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Question 134877This question is from textbook
: Factor:
3x^3+81
This question is from textbook
Answer by nycsharkman(136)   (Show Source):
You can put this solution on YOUR website! Factor:
3x^3+81
We have the sum of cubes.
We use this formula:
x^3 + y^3 = (x+y)(x^2-xy+y^2)
First thing you want to do is factor out 3.
3(x^3 + 27)...In other words, 3x^3 divided by 3 = x^3 and 81 divided by 3 = 27 (in case you are wondering where those numbers came from).
We now factor the sum of cubes listed in the parentheses.
Write 27 using base 3.
3^3 = 27
We now have:
3(x^3 + 3^3)
Base x and 3 = base x and y in the above formula.
Here comes the easy part.
We plug and chug.
x^3 + y^3 = (x+y)(x^2-xy+y^2)....This is the formula for the sum of cubes.
3(x^3 + 3^3) = 3(x + 3)(x^2 - 3y + 3^2)
3(x^3 + 3^3) = 3(x + 3)(x^2 - 3y + 9)
The original question actually breaks into 3 factors:
3, (x + 3) and (x^2 - 3y + 9)
Final answer:
3(x + 3)(x^2 - 3y + 9)
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