x, or x2 + 1, or -3x2 + 3x + 1
monomial – a number, a variable, or a product of numbers and variables.
Example: 3, 2x, -4x2 are all monomials.
binomial – the sum of two monomials that are unlike terms.
trinomial – the sum of three monomials that are unlike terms.
like terms – terms of a variable expression that have the same variable and the same exponent.
Example: 3x and 3x2 are unlike terms, but 3x and 2x are like terms.
factor – (in multiplication) a number being multiplied.
Example: What are the factors of 121? 1, 11, and 121.
121 = 11 X 11,
121 = 1 X 121
to factor a polynomial – to write a polynomial as a product of other polynomials
to factor a trinomial of the form ax2 + bx + c - to express the trinomial as the product of two binomials.
Example: x2 + 5x + 6 = (x+2)(x+3)
to factor by grouping – to group and factor terms in a polynomial in such a way that a common binomial
factor is found.
Example: 2x(x+1) – 3(x+1) = (x + 1)(2x – 3)
factor completely – to write a polynomial as a product of factors that are nonfactorable over the integers.
Example: 3x3-75x First factor out the GCF. 3x(x2-25).
The remaining polynomial is factorable 3x(x-5)(x+5) Now it’s completely factored
FOIL method – A method of finding the product of two binomials in which the sum of the products of the First
terms, of the Outer terms, of the Inner terms, and of the Last terms is found.
Example: (x+2)(x+3) =x2+ 3x + 2x + 2*3 = x2 + 5x + 6
Greatest Common Factor – the largest factor that is common to two or more numbers.
Example: What is the GCF of 12x2 and 16x5?
The GCF of 12 and 16 is 4. 4 is the biggest number that goes INTO both 12 and 16.
The GCF of x2 and x5 is x2. The variable term with the smallest exponent is the GCF.
The Greatest Common Factor of 12x2 and 16x5 is 4x2.
EXPRESSIONS
Examples: 3+2(1-4)2, -3x+x, (3x+2)2
Can be simplified
Ex1: Don’t forget order of operations!
3+2(1-4)2 can be simplified to
3+2(-3)2 which becomes
3+2(9) which becomes
3+18 which becomes 21.
Ex2:
-3x + x can be simplified to -2x
EQUATIONS
Examples: 3x + 2 = 5, 5x + 5y=10, x(x-5) = -6
Can be solved:
Ex2 :
Ex3 :
Ex1 :
3x = 5 − 2 = 3
3x 3
=
3 3
x =1
One solution
5 y = −5 x + 10
5 y − 5 x 10
=
+
5
5
5
y = −x + 2
Solution is all (x, y)
that make this linear
equation true.
When graphed
this is a line.
Can be evaluated: (3x+2)2 can be evaluated
at x= -1. (3(-1)+2)2 is (-3+2)2 which is (-1)2
which is 1.
Can be reduced:
Examples:
x( x − 5) = −6
x 2 − 5 x = −6
x 2 − 5 x+ 6 = 0
( x − 3)( x − 2) = 0
x = 3, x = 2
or {3,2}
Two solutions
Can be evaluated to see if a solution is true.
Is (3,4) a solution of y=-x+2?
4 = -3+2=-1 NO
20
4
⇒
25
5
2 3
25x y
5x
⇒
4
3y
15 xy
Rules for Variable Expressions:
Only like terms can be added, and when adding like terms, do not change the exponent of the variable.
(2x2 – x + 5) + (3x2 + 5x -2) = 2x2 + 3x2 -x + 5x + 5-2 = 5x2 + 4x + 3
When multiplying variable expressions, add exponents of like variables
(5xy3)(2y2)=10xy3+2 = 10xy5
When taking powers of variable expression that is a monomial (one term), multiply exponents of EVERY term
inside the parentheses.
(2x3y4)3 = 23x3*3y4*3 = 8x9y12
When dividiing variable expression that is a monomial (one term), subtract the exponents of like variables. If
the larger exponent is in the denominator, the positive result of the subtraction remains in the denominator.
25 x 2 y 3 5 2 −1 3− 4
5x
= x y = 5 xy −1 =
4
3
y
15 xy
A General Strategy for Factoring a Polynomial
1. Do all the terms in the polynomial have a common factor? If so, factor out the
Greatest Common Factor. Make sure that you don’t forget it in your final answer.
Example: Factor 24x4 - 6x2 GCF = 6x2 So this polynomial factors into 6x2(4x2 - 1). Also look to see
if the other polynomial factor and be factored more. (4x2-1)=(2x-1)(2x+1), so the final answer is
24x4 - 6x2 =6x2(2x-1)(2x+1),
2. After factoring out the GCF (if there is one) count the number of terms in the remaining polynomial.
Two terms: Is it a difference of squares? Factor by using: a2-b2 = (a+b)(a-b)
Example: 36x2 – 49 = (6x)2 – 72 = (6x-7)(6x+7)
If the polynomial can’t be factored, it is PRIME.
Three terms: Is it a perfect square trinomial?
If it is it would be in the form a2x2 + 2abx + b2 , which is factored as (a+b)2
or a2x2 + 2abx + b2 which is factored as (a-b)2
Example: 4x2 + 12x + 9 = (2x)2 + 2(2)(3)x + 32 = (2x + 3)2
Is it of the form x2 + bx + c?
Factor by finding two numbers that multiply to c and add to b.
Example: x2 -3x - 4 = (x+1)(x-4) because 1*-4 = -4 and 1 + -4 = -3
Can’t find the numbers? Maybe the polynomial is PRIME.
Is it of the form ax2 + bx + c?
Try factoring by the Grouping Method (or ac Method) ,Box or Snowflake Method
Example: 2x2 + 13x + 15 (the a*c method means multiply 2*15
which is 30.
Find factors of 30 that add up to the middle term’s coefficient, which in this
case is 13. 3*10=30 and 3+10 = 13. Split the middle term into two parts:
2x2 + 10x + 3x + 15 and then factor by grouping.
2x(x+5)+3(x+5) = (2x+3)(x+5)
Those methods don’t work? Maybe the polynomial is PRIME.
Four terms: Try Factoring by Grouping. Group the 1st two terms and the last two terms. Factor out
the Greatest Common Factor from each grouping. Then factor out the common binomial term.
3. Always factor completely. Double check that each of your factors can not be factored more.
4. Check your work by multiplying the factors together. Does it result in the original polynomial?
SOLVING POLYNOMIAL EQUATIONS: Put Equation in standard form, ax2 + bx + c = 0.
Then factor it and use the Zero Product Product to find the solutions. This says that if A*B=0, then
A=0 or B=0. Example: x2 -2x = -3 Standard form: x2 – 2x + 3 = 0. Factored: (x+1)(x-3) = 0
So x+1 = 0, which gives x = -1, or another possible solution is x-3 = 0, which gives x = 3.
Applications of Quadratic Equations
p. 430 Example 4: A community garden sits on a corner lot and is in the shape of a right triangle. One side is
10 ft longer than the shortest side, while the longest side is 20 ft longer than the shortest side. Find the lengths
of the sides of the garden.
Hint:
Step 2: Notice that the longest side is always the hypotenuse. Let variables represent the unknown quantities.
Use given information to put other uknowns in terms of one.
Step 3: Use given information and your variables to form an equation.
Because it is a right triangle, you can use the Pythagorean theorem.
x2 + (x+10)2 = (x + 20)2
Use distributive property and combine like terms. If it is a degree 2 equation, but it in
standard form by putting 0 on one side.
x2 + x2 + 20x + 100 = x2 + 40x + 400
2 x2 + 20x + 100 = x2 + 40x + 400
x2 - 20x -300 = 0
Step 4: Solve the equation. Now factor this polynomial and use the zero-product-rule.
(x - 30)(x + 10) = 0
x = 30 or x = -10
The length of a side of a triangle cannot be negative, so we cannot use x = -10 as an answer.
Therefore x = 30 feet = length of shortest side
x + 30 = 30 + 10 = 40 feet = length of second side
x + 20 = 30 + 20 = 50 feet = length of the longest side.
Step 5: Check your answer and state conclusion. Does 302 + 402 = 502? 900 + 1600 = 2500 ? Yes
CONCLUSION: The lengths of the sides of the garden are 30 ft, 40 ft and 50 ft.
Projectile Motion Example:
After the semester is over, Herman discovers that the math
department has changed textbooks (again) so the bookstore won't
buy back his nearly-new book. Herman goes to the roof of the math
building, which is 160 feet high, and chucks his book straight down
at 48 feet per second. How many seconds does it take his book to
strike the ground?
Use the formula h(t) = –16t2 – 48t + 160
I need to find the time for the book to reach a height of zero ("zero"
being "ground level"), so:
0 = –16t2 – 48t + 160
Factor out -16 so it is in standard form.
0 = -16(t2 + 3t – 10)
You can now divide both sides by -16 to simplify.
t2 + 3t – 10 = 0
(t + 5)(t – 2) = 0
t = –5 or t = 2
t = -5 does not make sense because time can’t be negative, so
Correct answer is t=2 seconds.
CONCLUSION: It takes 2 seconds for the book to strike the ground.
CHAPTER 8 RATIONAL EXPRESSION REVIEW
A rational expression is a fraction in which the numerator or denominator is a
variable expression (such as a polynomial). A rational expression is undefined if the
denominator has a value of 0.
A rational expression is in SIMPLEST form when the numerator and denominator
have no common factors other than 1.
6x
is not in simplest form.
9x2
2
is in simplest form.
3x
Reducing to simplest form – factor the numerator and denominator, then cancel out
any common factors in the numerator and denominator (not common factors that are
both in the numerator or both in the denominator, e.g. side by side).
2x2 + 2x
2 x(x + 1 )
2x
=
=
x 2 + 3 x + 2 (x + 1 )(x + 2 ) x + 2
Multiplying Rational Expressions – factor the numerators and denominators then
cancel out common factors as above, then multiply the numerators and multiply the
denominators.
2x2 + 2x x2 − 4
2 x(x + 1 ) ( x + 2)( x − 2) 2( x − 2)
⋅
=
⋅
=
2
2
x + 3 x + 2 x − x (x + 1 )(x + 2 )
x( x − 1)
x −1
Dividing Rational Expressions – change to a multiplication problem by changing
the DIVISOR into it’s RECIPROCAL.
2
Example: 25 − x ÷ x − 5
5x4
x 4 + 5x3
− 1( x 2 − 25) x 4 + 5 x 3
•
5x 4
x−5
3
− 1( x − 5)( x + 5) x ( x + 5)
•
5x4
x−5
− ( x + 5) 2
5x
Put each polynomial in standard form, flip divisor
(expression to the right of ÷) then completely
factor each polynomial
After factoring, cancel out common FACTORS
between numerators and denominators
Adding and Subtracting Rational Expressions –
1
4x2 −1 2x2 − x
x
+
Step 1: Factor the denominators, then find the LCM. The LCM of two polynomials is the
simplest polynomial that contains the factors of each polynomial. To find the LCM of two or
more polynomials, first factor each polynomial completely. The LCM is the product of each
factor the greater number of times it occurs in any one factorization.
1
+
(2 x − 1)(2 x + 1) x(2 x − 1)
x
1st denominator : (2 x − 1)(2 x + 1)
2 nd denominator : x(2 x − 1)
LCM = x(2 x − 1)(2 x + 1)
Step 2: Change each rational expression so that the new denominator will be the LCM. You
will multiply the numerator and denominator of each expression by whatever it takes to get
the LCM as the new denominator.
x
1
(2 x + 1)
x2
2x +1
⋅ +
⋅
=
+
(2 x − 1)(2 x + 1) x x(2 x − 1) (2 x + 1) x(2 x − 1)(2 x + 1) x(2 x − 1)(2 x + 1)
x
Step 3: Add the two new fractions by adding the numerators and keeping the
denominator (the LCM) the same.
x2 + 2x +1
x (2 x − 1)(2 x + 1)
Step 4: Now factor the resulting expression and cancel out any common factors in the
numerator and denominator.
x2 + 2x +1
( x + 1) 2
=
x (2 x − 1)(2 x + 1) x(2 x − 1)(2 x + 1)
Simplify Complex Fractions – Complex fractions are just rational expressions with
fractions within fractions. To simplify, find the LCM of all the denominators of every
fraction in the expression, then multiply the main numerator and denominator by that LCM.
Then simplify as usual.
1
1
1( x 3 ) + ( x 3 )
3
x
x3 + x2
x 2 ( x + 1)
x ⋅
x
=
=
=
= x2
1
1 x3
1
1
x +1
x +1
+ 3
(x3) + 3 (x3)
2
2
x
x
x
x
1+
LCD= x3
Solving Equations with Fractions – multiply BOTH SIDES of the equation by the LCM of
all denominators in the equation. Then solve as usual. Make sure you know the restrictions.
The solution cannot be a value that is not in the domain. That is, it cannot be a value that
would make any of the original fractions undefined.
3
x
+
= 1
LCD
= x ( x + 1)
x
x + 1
3
(x ( x + 1 ) ) + x (x ( x + 1 ) ) = x ( x + 1 )
x
x + 1
2
3 ( x + 1) + x = x ( x + 1)
x = 0 or x = -1
would make one of
these fractions
undefined.
2
3x + 3 + x
2 x = −3
x = − 3
2
+ x
2
:
CHECK
3
− 3
= x
− 3
+
2
− 3
2
2 = −2 + 3 = 1
= 1 = −2 +
− 3
−
1
+ 1
2
2
If the equation is one fraction set equal to another, this is called a PROPORTION. Solve by
CROSS-MULTIPLYING, then isolating the variable.
Example
x
=
3
18 x
18 x
18
WORK
:
12
18
= 3 (12 ) = 36
36
=
18
x = 2
Example
:
x
18
=
2
x + 5
x ( x + 5 ) = 36
Cross-multiply
Use dist. prop. to simplify.
If equation is degree 2, then
2
x + 5 x = 36
it is a quadratic equation.
Put it in standard form
(0 on one side).
x 2 + 5 x − 36 = 0
Factor and use
( x + 9 )( x − 4 ) = 0
zero-product rule
x + 9 = 0 or x − 4 = 0
to solve for x.
x = −9
or
x = 4
Rate of Work * Time Worked = Part of Tasked Completed
If someone can do a job in 60min, their rate of work is 1/60min.
If someone else can do the same job in 40minutes, their rate of work is
1
1
⋅t +
⋅t =1
1/40min.
60 min
40 min
The TIME to get the same job done TOGETHER can be found by
t
t
+
=1
Adding their parts together to make 1 whole job.
60 40