here are the answers: a. (3x-1)(9x^2+3x+1) b. (3x-1)(9x^2-3x-1) c. (3x-1)^3 d. (3x-1)(9x^2-3x+1)
2007-02-05 07:57:47 · 6 answers · asked by elona 1 in Science & Mathematics ➔ Mathematics
The correct answer is a! (3x-1)(9x^2-3x-1)
2007-02-05 08:08:31 · answer #1 · answered by Behy_unique 2 · 0⤊ 0⤋
the correct answer is a.(3x-1) (9x^2+3x+1)
2007-02-05 08:05:30 · answer #2 · answered by yosunn88 1 · 0⤊ 0⤋
a: is the correct answer: (3x - 1 ) (9^2 + 3x +1)
2007-02-05 08:06:03 · answer #3 · answered by Miss LaStrange 5 · 0⤊ 0⤋
27x^3 - 1
This is a difference of cubes.
To factor a difference of cubes, you have two sets of brackets.
(? + ?) (? + ? + ?)
In the first set of brackets, take the cube root of each term. Notice that the cube root of 27x^3 is 3x, and the cube root of -1 is -1. Therefore, we place 3x and -1 into the first set of brackets.
(3x - 1) (? + ? + ?)
In the second set of brackets, you have to do the following steps:
(a) Square the first
(b) Negative product
(c) Square the last
"Square the first" means to take the square of the first value in the first set of brackets. In this case, it's 3x. [3x]^2 is equal to 9x^2.
(3x - 1) (9x^2 + ? + ?)
"Negative product" means to take the product of both the values within the first set of brackets, and then multiply the product by -1. In this case, we have 3x and (-1). Multiplied together, they give you -3x. Multiply this by -1, and you get 3x (without the minus sign this time).
(3x - 1) (9x^2 + 3x + ?)
"Square the last" means to square the last value in the first set of brackets. In this case, we're squaring (-1), and (-1) squared is 1.
(3x - 1)(9x^2 + 3x + 1)
2007-02-05 08:05:46 · answer #4 · answered by Puggy 7 · 0⤊ 0⤋
Let y = 3X. Factor y^3 - 1. Substitute for y.
2007-02-05 08:06:03 · answer #5 · answered by Curt Monash 7 · 0⤊ 0⤋
(3x - 1)(9x² + 3x + 1)
2007-02-05 08:03:38 · answer #6 · answered by Philo 7 · 0⤊ 0⤋