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LMande
07-07-2006, 10:54 PM
y^3 - 64 / y-4

My guess is that it's simplified? Am I right or wrong?

stapel
07-07-2006, 11:39 PM
y^3 - 64 / y-4
The above means this:

. . . . .\L y^3 \,-\,\frac{64}{y} \,-\, 4

Is this what you meant? Or did you mean one of the following:

. . . . .\L y^3 \,-\, \frac{64}{y\,-\,4}

. . . . .\L \frac{y^3\,-\,64}{y\,-\,4}

...or something else?

Thank you.

Eliz.

skeeter
07-08-2006, 09:38 AM
please use proper grouping symbols ...
(y^3 - 64)/(y - 4)

y^3 - 64 = y^3 - 4^3, is the difference between two cubes ... which does factor.

fyi ... the factoring pattern is a^3 - b^3 = (a - b)(a^2 + ab + b^2)

also, the sum of two cubes is also factorable ...
a^3 + b^3 = (a + b)(a^2 - ab + b^2)

place these two algebra factoring facts into your algebra "toolbox" ... you will see them again.

LMande
07-08-2006, 10:19 AM
Okay... so I'm REALLY lost!

Eliz... it is the last one that you listed that's my question. I'm going to attempt to write it again (with brackets) and I hope it makes sense.

(y^3 - 64) / (y-4)

I hope that makes more sense.

pka
07-08-2006, 10:28 AM
\L
y^3 - 64 = \left( {y - 4} \right)\left( {y^2 + 4y + 16} \right)