OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

What are the factors of 10

5 years ago
OpenStudy (anonymous):

1 2 5 & 10 ?

5 years ago
jimthompson5910 (jim_thompson5910):

what are the factors of 2

5 years ago
OpenStudy (anonymous):

2 and 1 ...

5 years ago
jimthompson5910 (jim_thompson5910):

now divide each factor of 10 by each factor of 2

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

Now simply negate each possible positive rational zero to get the new list -1/2, -1, -2, -5/2, -5, -10

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

You can write this in a more compact way as shown below \[\Large \pm\frac{1}{2},\pm 1, \pm 2, \pm\frac{5}{2}, \pm 5, \pm 10\]

5 years ago
OpenStudy (anonymous):

omg thank you !

5 years ago
jimthompson5910 (jim_thompson5910):

you're welcome

5 years ago
OpenStudy (anonymous):

:))) your beauuutifullll !!

5 years ago
jimthompson5910 (jim_thompson5910):

lol why thank you

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

sure go for it

5 years ago
OpenStudy (anonymous):

actually i lied . two more questions ? lol :D

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

sure, I don't mind

5 years ago
jimthompson5910 (jim_thompson5910):

f(x) + g(x) (2x^2 – 4x + 5) + (2x – 2) 2x^2 – 4x + 5 + 2x – 2 Now combine like terms

5 years ago
jimthompson5910 (jim_thompson5910):

Tell me what you get

5 years ago
OpenStudy (anonymous):

2x2 – 2x + 3 ?

5 years ago
OpenStudy (anonymous):

2x^2

5 years ago
jimthompson5910 (jim_thompson5910):

you got it

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

probably the biggest hurdle you had to clear was the jump from f(x) + g(x) to (2x^2 – 4x + 5) + (2x – 2)

5 years ago
jimthompson5910 (jim_thompson5910):

since it's a bit of strange notation

5 years ago
OpenStudy (anonymous):

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)

5 years ago
OpenStudy (anonymous):

i got 4 -___-

5 years ago
jimthompson5910 (jim_thompson5910):

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)

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

idk what im doing over here lol

5 years ago
jimthompson5910 (jim_thompson5910):

that's ok

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

i help strangers all the time

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

go ahead and ask away

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

alright sounds like a great plan

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

thanks I'll keep that in mind

5 years ago
OpenStudy (anonymous):

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?

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

thats not one of the options :(

5 years ago
jimthompson5910 (jim_thompson5910):

what are your options?

5 years ago
OpenStudy (anonymous):

2x2 + x - 3 2x2 + x - 1 - 2x2 + x + 6 -x2 + 2x + 6

5 years ago
jimthompson5910 (jim_thompson5910):

The fraction is \[\Large \frac{2x^2+x-3}{6x}\] They just want the numerator (glossed over that part), so the answer is \[\Large 2x^2+x-3\]

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

it is an option

5 years ago
jimthompson5910 (jim_thompson5910):

it's choice A

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

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 )

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

The idea here is to factor as much as possible. Then cancel out any common terms. \[\Large \frac{y^2 + 2y - 8}{y^2 + y - 12}\] \[\Large \frac{(y+4)(y-2)}{(y+4)(y-3)}\] \[\Large \frac{\cancel{(y+4)}(y-2)}{\cancel{(y+4)}(y-3)}\] \[\Large \frac{y-2}{y-3}\]

5 years ago
OpenStudy (anonymous):

perfect .

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

it should be choice A

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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)

5 years ago
jimthompson5910 (jim_thompson5910):

So the LCD is 3x(x+1)

5 years ago
OpenStudy (anonymous):

your a genius .

5 years ago
jimthompson5910 (jim_thompson5910):

lol why thank you

5 years ago
OpenStudy (anonymous):

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))

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

but I'm glad you're understanding it

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

.........

5 years ago
jimthompson5910 (jim_thompson5910):

which part are you stuck on?

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

how much time do you have?

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

alright, go ahead and ask your last question

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

omg i love you .

5 years ago
jimthompson5910 (jim_thompson5910):

\[\Large \frac{\frac{x^2-x-20}{x-5}}{\frac{4}{10}}\] \[\Large \frac{x^2-x-20}{x-5} \times \frac{10}{4}\] \[\Large \frac{(x^2-x-20)*10}{(x-5)4}\] \[\Large \frac{10(x^2-x-20)}{4(x-5)}\] \[\Large \frac{2*5(x^2-x-20)}{2*2(x-5)}\] \[\Large \frac{2*5(x-5)(x+4)}{2*2(x-5)}\] \[\Large \frac{\cancel{2}*5\cancel{(x-5)}(x+4)}{\cancel{2}*2\cancel{(x-5)}}\] \[\Large \frac{5(x+4)}{2}\] \[\Large \frac{5x+20}{2}\] --------------------------------------------------------- So \[\Large \frac{\frac{x^2-x-20}{x-5}}{\frac{4}{10}}\] simplifies to \[\Large \frac{5x+20}{2}\]

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
jimthompson5910 (jim_thompson5910):

If I'm on OS, then the first two work If I'm not on OS, then you can email me

5 years ago
jimthompson5910 (jim_thompson5910):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
jimthompson5910 (jim_thompson5910):

alright go for it

5 years ago