View Full Version : A little algebra 2 help please

I've been stuck on this for awhile and i can't seem to figure this out. maybe you smart bunch could help me please?

factor x^4 - 13x^2 + 36

factor 27y^3 - 125

i have another one also

use the function f(x) = 2x^3 - 4x^2 - 10x +12

A. explain why x = 3 is a zero factor of f(x). Then find the remainder after x-3 is factored out of f(x)

B. Describe the end behavior.

I hope you guys can help me if you can. I hate factoring -_-

spikejones

16-01-09, 00:56

well.. Ive always found that if I couldnt do it, I had to ask for help. But to just have someone give me the answer - I would never learn it. Of course... if you are content to not know how to do it (as you will likely never need it in real life) just do the best you can and not worry about it. Drop Algebra 2 for an easier course if you're never gonna use the stuff in real life (that is unless it is necessary to have it to graduate - or it is a requisite for a course you want to take towards your future job). That said... Its been almost 7 years since I graduated (in 2002) and the last algebra class I took ended in '01 I believe. After that I too pre-cal which I think I even found to be a bit easier than factoring trinomials - i hated that crap. Anyhoo... if you dont understand it the way your book lays it out (and I certainly can't help you at this point in life without re-taking the class) try looking at different texts that may be able to explain it in a different way that you can understand better. try these sites:

http://www.jamesbrennan.org/algebra/polynomials/factoring_polynomials.htm

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm

(more if you google search "factoring polynomials")

thanx for the link. actually i think i just figured out the first two

factor x^4 - 13x^2 + 36 = (X^2 - 4) (x^2 - 9) ( x-2) (x+2) (x-3) (x+3) am i correct?

(sum and Difference)

factor 27y^3 - 125

3x^3 - 5^3= (3x-5) (4y^2 - 15y + 125) (difference of two cubes)

i think i did these right

that last one though. is still tricky. i must be missing a step/ I'm not really asking for an answer (although that would make it 10 times easier lol) i'm just hoping somebody can explain how to get the answer or how to start out

spikejones

16-01-09, 01:31

thanx for the link. actually i think i just figured out the first two

factor x^4 - 13x^2 + 36 = (X^2 - 4) (x^2 - 9) ( x-2) (x+2) (x-3) (x+3) am i correct?

(sum and Difference)

Just the part here:

( x-2) (x+2) (x-3) (x+3)

it threw me off at first when you had it all on one line, I thought "thats way too big to be correct" - just off the number of X's.

factor 27y^3 - 125

3x^3 - 5^3= (3x-5) (4y^2 - 15y + 125) (difference of two cubes)

im lost on what youre doing there ---- whats with having two different variables in there? If you rewrite your steps on separate lines, its easier for me to see what you are going on about. Then I should be able to better provide feedback.

okay... this part here:

(3x-5)

put you on the right track. you are correct in "difference of two cubes"

I dont know where you came up with the other part though. Its a whole different quadratic equation.

okay... this part here:

(3x-5)

put you on the right track. you are correct in "difference of two cubes"

I dont know where you came up with the other part though. Its a whole different quadratic equation.

difference of two cubes ( a^3 -b^3) = (a^2 + ab + b^2)

i then took 3x and -5 and substitute it into a and b with the equation that is given for it

p.s i just figure out the other problem i had to use synthetic division witha remainder of zero i believe. I would write the problem out, but it is very long lol.

spikejones

16-01-09, 02:03

difference of two cubes ( a^3 -b^3) = (a^2 + ab + b^2)

i then took 3x and -5 and substitute it into a and b with the equation that is given for it

p.s i just figure out the other problem i had to use synthetic division witha remainder of zero i believe. I would write the problem out, but it is very long lol.

i think you are wrong there.

(3x)-(5) = (a^3)-(b^3)

3x doesnt equal a, nor does 5 equal b.

3x IS equal to a^3, and 5 IS equal to b^3

if you do the math by substituting 3x and 5 into the spots of a and b, you wont arrive back at the original polynomial. How did you arrive at (3x-5)? No doubt you took the cube root of the original polinomial right? What does it mean to be a cubed root? What does it mean to cube something?

edit: it think im wrong. sorry.

need to hit the drawing board/

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anyhoo... I thought it was just supposed to be:

(3x-5)(3x-5)(3x-5)

would you not call that correct?

ahh... heres an example (http://www.purplemath.com/modules/specfact2.htm):

The other two special factoring formulas are two sides of the same coin: the sum and difference of cubes. These are the formulas:

a^3 + b^3 = (a + b)(a^2 – ab + b^2)

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

Factor x^3 – 8

This is x^3 – 2^3, so I get:

x^3 – 8 = x^3 – 2^3

= (x – 2)(x^2 + 2x + 2^2)

= (x – 2)(x^2 + 2x + 4)

but then wouldnt you need to factor out the second part?

sorry for the exponent part not coming out right.

CerebralAssassin

16-01-09, 02:19

difference of two cubes ( a^3 -b^3) = (a^2 + ab + b^2)

i then took 3x and -5 and substitute it into a and b with the equation that is given for it

p.s i just figure out the other problem i had to use synthetic division witha remainder of zero i believe. I would write the problem out, but it is very long lol.

a=3x, so you have a^3-b^3=(3x)^3-(5)^3=(3x-5)(9x^2+15x+25)

:)

spikejones

16-01-09, 02:23

there ya go^ with corrections. the original equation you wrote out was wrong-O. But I dont see why you need to do that much when it should just be (imo)

(3x-5)(3x-5)(3x-5) or just simply (3x-5)^2

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see... told you i couldnt hardly help at this point in my life. got me brushing off some freakin cob-webs.

ooooops sorry. made a mistake there lol. wait yes its 9x^2 but it wouldn't be +15 it would be - 15 right? because 3x x -5 would make it -15x correct?

o yes i do see your point. i don't know. i was looking through my algebra 2 book and i noticed that equation specifically made out the way i did it.my teacher is kinda picky, so i did it that way lol

spikejones

16-01-09, 03:18

yeah... you're correct. it IS -15 rather than positive. good catch. and I'll just have to trust that the mathemeticians know what they are doing. hehe.. it should multiply back out to the original polynomial. as per my crack pot idea of cubing the term (3x-5)... its a bust. I wrote it out on paper and it came out like:

welll .. NVM. i threw it out already. :vlol: lets just say it was wrong-o.

CerebralAssassin

16-01-09, 03:24

ooooops sorry. made a mistake there lol. wait yes its 9x^2 but it wouldn't be +15 it would be - 15 right? because 3x x -5 would make it -15x correct?

o yes i do see your point. i don't know. i was looking through my algebra 2 book and i noticed that equation specifically made out the way i did it.my teacher is kinda picky, so i did it that way lol

(3x-5)(9x^2+15x+25)

no man,it's a +.just expand everything in the expression I gave you,you should end up with the original expression :)

oh ahahaha i see. i'm not much a math person. never my strongest suit. i actually enjoy algebra 2 except for factoring lmao. i just hate it with a passion. thank you spikejones and cerebralassassin for all your help.

spikejones

16-01-09, 03:34

no man,it's a +.just expand everything in the expression I gave you,you should end up with the original expression :)

doh!! :hea::hea:

now i look back again and see it. right you are. methinks math is not my strong suit anymore. glad im not still in highschool.

freeze10108

16-01-09, 03:37

i actually enjoy algebra 2 except for factoring lmao. i just hate it with a passion.

For future reference, mathway.com (http://www.mathway.com) could probably lessen that hate.

PS: I would like to have helped you out too, but my computer locked up while I was typing.

CerebralAssassin

16-01-09, 03:42

doh!! :hea::hea:

now i look back again and see it. right you are. methinks math is not my strong suit anymore. glad im not still in highschool.

hehe,that's ok.:)I get stumped in math too lots of times.and in other subjects I suck monkey balls :vlol:

i still don't understand how it is +15. i mean wouldn't the negative 5 make it a negative. or do i omit the negative sign?

spikejones

16-01-09, 04:02

thats because it is (a) - (b) x (a^2 + ab + b^2)

the minus sign isnt a negative indicator in this case. your A is equal to 3x, and your B is equal to 5 (not negative 5) if the five was negative, you would end up with (3x+5) instead of (3x-5).

oh ok. i was confused there. so you do omit the negative sign. took me awhile to figure that out lol. i hate problems like these. they always used to tell me when there is a negative sign next to it use it. now you don't use it lol

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