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The notation tells you to simply multiply that number by every integer that is less than that number until you hit 1. So if , this means that .


In this case,




which simplifies to

and



So


Highlight the common terms.


Cancel out the common terms.


Simplify.


FOIL to get

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Start with the given equation


Factor the left side



Now set each factor equal to zero:
or

or Now solve for t in each case


So our answers are

or

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Start with the given expression

Group like terms


Factor out the GCF out of the first group. Factor out the GCF out of the second group


Since we have the common term , we can combine like terms

So factors to

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Well you can't solve it (since it's not an equation), but you can factor

Start with the given expression.


Rearrange the terms


Group like terms


Factor out the GCF out of the first group. Factor out the GCF out of the second group


Since we have the common term , we can combine like terms

So factors to

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A)
The compound interest formula is where A is the return, P is the principal, r is the interest rate, n is the compound frequency, and t is the time in years.


So in this case, the principal is , the rate is (note 7% is 0.07 in decimal form) and the compound frequency is (note: semiannually means that it is compounded twice a year)

So the equation is

==============================================================================

B)

Let's find out how much money there is in the account after 1 year


Start with given equation


Plug in t=1


Divide 0.07 by 2 to get 0.035


Multiply the exponents 2 and 1 to get 2


Add 1 and 0.035 to get 1.035


Raise 1.035 to the 2 th power to get 1.071225


Multiply 750 and 1.071225 to get 803.41875


So in 1 year, there is $803.42 (rounded to the nearest cent) in the account

------------------------------------------------------------


Let's find out how much money there is in the account after 6 years


Start with given equation.


Plug in t=6


Divide 0.07 by 2 to get 0.035


Multiply the exponents 2 and 6 to get 12


Add 1 and 0.035 to get 1.035


Raise 1.035 to the 12 th power to get 1.51106865734636


Multiply 750 and 1.51106865734636 to get 1133.30149300977


So in 6 years, there is $1133.30 (rounded to the nearest cent) in the account



------------------------------------------------------------


Let's find out how much money there is in the account after 10 years


Start with given equation.


Plug in t=10


Divide 0.07 by 2 to get 0.035


Multiply the exponents 2 and 10 to get 20


Add 1 and 0.035 to get 1.035


Raise 1.035 to the 20 th power to get 1.98978886346584


Multiply 750 and 1.98978886346584 to get 1492.34164759938



So in 10 years, there is $1492.34 (rounded to the nearest cent) in the account

------------------------------------------------------------


Let's find out how much money there is in the account after 15 years


Start with given equation.


Plug in t=15


Divide 0.07 by 2 to get 0.035


Multiply the exponents 2 and 15 to get 30


Add 1 and 0.035 to get 1.035


Raise 1.035 to the 30 th power to get 2.80679370470263


Multiply 750 and 2.80679370470263 to get 2105.09527852697


So in 15 years, there is $2105.10 (rounded to the nearest cent) in the account

------------------------------------------------------------

Let's find out how much money there is in the account after 15 years


Start with given equation.


Plug in t=25


Divide 0.07 by 2 to get 0.035


Multiply the exponents 2 and 25 to get 50


Add 1 and 0.035 to get 1.035


Raise 1.035 to the 50 th power to get 5.58492685566332


Multiply 750 and 5.58492685566332 to get 4188.69514174749



So in 25 years, there is $4188.69 (rounded to the nearest cent) in the account

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Start with the given inequality


Factor the left side


Set the left side equal to zero


Set each individual factor equal to zero:

or

Solve for x in each case:

or


So our critical values are and

Now set up a number line and plot the critical values on the number line





So let's pick some test points that are near the critical values and evaluate them.


Let's pick a test value that is less than (notice how it's to the left of the leftmost endpoint):

So let's pick

Start with the given inequality


Plug in


Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is ()





---------------------------------------------------------------------------------------------



Let's pick a test value that is in between and :

So let's pick

Start with the given inequality


Plug in


Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.


---------------------------------------------------------------------------------------------



Let's pick a test value that is greater than (notice how it's to the right of the rightmost endpoint):

So let's pick

Start with the given inequality


Plug in


Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.
So part our solution in interval notation is ()





---------------------------------------------------------------------------------------------





Summary:

So the solution in interval notation is:


() ()





Here's a graph to visually verify the answer

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Start with the area of a triangle formula



Plug in and


Multiply


Simplify the square root.


Reduce to get or just 3


So the area is

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Let's assume that for all values of a and b. So now let's just pick arbitrary values for a and b. So let's make and .


Start with the given equation.


Plug in and


Add 2 and 3 to get 5


Take the square root of 2 to get 1.41421. Take the square root of 3 to get 1.73205. Take the square root of 5 to get 2.23607.


Add 1.41421 and 1.73205 to get 3.14626


Since , this shows us that .


So this means that . There are some values that will make it true. For instance and are one pair of values, and are another. However, in general, the equation is not true.

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Let x=width of border.

First, let's draw the picture.



note: the lengths and are due to the subtraction of 2 "x" lengths. Take note that there are two dashed lines per side that are not part of the green tiles.


From the picture, we see that the actual counter top is by . So the length is and the width is


Now remember, the area of a rectangle is



Plug in the given area of the green tiles , and


FOIL


Subtract 704 from both sides.


Combine like terms.


Let's use the quadratic formula to solve for x


Start with the quadratic formula


Plug in , , and


Negate to get .


Square to get .


Multiply to get


Subtract from to get


Multiply and to get .


Take the square root of to get .


or Break up the expression.


or Combine like terms.


or Simplify.


So the possible answers are or


However, we need to see if they generate reasonable lengths and widths


Let's check the solution


Go back to the length equation


Plug in


Multiply


Subtract


Since a negative length is not possible, this means that the value is a reasonable solution.


----------------------------



Let's check the solution


Go back to the length equation


Plug in


Multiply


Subtract

So we get a reasonable length with

--------

Go back to the width equation


Plug in


Multiply


Subtract

So we also get a reasonable width with


===============================================

Answer:

So the only solution is


So the width of the border is 4 inches

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# 1

Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -3


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like

-3	|	2	7	0	-5
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)

-3	|	2	7	0	-5
	|
		2

Multiply -3 by 2 and place the product (which is -6) right underneath the second coefficient (which is 7)

-3	|	2	7	0	-5
	|		-6
		2

Add -6 and 7 to get 1. Place the sum right underneath -6.

-3	|	2	7	0	-5
	|		-6
		2	1

Multiply -3 by 1 and place the product (which is -3) right underneath the third coefficient (which is 0)

-3	|	2	7	0	-5
	|		-6	-3
		2	1

Add -3 and 0 to get -3. Place the sum right underneath -3.

-3	|	2	7	0	-5
	|		-6	-3
		2	1	-3

Multiply -3 by -3 and place the product (which is 9) right underneath the fourth coefficient (which is -5)

-3	|	2	7	0	-5
	|		-6	-3	9
		2	1	-3

Add 9 and -5 to get 4. Place the sum right underneath 9.

-3	|	2	7	0	-5
	|		-6	-3	9
		2	1	-3	4

Since the last column adds to 4, we have a remainder of 4. This means is not a factor of

Now lets look at the bottom row of coefficients:


The first 3 coefficients (2,1,-3) form the quotient



and the last coefficient 4, is the remainder, which is placed over like this





Putting this altogether, we get:



So

which looks like this in remainder form:
remainder 4



-----------------------------------
Answer:

So the quotient is and the remainder is 4

---

# 2


a)


Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of 8 (the last coefficient):



Now let's list the factors of 4 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur

---

b)


Now let's use synthetic division to test each possible zero:




Let's make the synthetic division table for the function given the possible zero :

1/2	|	4	5	7	-34	8
	|		2	7/2	21/4	-115/8
		4	7	21/2	-115/4	-51/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

1/4	|	4	5	7	-34	8
	|		1	3/2	17/8	-255/32
		4	6	17/2	-255/8	1/32

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

2	|	4	5	7	-34	8
	|		8	26	66	64
		4	13	33	32	72

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

4	|	4	5	7	-34	8
	|		16	84	364	1320
		4	21	91	330	1328

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

8	|	4	5	7	-34	8
	|		32	296	2424	19120
		4	37	303	2390	19128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1	|	4	5	7	-34	8
	|		-4	-1	-6	40
		4	1	6	-40	48

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1/2	|	4	5	7	-34	8
	|		-2	-3/2	-11/4	147/8
		4	3	11/2	-147/4	211/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1/4	|	4	5	7	-34	8
	|		-1	-1	-3/2	71/8
		4	4	6	-71/2	135/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-2	|	4	5	7	-34	8
	|		-8	6	-26	120
		4	-3	13	-60	128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-4	|	4	5	7	-34	8
	|		-16	44	-204	952
		4	-11	51	-238	960

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-8	|	4	5	7	-34	8
	|		-32	216	-1784	14544
		4	-27	223	-1818	14552

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of



====================================

Since none of the possible rational roots are actual roots, this means that the polynomial either has irrational roots or complex roots.

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Use the identity to rewrite the expression.


Rewrite tangent in terms of sine and cosine.


Rewrite as


Rewrite the expression using the identities and


Now, let's reference the unit circle




From the unit circle, we can see that at the angle , there is a point on the unit circle . So this tells us that and . Also at the angle , there is a point on the unit circle . So this tells us that and .

Take the cosine of to get -1. Take the sine of to get 0. Take the cosine of to get 1. Take the sine of to get 0.


Multiply


Add


Reduce


So

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# 1


Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 1


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.

1	|	3	-1	-4	1
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 3)

1	|	3	-1	-4	1
	|
		3

Multiply 1 by 3 and place the product (which is 3) right underneath the second coefficient (which is -1)

1	|	3	-1	-4	1
	|		3
		3

Add 3 and -1 to get 2. Place the sum right underneath 3.

1	|	3	-1	-4	1
	|		3
		3	2

Multiply 1 by 2 and place the product (which is 2) right underneath the third coefficient (which is -4)

1	|	3	-1	-4	1
	|		3	2
		3	2

Add 2 and -4 to get -2. Place the sum right underneath 2.

1	|	3	-1	-4	1
	|		3	2
		3	2	-2

Multiply 1 by -2 and place the product (which is -2) right underneath the fourth coefficient (which is 1)

1	|	3	-1	-4	1
	|		3	2	-2
		3	2	-2

Add -2 and 1 to get -1. Place the sum right underneath -2.

1	|	3	-1	-4	1
	|		3	2	-2
		3	2	-2	-1

Since the last column adds to -1, we have a remainder of -1. This means is not a factor of
Now lets look at the bottom row of coefficients:

The first 3 coefficients (3,2,-2) form the quotient



and the last coefficient -1, is the remainder, which is placed over like this





Putting this altogether, we get:



So

---

Start with the given expression.


Combine like terms.


Factor out the GCF



------------------------------------------------------------

Answer:
So factors to

---

# 3



Looking at we can see that the first term is and the last term is where the coefficients are 6 and -2 respectively.

Now multiply the first coefficient 6 and the last coefficient -2 to get -12. Now what two numbers multiply to -12 and add to the middle coefficient 1? Let's list all of the factors of -12:



Factors of -12:
1,2,3,4,6,12

-1,-2,-3,-4,-6,-12 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -12
(1)*(-12)
(2)*(-6)
(3)*(-4)
(-1)*(12)
(-2)*(6)
(-3)*(4)

note: remember, the product of a negative and a positive number is a negative number


Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1

First Number	Second Number	Sum
1	-12	1+(-12)=-11
2	-6	2+(-6)=-4
3	-4	3+(-4)=-1
-1	12	-1+12=11
-2	6	-2+6=4
-3	4	-3+4=1

From this list we can see that -3 and 4 add up to 1 and multiply to -12


Now looking at the expression , replace with (notice adds up to . So it is equivalent to )




Now let's factor by grouping:


Group like terms


Factor out the GCF of out of the first group. Factor out the GCF of out of the second group


Since we have a common term of , we can combine like terms

So factors to


So this also means that factors to (since is equivalent to )



------------------------------------------------------------



Answer:
So factors to

---

# 4



Start with the given expression


Factor out the GCF


Now let's focus on the inner expression




------------------------------------------------------------



Looking at we can see that the first term is and the last term is where the coefficients are 12 and 20 respectively.

Now multiply the first coefficient 12 and the last coefficient 20 to get 240. Now what two numbers multiply to 240 and add to the middle coefficient 31? Let's list all of the factors of 240:



Factors of 240:
1,2,3,4,5,6,8,10,12,15,16,20,24,30,40,48,60,80,120,240

-1,-2,-3,-4,-5,-6,-8,-10,-12,-15,-16,-20,-24,-30,-40,-48,-60,-80,-120,-240 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 240
1*240
2*120
3*80
4*60
5*48
6*40
8*30
10*24
12*20
15*16
(-1)*(-240)
(-2)*(-120)
(-3)*(-80)
(-4)*(-60)
(-5)*(-48)
(-6)*(-40)
(-8)*(-30)
(-10)*(-24)
(-12)*(-20)
(-15)*(-16)

note: remember two negative numbers multiplied together make a positive number


Now which of these pairs add to 31? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 31

First Number	Second Number	Sum
1	240	1+240=241
2	120	2+120=122
3	80	3+80=83
4	60	4+60=64
5	48	5+48=53
6	40	6+40=46
8	30	8+30=38
10	24	10+24=34
12	20	12+20=32
15	16	15+16=31
-1	-240	-1+(-240)=-241
-2	-120	-2+(-120)=-122
-3	-80	-3+(-80)=-83
-4	-60	-4+(-60)=-64
-5	-48	-5+(-48)=-53
-6	-40	-6+(-40)=-46
-8	-30	-8+(-30)=-38
-10	-24	-10+(-24)=-34
-12	-20	-12+(-20)=-32
-15	-16	-15+(-16)=-31

From this list we can see that 15 and 16 add up to 31 and multiply to 240


Now looking at the expression , replace with (notice adds up to . So it is equivalent to )




Now let's factor by grouping:


Group like terms


Factor out the GCF of out of the first group. Factor out the GCF of out of the second group


Since we have a common term of , we can combine like terms

So factors to


So this also means that factors to (since is equivalent to )



------------------------------------------------------------




So our expression goes from and factors further to


------------------
Answer:

So factors to

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First, let's graph the function to get



From the graph, we can see that the graph has the approximate zeros and

==============================================================

Now let's use the Rational Root Theorem to list all of the possible rational roots

Rational Root Theorem:

where p and q are the factors of the last and first coefficients


So let's list the factors of -2 (the last coefficient):



Now let's list the factors of 1 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient





Now simplify

These are all the distinct rational zeros of the function that could occur




---------------------------------------------


Now let's use synthetic division to test each possible zero




Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

1	|	1	1	-3	-5	-2
	|		1	2	-1	-6
		1	2	-1	-6	-8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

2	|	1	1	-3	-5	-2
	|		2	6	6	2
		1	3	3	1	0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

-1	|	1	1	-3	-5	-2
	|		-1	0	3	2
		1	0	-3	-2	0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

-2	|	1	1	-3	-5	-2
	|		-2	2	2	6
		1	-1	-1	-3	4

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of



===========================================
Summary:

So only and are actually rational roots.


Now looking back at the table for the test zero , we see

-1	|	1	1	-3	-5	-2
	|		-1	0	3	2
		1	0	-3	-2	0

The bottom row of coefficients (minus the last one) form the quotient




Now let's perform synthetic division using the other zero on the function

2	|	1	0	-3	-2
	|		2	4	2
		1	2	1	0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of

Once again the first three coefficients in the bottom row form the quotient




Let's use the quadratic formula to find the zeros of


Start with the quadratic formula


Plug in , , and


Square to get .


Multiply to get


Subtract from to get


Multiply and to get .


Take the square root of to get .


or Break up the expression.


or Combine like terms.


or Simplify.


So the zeros of are or or just with a multiplicity of 2


Now there are 3 instances where we get a zero of . So this tells us that the zero has a multiplicity of 3

================================================

Answer:

So the zeros of are (with a multiplicity of 3) and

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Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 1


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for , , and . This is simply because really looks like

1	|	1	-1	1	-1	0	0	0	4
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

1	|	1	-1	1	-1	0	0	0	4
	|
		1

Multiply 1 by 1 and place the product (which is 1) right underneath the second coefficient (which is -1)

1	|	1	-1	1	-1	0	0	0	4
	|		1
		1

Add 1 and -1 to get 0. Place the sum right underneath 1.

1	|	1	-1	1	-1	0	0	0	4
	|		1
		1	0

Multiply 1 by 0 and place the product (which is 0) right underneath the third coefficient (which is 1)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0
		1	0

Add 0 and 1 to get 1. Place the sum right underneath 0.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0
		1	0	1

Multiply 1 by 1 and place the product (which is 1) right underneath the fourth coefficient (which is -1)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1
		1	0	1

Add 1 and -1 to get 0. Place the sum right underneath 1.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1
		1	0	1	0

Multiply 1 by 0 and place the product (which is 0) right underneath the fifth coefficient (which is 0)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0
		1	0	1	0

Add 0 and 0 to get 0. Place the sum right underneath 0.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0
		1	0	1	0	0

Multiply 1 by 0 and place the product (which is 0) right underneath the sixth coefficient (which is 0)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0
		1	0	1	0	0

Add 0 and 0 to get 0. Place the sum right underneath 0.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0
		1	0	1	0	0	0

Multiply 1 by 0 and place the product (which is 0) right underneath the seventh coefficient (which is 0)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0	0
		1	0	1	0	0	0

Add 0 and 0 to get 0. Place the sum right underneath 0.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0	0
		1	0	1	0	0	0	0

Multiply 1 by 0 and place the product (which is 0) right underneath the eighth coefficient (which is 4)

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0	0	0
		1	0	1	0	0	0	0

Add 0 and 4 to get 4. Place the sum right underneath 0.

1	|	1	-1	1	-1	0	0	0	4
	|		1	0	1	0	0	0	0
		1	0	1	0	0	0	0	4

Since the last column adds to 4, we have a remainder of 4.


--------------------------------------------

Answer:

So when is divided by , the remainder is 4

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Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -2


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for , , , and . This is simply because really looks like

-2	|	1	0	0	0	0	32
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

-2	|	1	0	0	0	0	32
	|
		1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 0)

-2	|	1	0	0	0	0	32
	|		-2
		1

Add -2 and 0 to get -2. Place the sum right underneath -2.

-2	|	1	0	0	0	0	32
	|		-2
		1	-2

Multiply -2 by -2 and place the product (which is 4) right underneath the third coefficient (which is 0)

-2	|	1	0	0	0	0	32
	|		-2	4
		1	-2

Add 4 and 0 to get 4. Place the sum right underneath 4.

-2	|	1	0	0	0	0	32
	|		-2	4
		1	-2	4

Multiply -2 by 4 and place the product (which is -8) right underneath the fourth coefficient (which is 0)

-2	|	1	0	0	0	0	32
	|		-2	4	-8
		1	-2	4

Add -8 and 0 to get -8. Place the sum right underneath -8.

-2	|	1	0	0	0	0	32
	|		-2	4	-8
		1	-2	4	-8

Multiply -2 by -8 and place the product (which is 16) right underneath the fifth coefficient (which is 0)

-2	|	1	0	0	0	0	32
	|		-2	4	-8	16
		1	-2	4	-8

Add 16 and 0 to get 16. Place the sum right underneath 16.

-2	|	1	0	0	0	0	32
	|		-2	4	-8	16
		1	-2	4	-8	16

Multiply -2 by 16 and place the product (which is -32) right underneath the sixth coefficient (which is 32)

-2	|	1	0	0	0	0	32
	|		-2	4	-8	16	-32
		1	-2	4	-8	16

Add -32 and 32 to get 0. Place the sum right underneath -32.

-2	|	1	0	0	0	0	32
	|		-2	4	-8	16	-32
		1	-2	4	-8	16	0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 5 coefficients (1,-2,4,-8,16) form the quotient




------------------------------------------------
Answer:

So which means that the quotient is and the remainder is 0.

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To use the intermediate value theorem, simply evaluate the endpoints of the interval.


Let's evaluate the left endpoint


Start with the given equation.


Plug in .


Raise to the 5th power to get .


Multiply and to get .


Multiply and to get .


Combine like terms.


So the function value at is . This makes the point (1,-4) which tells us that the y-coordinate is negative.

--------------------------------------

Now let's evaluate the right endpoint


Start with the given equation.


Plug in .


Raise to the 5th power to get .


Multiply and to get .


Multiply and to get .


Combine like terms.



So the function value at is . This makes the point (2,51) which tells us that the y-coordinate is positive.



Since the sign of the y-coordinate transitioned from negative to positive on the interval [1,2], this means that the graph must have crossed the x-axis somewhere in the interval. So there is a zero in the interval [1,2].

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First, let's list all of the possible rational zeros.

Any rational zero can be found through this formula

where p and q are the factors of the last and first coefficients


So let's list the factors of 2 (the last coefficient):



Now let's list the factors of 1 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient






Now simplify

These are all the distinct rational zeros of the function that could occur




------------------------


Now let's test each possible rational root with use of synthetic division




Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

1	|	1	-1	-2	2
	|		1	0	-2
		1	0	-2	0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

2	|	1	-1	-2	2
	|		2	2	0
		1	1	0	2

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

-1	|	1	-1	-2	2
	|		-1	2	0
		1	-2	0	2

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's see if the possible zero is really a root for the function


So let's make the synthetic division table for the function given the possible zero :

-2	|	1	-1	-2	2
	|		-2	6	-8
		1	-3	4	-6

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of



===========================================

So to recap, we only found one rational zero. So the only rational zero for the function is

Now if we go back to the corresponding synthetic division table for the test zero , we get

1	|	1	-1	-2	2
	|		1	0	-2
		1	0	-2	0

Now looking at the bottom row of coefficients, we see the first three numbers: 1, 0 and -2

These coefficients form the quotient or just

This means that or

Set the quotient equal to zero


Add 2 to both sides.


Take the square root of both sides.


or Break up the expression.


So the remaining two zeros are or


===============================================

Answer:

So the zeros of are:

, or where each zero has a multiplicity of 1.

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Let x=length of first side.


Since "one side is twice as long as the first", this means that


Also, because "another is half as long as the first side", this means that



However, there's a problem. In order for any triangle to be constructed, the lengths of any two sides must be greater than the length of the third side. So for instance, this means that .

Plug in and


Add


Since this inequality is never true for any positive x values, this means that you cannot construct a triangle with sides of x, 2x, and (go ahead and try it out on paper if you don't believe me).


So I would double check the problem.

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Start with the given expression.


Rewrite the cotangent in terms of tangent.


Use the identity to rewrite the expression.


Reduce.


Rewrite the tangent function in terms of sine and cosine.


Multiply -1 by the reciprocal of the dividing fraction.


Multiply.



Now, let's reference the unit circle




From the unit circle, we can see that at the angle , there is a point on the unit circle . So this tells us that and



So this means that then becomes


Multiply the first fraction by the reciprocal of the second fraction.


Cancel like terms.


Multiply and simplify.


So

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Start with the given expression.


Factor to get .


Factor to get .


Highlight the common terms.


Cancel out the common terms.


Simplify.


So simplifies to .


In other words, where or

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Start with the given expression.


Combine the fractions.


Multiply.


Expand. Remember, and


Highlight the common terms.


Cancel out the common terms.


Simplify.


Regroup.


So simplifies to .


In other words, where

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Start with the given equation.


Divide both sides by .


Square both sides. This will eliminate the square root.


Multiply both sides by .


Distribute the outer exponent.


Square to get .


Multiply both sides by .


Divide both sides by .


Take the cube root of both sides.

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Start with the given expression.


Rewrite as using the identity


Combine the logs using the identity


Combine the logs using the identity


Simplify.


So simplifies to

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Start with the given equation.


Combine the logs using the identity


Rewrite the equation using the property: ====>


Raise 10 to the 5th power to get 100,000


Divide both sides by 2.


So the answer is

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a)


Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of 8 (the last coefficient):



Now let's list the factors of 4 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur

---

b)


Now let's use synthetic division to test each possible zero:




Let's make the synthetic division table for the function given the possible zero :

1/2	|	4	5	7	-34	8
	|		2	7/2	21/4	-115/8
		4	7	21/2	-115/4	-51/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

1/4	|	4	5	7	-34	8
	|		1	3/2	17/8	-255/32
		4	6	17/2	-255/8	1/32

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

2	|	4	5	7	-34	8
	|		8	26	66	64
		4	13	33	32	72

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

4	|	4	5	7	-34	8
	|		16	84	364	1320
		4	21	91	330	1328

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

8	|	4	5	7	-34	8
	|		32	296	2424	19120
		4	37	303	2390	19128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1	|	4	5	7	-34	8
	|		-4	-1	-6	40
		4	1	6	-40	48

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1/2	|	4	5	7	-34	8
	|		-2	-3/2	-11/4	147/8
		4	3	11/2	-147/4	211/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-1/4	|	4	5	7	-34	8
	|		-1	-1	-3/2	71/8
		4	4	6	-71/2	135/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-2	|	4	5	7	-34	8
	|		-8	6	-26	120
		4	-3	13	-60	128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-4	|	4	5	7	-34	8
	|		-16	44	-204	952
		4	-11	51	-238	960

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of


------------------------------------------------------


Let's make the synthetic division table for the function given the possible zero :

-8	|	4	5	7	-34	8
	|		-32	216	-1784	14544
		4	-27	223	-1818	14552

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of



====================================

Since none of the possible rational roots are actual roots, this means that the polynomial either has irrational roots or complex roots.

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Start with the given inequality.


Add to both sides.


Combine like terms on the right side.


Divide both sides by to isolate . note: Remember, the inequality sign flips when we divide both sides by a negative number.


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is


So the answer in interval notation is


Also, the answer in set-builder notation is



Here's the graph of the solution set

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Start with the given equation.


Subtract from both sides.


Rearrange the terms.


Divide both sides by to isolate .


Break up the fraction and simplify.

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Start with the given inequality


Break up the absolute value (remember, if you have , then or )

or Break up the absolute value inequality using the given rule




Now lets focus on the first inequality


Start with the given inequality


Subtract 1 from both sides


Combine like terms on the right side


Divide both sides by 2 to isolate x



Divide


Now lets focus on the second inequality


Start with the given inequality


Subtract 1 from both sides


Combine like terms on the right side


Divide both sides by 2 to isolate x



Divide



----------------------------------------------------

Answer:

So our answer is

or



So the solution in interval notation is: (] [)




Here's the graph of the solution set




Note:
There is a closed circle at which means that we're including that value from the solution set.


Also, there is a closed circle at which means that we're including that value from the solution set.

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Start with the given equation


Break up the absolute value (remember, if you have , then or )

or Set the expression equal to the original value 10 and it's opposite -10




Now lets focus on the first equation


Add 3 to both sides


Combine like terms on the right side


Divide both sides by 5 to isolate x



Reduce

---

Now lets focus on the second equation



Add 3 to both sides


Combine like terms on the right side


Divide both sides by 5 to isolate x



Reduce





So the solutions to are:

and

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Let x=amount of money in your pocket.


So when you have "at least $25", this means that you have 25 or more. Mathematically speaking, this means that some unknown quantity "x" is greater than or equal to 25.


So symbolically, this inequality is