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Old 10-12-2008, 02:20 PM   #1
slothonaleash
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honors precalculus question

Hey I have a graded set due in precalc tomorrow and im stuck on one of the questions. The problem is that ive never had to expand multiple binomials at the same time, and im rather confused. I don't need a solution to the problem, I'd just like to know how to go about starting it. Thanks.

Problem: When ((x-a)^3)((x-b)^6) is completely expanded, the coefficient of x^7 is -9 and there is no x^8 term. Algebraically solve for the values of a and b.
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Old 10-12-2008, 02:24 PM   #2
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wow precal on zilvia. Thats a first. I use to know this stuff but forgot it once i figured that hardly anyone uses this in the real world. But good luck my man!
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Old 10-12-2008, 02:25 PM   #3
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Old 10-12-2008, 02:27 PM   #4
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lollllsfsfsfsfsfsf
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Old 10-12-2008, 02:33 PM   #5
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hahahah

zilvia: helpful for mathematical questions
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Old 10-12-2008, 02:43 PM   #6
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LMAO picture is just too funny, wish I could help you though don't remember precal anymore.
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Old 10-12-2008, 02:51 PM   #7
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Old 10-12-2008, 11:29 PM   #8
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(x-a)(x-a) (x-a) Foil out the first pair, then multiply by (x-a) again.

(x^2-2ax+a^2)(x-a) = (x^3-2ax^2+a^2x) from distributing the x through, then distribute the -a through and get (-ax^2+2a^2x-a^3)
(x-a)^3=x^3-3ax^2+3a^2x-a^3

now get ready to do the same for (x-b)^6. Just cause I'm nice I'll expand that one, but I don't want to go through each step, I'd probably foil it 3 times then start through the tedious shit. There's probably some kind of formula for expanding these, but my calculator does things for me. (x^6-6bx^5+15b^2x^4-20b^3x^3+15b^4x^2-6B^5x+b^6)

Then you go and get extra tedious, multiply everything from the (x-a)^3 to all that. you get

x^9-3ax^8-6bx^8+3a^2x^7+18abx^7+15b^2x^7 etc... those are the only terms you need to answer your question so I would stop there for expanding.

so when you combine the x^8 terms you should end up with 0*x^8 so -3a-6b=0 and for the x^7 you get3a^2+18bx+15b^2=-9

Now I'm tired and its pretty easy to go at it from there anyways.

And don't feel bad, I don't think I knew how to expand anything bigger than (a+-b)^2 until sometime in calculus. I understood "FOIL" really well, but I didn't realize that you just multiply everything in one part of the expression by everything in the other part.
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Old 10-13-2008, 01:36 AM   #9
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^wow. yeah.

the answer is 43.67 ....ok not really. but dan there got to it. stupid math....good old business math can be done by calculators..no hard using of the brain stuff.
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Old 10-13-2008, 10:27 AM   #10
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Algebra and lots of calculus can be done on calculators too. TI-89 was a great friend in pre-calc and calculus during high school. TI-84 plus got me through calc 1 last semester, its a big help in calc 2 but to really save time I got the TI-89 all over again. Some of the big stuff slows it down but it can do a LOT for you. for high school the 92 and voyage 200 aren't practical because you can't use them on any of the big tests.

Just don't get to the point where you need the calculator to do everything, there's tests with no graphing calculator allowed. In those cases a scientific calculator that can give you exact answers is awesome. I use a casio fx-115es. Does definite integration and will give you values for a derivative at specifc points on a function.
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Old 10-13-2008, 10:31 AM   #11
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Originally Posted by slothonaleash View Post

Problem: When ((x-a)^3)((x-b)^6) is completely expanded, the coefficient of x^7 is -9 and there is no x^8 term. Algebraically solve for the values of a and b.
7. That is my default math answer.

That should help.
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Old 10-13-2008, 03:27 PM   #12
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7. That is my default math answer.

That should help.
lol I had a teacher last semester for physics 3, this guy 90% of the questions he would ask in class the answer was zero. lol people would be stumped and I would be like zero and he would go correct!
hahaha

but looking back at precalc, its funny how much diffy q calls on all your previous math, so understand this shit if your trying to go far in math.
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