Danny Ingram
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Can this polynominal be factored further than what I have.
The problem 8x^3 - 27.
I can only think of 2x^3 - 3^3
The problem 8x^3 - 27.
I can only think of 2x^3 - 3^3
Factoring: Is 8x^3 - 27 factorable?
- Thread starter Danny Ingram
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In the future, please post corrections to the original thread, rather than starting a new thread on the same exercise. Thank you.
To factor this difference of cubes, apply the difference-of-cubes formula, a<sup>3</sup> - b<sup>3</sup> = (a - b)(a<sup>2</sup> + ab + b<sup>2</sup>), from your book and/or class notes. (Note: 8 does not equal 2, so 2x<sup>3</sup> cannot equal 8x<sup>3</sup>.)
Before the next test, it would probably be wise to memorize this and related formulas.
Eliz.
To factor this difference of cubes, apply the difference-of-cubes formula, a<sup>3</sup> - b<sup>3</sup> = (a - b)(a<sup>2</sup> + ab + b<sup>2</sup>), from your book and/or class notes. (Note: 8 does not equal 2, so 2x<sup>3</sup> cannot equal 8x<sup>3</sup>.)
Before the next test, it would probably be wise to memorize this and related formulas.
Eliz.
arthur ohlsten
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8x^3-27=0
how many roots 3
x^3=27/8
x=3/2
let x=3/2
8[x^3-27/8] / [x-3/2]=8[x^2+3/2x+9/4]
8x^3-27 factor out 8
8[x^3-27/8] factor out [x-3/2]
8[x-3/2][x^2+3/2x+9/4] answer for real factors answer
x^2+3/2x +9/4
x=-3/2+/-[9/4-9]^1/2 all over 2 not real factors
Arthur
how many roots 3
x^3=27/8
x=3/2
let x=3/2
8[x^3-27/8] / [x-3/2]=8[x^2+3/2x+9/4]
8x^3-27 factor out 8
8[x^3-27/8] factor out [x-3/2]
8[x-3/2][x^2+3/2x+9/4] answer for real factors answer
x^2+3/2x +9/4
x=-3/2+/-[9/4-9]^1/2 all over 2 not real factors
Arthur