How do I simplify:
(9x^2 - 4) (9x^4 - 6x^3 + 4x^2)
------------------- x --------------------------
(3x^2 - 5x + 2) (27x^4 + 8x)
2007-08-05 10:22:32 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It's x / (x - 1). Here's how to find this:
Factor the various terms, first.
The numerator = (3x - 2)(3x + 2) x^2 (9 x^2 - 6 x + 4).
[The final quadratic term doesn't factorize, as you can check by attempting to put it equal to zero. Its "discriminant,"
"b^2 - 4ac" is < 0, meaning the roots would be complex. But in any case, looking ahead, I suspect that this quadratic factor will cancel with a factor that will emerge in the denominator. That's because it is itself a factor of the sum of two cubes, (27x^3 + 8).]
The denominator is (3x - 2 )(x - 1)*x*(27x^3 + 8)
= (3x - 2 )(x - 1)*x*(3x + 2)(9x^2 - 6x + 4).
So the entire thing is:
(3x - 2)(3x + 2) x^2 (9 x^2 - 6 x + 4) divided by
[(3x - 2 )(x - 1)*x*(3x + 2)(9x^2 - 6x + 4)].
The quadratic term cancels (as anticipated), as do the
(3x - 2) and (3x + 2) terms, leaving just:
x^2 / [(x - 1)x] = x / (x - 1).
QED!
Live long and prosper.
2007-08-05 10:32:58 · answer #1 · answered by Dr Spock 6 · 0⤊ 0⤋
IN THE NUMERATOR the first term is a diff of squares so it factors to (3x-2)(3x+2)
IN THE DENOMINATOR we factor the first expression to
(3x-2)(x-1) and the second is x(27x^3+8) which is a sum of cubes and factors further to x(3x+2)(9x^4 - 6x^3 + 4x^2) write all those terms in a big fraction and cancel like terms top and bottom to be left with:
x in the numerator and (x-1) on the bottom
final answer = x / (x-1)
the answer above mine is correct up to the last step where the factoring is in error.
2007-08-05 17:43:10 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
the first fraction, first you factor...
(3x + 2)(3x - 2)
------------------- .... the (3x - 2)'s cancel... leaving you with...
(3x - 2)(x - 1)
(3x+2)
--------- ... that's for the 1st fraction...
(x - 1)
For the 2nd fraction... factor out an x^2 in the numerator... and factor out x in the denominator... you then have...
x^2(9x^2 - 6x + 4)
---------------------- for the 2nd fraction...
x(27x^3 + 8)
(27x^3 + 8) you have to realize is "the sum of two cubes"... so...
as the formula goes for factoring the sum of two cubes (see the link that I'm including).... when you factor you get...
a3 + b3 = (a + b)(a2 – ab + b2)
so in our case of 27x^3 + 8.... "a" is "3x" and "b" is "2"....
so... 27x^3 + 8 = (3x + 2)(9x^2 - 6x + 4)... so that is the denominator for the 2nd fraction... oh, and don't forget the "x" in front... because for the 2nd fraction you had x(27x^3 + 8), remember?
(3x+2) (x^2)(9x^2 - 6x + 4)
--------- * --------------------------
(x - 1) ........x(3x + 2)(9x^2 - 6x + 4)
if you notice above... the (3x + 2)'s cancel and the (9x^2 - 6x + 4)'s also cancel...
.... so you are left with...
x^2 ......... x
--------- = ---------
x(x - 1) .....(x-1)
2007-08-05 18:12:16 · answer #3 · answered by blueskies 7 · 0⤊ 0⤋
foil it: 9x^2 times 9x^4 and so on minus 4 times the rest of the expression
2007-08-05 17:31:19 · answer #4 · answered by Hectorin 1 · 0⤊ 0⤋
(9x^2 - 4) (9x^4 - 6x^3 + 4x^2)
------------------- --------------------------
(3x^2 - 5x + 2) (27x^4 + 8x)
=(3x - 2)(3x+2) (x^2)(9x^2 - 6x + 4)
------------------- ----------------------
(x - 2)(x - 3) x((3x)^3 + 2^3)
=(3x - 2)(3x+2) (x^2)(9x^2 - 6x + 4)
------------------- ----------------------
(x - 2)(x - 3) x(3x+2)(9x^2 -6x +4)
=(3x - 2)x
----------------
(x - 2)(x - 3)
2007-08-05 17:37:26 · answer #5 · answered by fofo m 3 · 0⤊ 0⤋