factor completely: 3x to the third power- 24
- Thread starter art2ista
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\(\displaystyle \L 3x^3 - 24\)
Start by factoring out a 3
\(\displaystyle \L 3(x^3 - 8)\)
now use the difference of cubes theorm to factor into a binomial and a trinomial:
\(\displaystyle \L a^3-b^3 = (a-b)(a^2+ab+b^2)\)
now what to the third power, equals 8? 2^3 = 8, so we can write this as
\(\displaystyle \L 3(x^3 -2^3)\)
now use the difference of cubes, as I showed, above, and factor:
You end up with:
\(\displaystyle \L 3(x-2)(x^2+2x+4)\)
And that's your answer!
See how I did it? In this case, a = x and b = 2
Start by factoring out a 3
\(\displaystyle \L 3(x^3 - 8)\)
now use the difference of cubes theorm to factor into a binomial and a trinomial:
\(\displaystyle \L a^3-b^3 = (a-b)(a^2+ab+b^2)\)
now what to the third power, equals 8? 2^3 = 8, so we can write this as
\(\displaystyle \L 3(x^3 -2^3)\)
now use the difference of cubes, as I showed, above, and factor:
You end up with:
\(\displaystyle \L 3(x-2)(x^2+2x+4)\)
And that's your answer!
See how I did it? In this case, a = x and b = 2
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- Feb 4, 2004
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- 16,582
Which part didn't you get? Are you not understanding the prerequisite material (how to factor out the 3)? Or did you forget to memorize the difference-of-cubes formula? Or something else? (And why can't skeeter, an able tutor and a nice guy, not provide this further assistance?)art2ista said:
We'll be glad to help, provide hints, give you links to lessons, etc (because most tutors won't give you the complete solution, preferring instead that you learn how to stand on your own). I'm afraid that, when you've been given a fairly straightforward and complete reply, saying "i still dont get it" simply doesn't give us enough to work with.
Thank you for your understanding.
Eliz.