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02-18-2004, 07:40 PM #1Web Hosting Master
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Algebra Question: How do you factor this?
In my Algebra 2 class we had a problem that not even the teacher knew the method that was used to factor down the problem.
Here's the final step before the factoring;
0 = X3 – 18x2 + 107x - 216
Now this is what the book said the answer was;
0 = X3 – 18x2 + 107x - 216
0 = (X-8)(X2-10x+27)
0 = X-8
X = 8
So, could anyone help as to figuring out what method was used, and how the problem was factored down?
And yes, I couldn't make any of the text superscript, so I made the exponents smaller and red.
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02-18-2004, 07:43 PM #2Web Hosting Master
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looking through my notes...
Isn't it just grouping? I'll solve it, but it looks like regular grouping.
Nope... I'll work on it
TO THE POST UNDER ME: Doesnt fit into the quadratic equation.Last edited by Xshare; 02-18-2004 at 07:50 PM.
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02-18-2004, 07:47 PM #3Tells All!
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Nope, I was wrong I'll have a look at it properly now!
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02-18-2004, 07:48 PM #4Web Hosting Guru
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You would have to substiture X with numbers so you equation would be equal to 0. In this case, 8 is the number, when you plug in 8 in place of X, that would wind up to 8^3 - 18(8)^2 + 107(8) - 216 = 0.
Then you divide the given equation: X3 – 18x2 + 107x - 216
by (x-8), which would lead to X2-10x+27.
Hope that helps, If you need more explantion, I would be more than glad to help you out.
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02-18-2004, 07:50 PM #5Tells All!
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Del, that method is only good if you know the answer
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02-18-2004, 07:51 PM #6Web Hosting Master
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Yeah, that's just checking to make sure that the answer is correct. I need to figure out how they got that answer.
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02-18-2004, 07:54 PM #7Web Hosting Guru
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Originally posted by Joseph_M
Del, that method is only good if you know the answer
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02-18-2004, 08:04 PM #8Web Hosting Master
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Hmm.. well 107 is prime. I'm always hitting that stumbling block.
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02-18-2004, 08:05 PM #9Tells All!
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Easy numbers, we're talking it can be done reasonably quickly up to 4 D.Ps, but its time-consuming for anything over that, there is a formula to work it out with, I just don't have it to hand, I'll ask my mathematician friend, for he shall surely know the formula, or will have it on a piece of paper at his place.
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02-18-2004, 08:09 PM #10Registered User
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You're still using the FOIL method except for a longer equation...
X3 – 18x2 + 107x - 216
0 = (x - 8) (x2 - 10x + 27)
F: x * x2 = x3
O: x * 27 = 27x
O: x * -10x = -10x2
I: -8 * 10x = 80x
I: -8 * x2 = -8x2
(Add your outer/inners together to get +107x and -18x2)
L: 8 * 27 = 216
Put it back together and you get...
X3 – 18x2 + 107x - 216
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02-18-2004, 08:14 PM #11Web Hosting Master
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Thats still only knowing the answer.. we're talking without knowing the answer.
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02-18-2004, 08:23 PM #12Web Hosting Guru
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(x^3-y^3)=(x-y)(x^2+xy+y^2)
(x+y)^3=x^3+3x^2y+3x^2y+y^3
(x-y)^3=x^3-3x^2y+3x^2y-y^3
you don't use that much foiling when you get to pre-calc/calc. just understand the major concepts.
after this ask the teacher this:
if (x)^2=(y)^2, then x=y
(1)^2=2
(-1)^2=2
2=2, therefore (1)^2=(-1)^2), therefore ...
1=-1?
good luck.
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02-18-2004, 08:28 PM #13Web Hosting Master
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Umm.. what y?
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02-18-2004, 08:29 PM #14Web Hosting Master
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0 = X3 – 18x2 + 107x - 216
Factor out one x
x (x2-18x+107-216)
Combine like terms
x (x2-18x-109)
Take half of the middle term, then square it, add it in the middle, then subtract it on the end
x (x2-18x+81) -81-109
Ask, what multiples to get 81 and adds to get -18?
x (x-9)(x-9) - 190
190 = x(x-9)(x-9)
Hmm... that didn't work... maybe this is a synthetic division problem... or maybe you should graph it and just find the zeros...Last edited by brandonk; 02-18-2004 at 08:39 PM.
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02-18-2004, 08:31 PM #15Web Hosting Master
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Originally posted by ToaD
2=2, therefore (1)^2=(-1)^2), therefore ...
1=-1?
good luck.
TO BRANDON:
No, you messed up. You can't factor an X out of 216. I've found out how you get the answer, I just need to work out the specifics.Ask about custom logo design! :: TalkSka.com - Your source for Ska News and Forums. Join today!
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02-18-2004, 08:32 PM #16Web Hosting Master
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Use synthetic prositution oh wait I mean subsitution, it works every time.
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02-18-2004, 08:33 PM #17Web Hosting Master
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Originally posted by brandonk
x (x2-18x+107-216)
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02-18-2004, 08:38 PM #18Web Hosting Master
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using synthetic subsititution, sure it would, you just have to know how to use it.
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02-18-2004, 08:50 PM #19Web Hosting Master
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Basically, you need to factor out 18x2 + 107x - 216 first. I know theres a solution, i just am too lazy to find it.
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02-18-2004, 08:53 PM #20Newbie
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Originally posted by ToaD
if (x)^2=(y)^2, then x=y
x does not necessarily equal y, as you've shown in your example. if x is -1, and y is 1, then x^2=y^2=1. This proves either that you either missed the logic, or gave us an absurdly easy brainteaser.
It seems that the superscript tag for vBcode has been disabled. Any particular reason for that?
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02-18-2004, 08:54 PM #21Web Hosting Master
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Originally posted by Xshare
Basically, you need to factor out 18x2 + 107x - 216 first. I know theres a solution, i just am too lazy to find it.
Send a complaint to the author of the book damn it! They charge so much for those books, they should be perfect!
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02-18-2004, 08:57 PM #22Web Hosting Master
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Nevermind, you can't it's unfactorable.
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02-18-2004, 08:58 PM #23Web Hosting Master
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Originally posted by Reuven
if x^2=y^2,
x does not necessarily equal y, as you've shown in your example. if x is -1, and y is 1, then x^2=y^2=1. This proves either that you either missed the logic, or gave us an absurdly easy brainteaser.
It seems that the superscript tag for vBcode has been disabled. Any particular reason for that?
2^x=2^y then x=y
If the bases are the same, then the exponents drop down and equal each other... but the bases don't equal each other
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02-18-2004, 09:32 PM #24Web Hosting Master
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Re: Algebra Question: How do you factor this?
Originally posted by inogenius
0 = X3 – 18x2 + 107x - 216
0 = (X-8)(X2-10x+27)
0 = X-8
X = 8
Let f(x)=x^3 - 18x^2 + 107x - 216
You need to realize that the root of f(x) will make f(x)=0. Thus you can plot the graph on your graphing calculator and see at what x values does f(x) = 0. Say, f(x) = 0 at three values, x=a, b, c then factors of f(x) = (x-a)(x-b)(x-c).
Alternatively, if you do not have graphing calculator or if it's not allowed, then you can try plugging numbers into the original equation f(x), say x=1. Does f(1) = 0? If not, then (x-1) is not a factor of f(x). Try x=2. Does f(2) = 0? If not, then (x-2) is not a factor of f(x). You will see that x=8 will make f(x) = 0. Thus (x-8) is one of the factor of f(x). After this, find g(x)=f(x)/(x-a). Then find a factor of g(x) with the same method (find x such that g(x)=0).
But once g(x) has been reduced to a polynomial of degree 2, then it's very easy to find its remaing roots/factors of f(x).Last edited by FHDave; 02-18-2004 at 09:43 PM.
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02-18-2004, 09:37 PM #25Web Hosting Master
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Originally posted by brandonk
0 = X3 – 18x2 + 107x - 216
Factor out one x
x (x2-18x+107-216)Fluid Hosting, LLC - Enterprise Cloud Infrastructure: Cloud Shared and Reseller, Cloud VPS, and Cloud Hybrid Server