waiting for a quick respond need urgent help... factoringg!!
2006-08-28 16:39:20 · 7 answers · asked by ksk_gt 1 in Science & Mathematics ➔ Mathematics
(3x-5)(9x^2-3x+7)
thats the answere but i dont know how to approach it???
2006-08-28 16:46:01 · update #1
a^3 - b^3 = (a-b)(a^2 +ab + b^2)
a = 3x-2 and b = 3 (since 27 = 3^3)
(3x-2)^3 -27
= ((3x-2)-(3))((3x-2)^2 + (3x-2)(3) + (3)^2)
= (3x-5)(9x^2 - 6x - 6x + 4 + 9x - 6 + 9)
= (3x-5)(9x^2 - 3x + 7)
2006-08-28 16:46:27 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
Remember that a^3 - b^3 = (a-b)(a^2 + ab + b^2)
(3x - 2)^3 - 27
= (3x - 2)^3 - 3^3
= (3x - 2 - 3)((3x - 2)^2 + 3(3x - 2) + 3^2)
= (3x - 5)(9x^2 - 12x + 4 + 9x - 6 + 9)
= (3x - 5)(9x^2 - 3x + 7)
We cannot factorize further, because the value of b^2 - 4ac for (9x^2 - 3x + 7) is negative, 9 - 4x9x7 < 0
2006-08-28 23:47:07 · answer #2 · answered by dactylifera001 3 · 0⤊ 0⤋
Find a root, i.e. what is x to have (3x-2)^3 - 27 = 0
3x-2 = cuberoot of 27
3x-2 = 3, x = 5/3
Thus, (3x - 5) is a factor.
Another way, let z = 3x-2, so z^3 - 27 = 0
z = 3 is a root, so (z-3) is a factor, and
(z-3)(z^2+3z+9) = 0
(3x-2-3)((3x-2)^2 + 3(3x-2) + 9) = 0
...
2006-08-29 00:45:11 · answer #3 · answered by back2nature 4 · 0⤊ 0⤋
(3x - 2)³ - 27
Since 27 is the cube of 3,
= (3x - 2)³ - 3³
Use special factoring for sum/difference of cubes
= [(3x - 2) - 3][(3x - 2)² + 3(3x - 2) + (3)²]
= (3x - 2 - 3)(9x² - 12x + 4 + 9x - 6 + 9)
= (3x - 5)(9x² - 3x + 7)
^_^
2006-08-29 08:06:59 · answer #4 · answered by kevin! 5 · 0⤊ 0⤋
22
2006-08-28 23:44:07 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(3x-2)^3 = 3^3
3x-2 = 3
3x = 5
x = 5/3
2006-08-28 23:42:52 · answer #6 · answered by ArcherOmega 4 · 0⤊ 0⤋
(3x-2)^3-27
((3x-2)-3)((3x-2)+3)((3x-2)-3)
2006-08-28 23:47:25 · answer #7 · answered by ronw 4 · 0⤊ 0⤋