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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
e.g. X^16 – Y^8 + 345Y^2
Thanks,
Aztec
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25 Jul 2009, 12:01
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Aztec
PLZ post ur Question then it will be appropriate to see in that context
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!!!
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!!!
