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OpenStudy (anonymous):
OpenStudy (anonymous):
Do I use the foil method... or?
OpenStudy (anonymous):
you should know that (a+b)(a-b)=a^2-b^2.
now try to solve the question.
OpenStudy (anonymous):
for (t^2-n^2) do I simplify it to (t-n)(t+n) ?
Directrix (directrix):
Multiply (t^2 - n^2) times (t^2 - n^2) and match the simplified product with one of the answer options. @RH
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OpenStudy (rane):
|dw:1373192972923:dw|
OpenStudy (anonymous):
@RANE its not (t-n) on the second one, its (t+n)
Directrix (directrix):
@RANE t^4 - n^4 is NOT correct. Try again. You lost the middle term.
(t^2 - n^2) times (t^2 - n^2) = ?
OpenStudy (rane):
there r no bracket once u simplify the ist 2 bracket directrix
OpenStudy (rane):
ops my bad RH
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OpenStudy (rane):
NO W8!
OpenStudy (anonymous):
I keep getting wrong answers!!!:( can someone explain?!
OpenStudy (rane):
@RH it becomes -ve in the 2nd equation becoz when u times -ve and +ve = -ve
OpenStudy (rane):
so there is no +ve left
Directrix (directrix):
@RH
The original problem is (t-n)(t+n)(t^2-n^2).
As some body above posted, the product of the first two is (t^2 - n^2).
So, (t-n)(t+n)(t^2-n^2) = (t^2 - n^2) (t^2 - n^2)
So, @RH, crank out (t^2 - n^2) (t^2 - n^2) and that's the answer.
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OpenStudy (rane):
hun?
Directrix (directrix):
@RANE What is a "+ve "
OpenStudy (anonymous):
I got t^4 - t^2 n^2 - n^2 t^2 - n^4
OpenStudy (anonymous):
@Directrix
OpenStudy (anonymous):
Is A correct?
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