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bloodycolouredrose
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So, I've always been fairly decent at factoring. But the answer key for a standard (4-t)^2 isn't making sense. Per the rule - (a-b)^2 should be: a^2 - 2ab + b^2, correct? It follows then that (4-t)^2 should equal 16 - 8t + t^2, however the answer key indicates the answer it is looking for is 16 - 4t + t^2. Am I missing something? The original problem is as follows: Reduce to the lowest term: [math]64 + t^3 / 16 - 4t + t^2[/math] Reducing to the lowest terms SHOULD bring you to (4+t)(4-t)^2 over 16 - 4t + t^2 - except that (4-t)^2 should be 16 - 8t +t^2, not 16 - 4t + t^2 - correct? According to the answer key, the question is solved this way: [math]64 + t^3 / 16 - 4t + t^2[/math]
=
[math](4+t)(16 - 4t + t^2) / 16 - 4t + t^2[/math]
=
[math](4+t)[/math] as the final reduction. This is not the only question it shows this as. Another has (x^3-8) = (x-2)(x^2 +2x +4).
Is there something I'm missing when factoring cubed binomials? Or could it be possible that my answer key is incorrect? I've been bashing my brains out on these for most of the day. (sorry it's not more complex guys - I'm a re-entry into college and had to start all over at the age of 30, so I'm only in Intermediate Algebra after over a decade out of school.) I have read and re-read my notes and the entire chapter on factoring so many times today I've lost track. I do plan on asking my professor this week, but he's got an extremely thick accent and takes FOREVER to explain anything, that I'm afraid he'll lose me after the first couple of minutes (or, as typical, misunderstand the question and take forever to explain something completely different!).

Emily

phlip
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Your expansion of (4-t)2 is correct... but your factorisation of 64 + t3 is not. Try expanding (4+t)(4-t)2 and see what you get... because it's not 64 + t3.

The factorisation rule is: (a3 + b3) = (a + b)(a2 - ab + b2) - note that there isn't a 2 on the ab term. Try expanding that out too, to check.
Last edited by phlip on Mon Oct 12, 2009 6:41 am UTC, edited 1 time in total.

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bloodycolouredrose
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Thank you. For some unknown reason, my algebra text has the factorization of cubes rules in the NEXT chapter instead of the one we are working on or any prior (I looked after you replied). This makes infinitely more sense now!

Problem solved.

Although - can someone explain WHY the cube factorization does not have a 2 on the middle term? ie: (a^3+b^3) = (a+b)(a^2 - ab + b^2). Skimming through the chapter, it doesn't explain the why.... If not, I won't worry about it and just accept the rule as fact.

Thanks again!

Kalathalan
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bloodycolouredrose wrote:Although - can someone explain WHY the cube factorization does not have a 2 on the middle term? ie: (a^3+b^3) = (a+b)(a^2 - ab + b^2). Skimming through the chapter, it doesn't explain the why.... If not, I won't worry about it and just accept the rule as fact.

If you've learned synthetic division, (polynomial division) try dividing the polynomial [imath](x^3+b^3)[/imath] by the term [imath](x+b)[/imath].

Alternatively, just multiply it out and check. With [imath](a+b)(a^2 - ab + b^2)[/imath], things cancel nicely, but with [imath](a+b)(a^2 - 2ab + b^2)[/imath], they wouldn't.
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Talith
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I've always just stuck to remembering that the difference of two nth exponents a^n - b^n always has a factor (a-b) and then do the necessary polynomial long division.

PM 2Ring
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Talith wrote:I've always just stuck to remembering that the difference of two nth exponents a^n - b^n always has a factor (a-b) and then do the necessary polynomial long division.
And I just remember the second factor is the sum of a geometric progression.

Yakk
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I'm not sure what to make of your question "why doesn't it have this term".

Do you mean why is the pattern different from what you expect (in which case, why do you expect your pattern to be true)? Or why is it -- that one statement -- true?

It is true because factoring is the opposite of multiplication -- X factors into Y*Z if and only if Z*Y multiplies out to X. And clearly this works in this case.

Now, note that you can do long division of your terms without any patterns.

t^2 -4t +16 dividing into t^3 + 0 t^2 + 0 t + 64
Take the highest term -- the t^2 -- on the left, and the highest term -- t^3 -- on the right. Well, t * (t^2) = t^3, so we start with that:

[t^3 + 0 t^2 + 0 t + 64] - [t^2 -4t +16] * t
= [t^3 + 0 t^2 + 0 t + 64] - [t^3 -4t^2 +16t]
= [4t^2 -16t +64]

Now do the same by dividing t^2 -4t +16 into 4t^2 -16t +64. Look at the highest coefficient -- t^2 and 4t^2. 4*t^2 = 4t^2.
[4t^2 -16t +64] - [t^2 - 4t +16] * 4
= [4t^2 -16t +64] - [4t^2 -16t +64]
= 0

Now why did we do this? Because we are doing long division. Remember long division?

[t^3 + 0 t^2 + 0 t + 64] - [t^2 -4t +16] * t = [4t^2 -16t +64]
[4t^2 -16t +64] - [t^2 - 4t +16] * 4 = 0
Perform the italic substitution and get this:
[t^3 + 0 t^2 + 0 t + 64] - [t^2 -4t +16] * t - [t^2 - 4t +16] * 4 = 0
Now, group the [t^2 -4t +16] terms:
[t^3 + 0 t^2 + 0 t + 64] - [t^2 -4t +16] * (t + 4) = 0
Turn it into an equality:
[t^3 + 0 t^2 + 0 t + 64] = [t^2 -4t +16] * (t + 4)
Now divide by t^2 -4t +16:
[t^3 + 0 t^2 + 0 t + 64]/ [t^2 -4t +16] = (t + 4)
Drop the zero terms:
[t^3 + 64]/ [t^2 -4t +16] = (t + 4)
See how that works?

What I just did there was long division, but I did it backwards. As it happens, in this case, the remainder was 0, so the division was exact.

If you have (x^3 + 7) / (x+1), we can do the same thing:

x^3 + 7 - x^2(x+1) = -x^2 + 7
-x^2 + 7 + x(x+1) = x + 7
x+7 - 1(x+1) = 6
Each time, I am taking the remainder of the previous step, and dividing it again by the denominator.
We then get this:
(x+1)(x^2 - x + 1) = (x^3 + 7) - 6
which we can then check (my multiplying) to make sure we didn't do a mistake.

(x^2 - x + 1) = (x^3 + 7)/(x+1) - 6/(x+1)

In this case, we ended up with a remainder of 6. Had we ended up with a remainder of 0, it would have looked like the first case. Does this make sense?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

bloodycolouredrose
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Joined: Mon Oct 12, 2009 5:58 am UTC

Yes, it makes sense now. It's just that when I was initially trying to factor it out, I was trying (unsuccessfully) to factor x^3+y^3 as (x+y)(x-y)^2. Which made sense at the time, until I got further into the problems. Tonight my prof addressed factoring cubed binomials as soon as he walked in the door. It'd be nice to get the instruction on something within the homework at the same time that the homework is assigned. Thankfully, I only have to redo two pages instead of eight....

Thank you to everyone for your assistance!

~emily

Yakk
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So, I don't think you get it.

Your misremembered factoring doesn't work, because you can check it without being taught anything -- just multiply it out.

You can solve this question without a memorized formula for facatoring the sum of cubes -- polynomial division works quite fine, and gives the result you want.

Most, if not all, memorized tricks in mathematics isn't what mathematics is about -- those are just particularly pretty gems that stand out and can be useful short cuts. At the moment you factored x^3+y^3 = (x+y)(x-y)^2, you screwed up, because (x+y)(x-y)(x-y) = (x^2-y^2)(x-y) = x^3 - x^2y - xy^2 + y^3, which is a quite different value; and you didn't need to be taught this to figure this out, you just needed to check your step when you where not absolutely certain how to factor the sum of polynomial cubes.

Thinking "x^3 + y^3 -- maybe that is (x+y)(x-y)^2" -- that was a good step. Not checking it? That is where you made your fundamental mistake.

In short: you had everything you needed to solve the question when it was assigned. The professor did tell you a short cut after the question was assigned that happens to make the problem a tad easier. And when you come up with a "maybe this will work", check it -- the most beautiful part of mathematics is the ability to check your (or someone else's) work.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.