Question

i neeedddd help pleasee!
thankyou & godbless!
What are the possible rational zeros of f(x) = 2x3 – 4x2 – 7x + 10

Answer 1

What are the factors of 10

Answer 2

1 2 5 & 10 ?

Answer 3

what are the factors of 2

Answer 4

2 and 1 ...

Answer 5

now divide each factor of 10 by each factor of 2

Answer 6

1/1 = 1
1/2 = 1/2
2/1 = 2
2/2 = 1
5/1 = 5
5/2 = 5/2
10/1 = 10
10/2 = 5


So the possible positive rational zeros are:

1, 1/2, 2, 1, 5, 5/2, 10, 5

which sorts to

1/2, 1, 2, 5/2, 5, 10

Answer 7

Now simply negate each possible positive rational zero to get the new list


-1/2, -1, -2, -5/2, -5, -10

Answer 8

so the total list of all possible rational zeros is


1/2, 1, 2, 5/2, 5, 10
-1/2, -1, -2, -5/2, -5, -10

Answer 9

You can write this in a more compact way as shown below

\[\Large \pm\frac{1}{2},\pm 1, \pm 2, \pm\frac{5}{2}, \pm 5, \pm 10\]

Answer 10

omg thank you !

Answer 11

you're welcome

Answer 12

:))) your beauuutifullll !!

Answer 13

lol why thank you

Answer 14

can i ask you for help on another problem ? :/

Answer 15

sure go for it

Answer 16

actually i lied . two more questions ? lol :D

Answer 17

If f(x) = 2x2 – 4x + 5 and g(x) = 2x – 2, what is f(x) + g(x) ?

Answer 18

sure, I don't mind

Answer 19

f(x) + g(x)


(2x^2 – 4x + 5) + (2x – 2)


2x^2 – 4x + 5 + 2x – 2


Now combine like terms

Answer 20

Tell me what you get

Answer 21

2x2 – 2x + 3 ?

Answer 22

2x^2

Answer 23

you got it

Answer 24

really ?!?!?! omg <3 i love you jim!

Answer 25

yes, really...you know what you're doing, which is great

Answer 26

probably the biggest hurdle you had to clear was the jump from

f(x) + g(x)

to

(2x^2 – 4x + 5) + (2x – 2)

Answer 27

since it's a bit of strange notation

Answer 28

yeah idk how to do that -___- lol but its okay :) & this is the last one !


What is the remainder when (2x3 – 8x2 + 6x – 36) ÷ (x – 4)

Answer 29

i got 4 -___-

Answer 30

Use the remainder theorem, which basically says

if you divide some polynomial function P(x) into x-k like so

P(x)/(x-k)

then the remainder will be equal to P(k)

Answer 31

So this means the remainder is

2x^3 - 8x^2 + 6x - 36

2(4)^3 - 8(4)^2 + 6(4) - 36

2(64) - 8(16) + 6(4) - 36

128 - 128 + 24 -36

0 + 24 - 36

24 - 36

-12

So 2x^3 - 8x^2 + 6x - 36 = -12 when x = 4

This means that P(4) = -12, which makes the remainder equal to -12

Answer 32

idk what im doing over here lol

Answer 33

that's ok

Answer 34

I can help, but I don't have face time

Answer 35

i help strangers all the time

Answer 36

and i don't care what you look like, its the brain that matters when it comes to taking the test anyway

Answer 37

go ahead and ask away

Answer 38

okay im gonna try to solve them if i need help or get stuck im comming to you !! lol

Answer 39

alright sounds like a great plan

Answer 40

p.s. if you ever need help in history i gotchu ! haha

Answer 41

thanks I'll keep that in mind

Answer 42

omg ... im comlpetely lost ..After the fraction ((x+2)/3)-(x+1)/(2x) have been combined using the Least Common Denominator of 6x, what is the numerator?

Answer 43

You need to get each denominator equal to the LCD 6x.


So you need to multiply the first fraction by 2x/2x


Then you need to multiply the second fraction by 3/3


Once the denominators are equal, you can combine the numerators and simplify.



\[\Large \frac{x+2}{3} - \frac{x+1}{2x}\]


\[\Large \frac{2x(x+2)}{2x*3} - \frac{x+1}{2x}\]


\[\Large \frac{2x^2+4x}{6x} - \frac{x+1}{2x}\]


\[\Large \frac{2x^2+4x}{6x} - \frac{3(x+1)}{3*2x}\]


\[\Large \frac{2x^2+4x}{6x} - \frac{3x+3}{6x}\]


\[\Large \frac{2x^2+4x-(3x+3)}{6x}\]


\[\Large \frac{2x^2+4x-3x-3}{6x}\]


\[\Large \frac{2x^2+x-3}{6x}\]


So

\[\Large \frac{x+2}{3} - \frac{x+1}{2x}\]

simplifies to

\[\Large \frac{2x^2+x-3}{6x}\]

Answer 44

thats not one of the options :(

Answer 45

what are your options?

Answer 46

2x2 + x - 3

2x2 + x - 1

- 2x2 + x + 6

-x2 + 2x + 6

Answer 47

The fraction is

\[\Large \frac{2x^2+x-3}{6x}\]

They just want the numerator (glossed over that part), so the answer is

\[\Large 2x^2+x-3\]

Answer 48

Hopefully you saw/understand how I got \[\Large \frac{2x^2+x-3}{6x}\]

Answer 49

yeahh i deff understood, but idk why thats not an option .. w.e. ill leave that one for later .

Answer 50

it is an option

Answer 51

it's choice A

Answer 52

2x2 + x - 3 is 2x^2 + x - 3

Answer 53

simplify (2x^2+3x+1)/(2x+1)

i got (x+1)

then i have to use complete sentences, explain why f(1) = 2, f(0) = 1, and f(-1) = 0, yet f(-one¨Chalf) is undefined. Be sure to show your work. (this will be on my exam )

Answer 54

another question is

i have to show all my work in dividing that . and explain how to simplify it -____________-

Answer 55

Do you see how choice A (from the prev problem) matches the numerator?

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

(2x^2+3x+1)/(2x+1)


(2x^2+2x+x+1)/(2x+1)


((2x^2+2x)+(x+1))/(2x+1)


(2x(x+1)+1(x+1))/(2x+1)


((2x+1)(x+1))/(2x+1)


x+1


So you have the correct simplification


f(x) = (2x^2+3x+1)/(2x+1)


f(1) = (2(1)^2+3(1)+1)/(2(1)+1)


f(1) = (2(1)+3(1)+1)/(2(1)+1)


f(1) = (2+3+1)/(2+1)


f(1) = (6)/(3)


f(1) = 2

-----------

f(x) = (2x^2+3x+1)/(2x+1)


f(0) = (2(0)^2+3(0)+1)/(2(0)+1)


f(0) = (2(0)+3(0)+1)/(2(0)+1)


f(0) = (0+0+1)/(0+1)


f(0) = (1)/(1)


f(0) = 1

-----------

f(x) = (2x^2+3x+1)/(2x+1)


f(-1) = (2(-1)^2+3(-1)+1)/(2(-1)+1)


f(-1) = (2(1)+3(-1)+1)/(2(-1)+1)


f(-1) = (2-3+1)/(-2+1)


f(-1) = (0)/(-1)


f(-1) = 0

Answer 56

yet...

f(x) = (2x^2+3x+1)/(2x+1)


f(-1/2) = (2(-1/2)^2+3(-1/2)+1)/(2(-1/2)+1)


f(-1/2) = (2(1/4)+3(-1/2)+1)/(2(-1/2)+1)


f(-1/2) = (2/4-3/2+1)/(-2/2+1)


f(-1/2) = (2/4-3/2+1)/(-1+1)


f(-1/2) = (0)/(0)


but you can't divide by zero, so f(-1/2) is undefined

Answer 57

The idea here is to factor as much as possible. Then cancel out any common terms.


\[\Large \frac{y^2 + 2y - 8}{y^2 + y - 12}\]


\[\Large \frac{(y+4)(y-2)}{(y+4)(y-3)}\]


\[\Large \frac{\cancel{(y+4)}(y-2)}{\cancel{(y+4)}(y-3)}\]


\[\Large \frac{y-2}{y-3}\]

Answer 58

perfect .

Answer 59

glad it's making sense, did you see my prev message about the one that didn't have an option?

Answer 60

it should be choice A

Answer 61

yess i got it thanks !:) &
What is the Least Common Denominator of

Answer 62

List out the factorization of each number:


x: 1*x

x+1: 1*(x+1)

3x: 1*3*x

Now multiply out the unique and most occurring factors. So multiply 1, x, x+1, and 3 to get


1*x*(x+1)*3

3x(x+1)

Answer 63

So the LCD is 3x(x+1)

Answer 64

your a genius .

Answer 65

lol why thank you

Answer 66

the next one says show all your work to simplify -__- explain how to simplify it & list restrictions .
(2/x)-(2/(x-1))+(2/(x-2))

Answer 67

after that one ,
theres two more questions for today THANKGOD!

Answer 68

\[\Large \frac{2}{x}-\frac{2}{x-1}+\frac{2}{x-2}\]


\[\Large \frac{2(x-1)(x-2)}{x(x-1)(x-2)}-\frac{2}{x-1}+\frac{2}{x-2}\]


\[\Large \frac{2(x^2 - 3x + 2)}{x(x-1)(x-2)}-\frac{2}{x-1}+\frac{2}{x-2}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2}{x-1}+\frac{2}{x-2}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x(x-2)}{(x-1)*x(x-2)}+\frac{2}{x-2}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x^2 - 4x}{(x-1)*x(x-2)}+\frac{2}{x-2}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x^2 - 4x}{x(x-1)(x-2)}+\frac{2}{x-2}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x^2 - 4x}{x(x-1)(x-2)}+\frac{2x(x-1)}{(x-2)*x(x-1)}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x^2 - 4x}{x(x-1)(x-2)}+\frac{2x^2 - 2x}{(x-2)*x(x-1)}\]


\[\Large \frac{2x^2 - 6x + 4}{x(x-1)(x-2)}-\frac{2x^2 - 4x}{x(x-1)(x-2)}+\frac{2x^2 - 2x}{x(x-1)(x-2)}\]


\[\Large \frac{2x^2 - 6x + 4-(2x^2 - 4x)+(2x^2 - 2x)}{x(x-1)(x-2)}\]


\[\Large \frac{2x^2 - 6x + 4-2x^2 + 4x+2x^2 - 2x}{x(x-1)(x-2)}\]


\[\Large \frac{2x^2 - 4x + 4}{x(x-1)(x-2)}\]


and that's as simple as it gets


Note: you can expand out the denominator if you want, but it depends on your book really (and what it wants)

Answer 69

The steps are the same: you need to get each denominator equal to the LCD. You do this by multiplying top/bottom of each fraction by the missing piece of the LCD

Once the denominators are the same, you can combine the fractions.

Answer 70

that was alot to take in ... lol but i got it

Answer 71

yes it is a lot since the LCD has 3 parts to it

Answer 72

but I'm glad you're understanding it

Answer 73

Simplify the complex fractions & are they equivalent why or why not ?

Answer 74

Let's start with the first complex fraction and simplify it

\[\Large \frac{\frac{x^2-x-20}{4}}{\frac{x-5}{10}}\]


\[\Large \frac{x^2-x-20}{4} \times \frac{10}{x-5}\]


\[\Large \frac{(x^2-x-20)*10}{4(x-5)}\]


\[\Large \frac{10(x^2-x-20)}{4(x-5)}\]


\[\Large \frac{2*5(x^2-x-20)}{2*2(x-5)}\]


\[\Large \frac{2*5(x-5)(x+4)}{2*2(x-5)}\]


\[\Large \frac{\cancel{2}*5\cancel{(x-5)}(x+4)}{\cancel{2}*2\cancel{(x-5)}}\]


\[\Large \frac{5(x+4)}{2}\]


\[\Large \frac{5x+20}{2}\]


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

So the first complex fraction

\[\Large \frac{\frac{x^2-x-20}{4}}{\frac{x-5}{10}}\]

simplifies to

\[\Large \frac{5x+20}{2}\]

Answer 75

Use these steps/techniques to simplify the second complex fraction and tell me what you get

Answer 76

.........

Answer 77

which part are you stuck on?

Answer 78

the second complex fraction my moms rushing me off the computer omg i have another question to ask D:

Answer 79

how much time do you have?

Answer 80

like 5 minutes -___- my mom wants to leave and take me out to dinner .

Answer 81

alright, go ahead and ask your last question

Answer 82

I'll be writing up the steps for the second complex fraction

Answer 83

show work to simplify (2/x)+(1/9)=(1/3)

Answer 84

omg i love you .

Answer 85

\[\Large \frac{\frac{x^2-x-20}{x-5}}{\frac{4}{10}}\]


\[\Large \frac{x^2-x-20}{x-5} \times \frac{10}{4}\]


\[\Large \frac{(x^2-x-20)*10}{(x-5)4}\]


\[\Large \frac{10(x^2-x-20)}{4(x-5)}\]


\[\Large \frac{2*5(x^2-x-20)}{2*2(x-5)}\]


\[\Large \frac{2*5(x-5)(x+4)}{2*2(x-5)}\]


\[\Large \frac{\cancel{2}*5\cancel{(x-5)}(x+4)}{\cancel{2}*2\cancel{(x-5)}}\]


\[\Large \frac{5(x+4)}{2}\]


\[\Large \frac{5x+20}{2}\]


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

So

\[\Large \frac{\frac{x^2-x-20}{x-5}}{\frac{4}{10}}\]

simplifies to

\[\Large \frac{5x+20}{2}\]

Answer 86

Since the two complex fractions simplify to the same thing, this means that the two complex fractions are equivalent

Answer 87

Keep in mind that each step is equivalent to any other step (in each separate part)

Answer 88

2/x + 1/9 = 1/3


(2/x)*(9/9) + 1/9 = 1/3


18/(9x) + 1/9 = 1/3


18/(9x) + (1/9)*(x/x) = 1/3


18/(9x) + x/(9x) = 1/3


18/(9x) + x/(9x) = (1/3)*((3x)/(3x))


18/(9x) + x/(9x) = (3x)/(9x)


(18 + x)/(9x) = (3x)/(9x)


Since the denominators are equal, the numerators must be equal, so..


18 + x = 3x


18 = 3x - x


18 = 2x


18/2 = 2x/2


9 = x


x = 9


So the solution is x = 9

Answer 89

im sorry i couldnt thankyou the other day i couldnt get back on until just now :/
but hopefully your still willing to help me ?

Answer 90

sure I can help you out. You can either message me directly, you can start a problem and tag me in it, or you can email me

Answer 91

If I'm on OS, then the first two work

If I'm not on OS, then you can email me

Answer 92

that's not the direct message feature, you just posted a testimonial

Answer 93

uhhh idk what im doing and i forgot my password to my email, im just gonna ask a new question

Answer 94

alright go for it