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# Method for factoring equations with large powers

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Joined: 20 May 2009
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Method for factoring equations with large powers [#permalink]

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25 Jul 2009, 06:05
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Does anyone have a method for factoring equations with large powers

e.g. X^16 – Y^8 + 345Y^2

Thanks,

Aztec

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Manager
Joined: 18 Jul 2009
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Location: India
Schools: South Asian B-schools
Re: Method for factoring equations with large powers [#permalink]

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25 Jul 2009, 12:01
Aztec wrote:
Does anyone have a method for factoring equations with large powers

e.g. X^16 – Y^8 + 345Y^2

Thanks,

Aztec

PLZ post ur Question then it will be appropriate to see in that context
_________________

Bhushan S.
If you like my post....Consider it for Kudos

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Manager
Joined: 03 Jul 2009
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Location: Brazil
Re: Method for factoring equations with large powers [#permalink]

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25 Jul 2009, 12:49
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Aztec, I would give you three suggestions:

First: Memorize those products with notable results:
$$(x + a)^2 = x^2 + 2ax + a^2$$
$$(x - a)^2 = x^2 - 2ax + a^2$$
$$(x - a)(x + a) = x^2 - a^2$$

You need not only to memorize them, but also to recognize the inverted process. For example, you must be so familiar with those equations, that when you see something like
$$x^2 + 14x + 49$$, you know that in fact this can be $$(x+7)^2$$.

Second: Zeros of the equations:
Remember that the zeros of the equation, are the opposite terms inside the parentheses of those equations.
For instance, if you see the equation $$m^2 - 3m -28$$, you can discover the zeros, which are $$-4$$ and $$7$$, and reconstruct the equation with the opposite of the solutions: $$4$$ and $$-7$$
Thus $$(x + 4)(x - 7)$$

Third: Recognize equal terms.
If you see the equation $$3z^5 + 36z^4 - 84z^3$$, you can notice that $$z^3$$ is a common term in all the terms. Additionally, all the coefficients are divisible by $$3$$. Thus the term $$3z^3$$ is in all the terms of the equation. If you take off this term, the new equation is:
$$3z^3 (z^2 + 12z - 28)$$
You can go farther, and notice that the equation inside the parenthesis has the zeros $$2$$ and $$-14$$. Using the second advice listed you will have
$$3z^3 (z - 2) (z +14)$$, much more simple than $$3z^5 + 36z^4 - 84z^3$$
Now, coming back to your equation, we could rewrite as $$X^16 - Y^2(Y^6-345)$$. But there are other possibilities. If this is above some fraction, or even in some other context, it is a little bit easier to realize what factoring we should do.

Good studies!!!

PS.: If you liked the explanation, consider a kudo. I want to access those GMATClub tests!!!

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Intern
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Re: Method for factoring equations with large powers [#permalink]

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27 Jul 2009, 14:05
Thanks a lot coelholds
These tips are useful.

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Manager
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Posts: 106

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Location: Brazil
Re: Method for factoring equations with large powers [#permalink]

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27 Jul 2009, 16:28
You are welcome

Kudos [?]: 93 [0], given: 13

Re: Method for factoring equations with large powers   [#permalink] 27 Jul 2009, 16:28
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# Method for factoring equations with large powers

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