12x^3-16x^2+3x-4
8a^6+27c^3
2006-10-25 20:20:45 · 10 answers · asked by avenger 2 in Science & Mathematics ➔ Mathematics
how about 4a^4b^2+a^2b-4
2006-10-25 20:21:57 · update #1
could you pls. tell me how did you simplify those?
2006-10-25 20:31:16 · update #2
12 x^3 - 16^2 + 3x^-4
12X^3 - 16^2 + ( 3 / x^4 )
( 12X^7 - 16X^6 + 3 ) / ( x^4 )
4a^4 b^2 + a^2 b^-4
4 a^4 b^2 + a^2 (1 / b^4 )
(4 b^4 a^4 b^4 b^6 + a^2) / (b^4)
4 a^4 b^2 + a^2
a^2 (4 a^2 b^2 +1)
2006-10-25 20:27:02 · answer #1 · answered by c00kies 5 · 0⤊ 0⤋
Answer 1:
I suppose you mean: 12x^3-16x^2+3x-4 = 0
In that case there are two solutions
First, you need to factorise the cubic polynomial
12x^3-16x^2+3x-4 = 0
(4x^2+1)(3x-4) = 0
Either 4x^2+1 =0, or 3x-4 =0
Putting 4x^2+1 = 0
4x^2 = -1
x^2 = -1/4
x = Root of -1/4
Put 3x -4 =0
3x = 4
x = 4/3
Answer 2:
8a^6+27c^3
Here, the formula a^3 + b^3 = (a+b)(a^2+b^2-ab) can be applied
8a^6 = (2a^2)^3
27c^3 = (3c)^3
Here, a = 2a^2 and b = 3c
(2a^2)^3 + (3c)^3 = (2a^2 + 3c)[(2a^2)^2 + (3c)^2-(2a^2)(3c)]
= (2a^2 + 3c)(4a^4 + 9c^2 - 6ca^2)
2006-10-28 02:22:07 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
12x^3 - 16x^2 + 3x - 4 =
4(3x^3 - 4x^2) + 3x - 4 =
4x^2(3x - 4) + (3x - 4) =
(4x^2 + 1)(3x - 4)
8a^6+27c^3 = (2a^2)^3 + (3c)^3
4a^4b^2+a^2b-4
(2a^2b + 2)(2a^2b - 2) + a^2b or
let x=a^2b
4x^2 + x - 4 =
4(x^2 + (1/4)x - 1) =
4(x^2 + (1/4)x + 1/64 - 1 - 1/64) =
4(x + 1/8)^2 - 65/16 =
(2a^4b + 1/4)^2 - 65/16
2006-10-26 03:58:01 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
simple
12x^3-16x^2+3x-4
= 4x^2(3x-4) + 3x-4
= (3x-4)(4x^2+1)
realising 12: 16 = 3: 4
2nd one
8a^6 + 27c^3 = (2a^2)^3 + (3c)^3 and using a^3+b^3 formula
= (2a^2+3c)(4a^4-6a^2c+9c^2)
2006-10-26 03:29:20 · answer #4 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
12 x^3 - 16^2 + 3x^-4
12X^3 - 16^2 + ( 3 / x^4 )
( 12X^7 - 16X^6 + 3 ) / ( x^4 )
YUP
2006-10-26 03:28:07 · answer #5 · answered by B. 4 · 0⤊ 0⤋
12x^3-16x^2+3x-4
=4x^2(3x-4)+1(3x-4)
=(4x^2+1)(3x-4)
if the value of the given expression is given, then only we can solve....
otherwise how can we solve?
asfar as i know.,we can solve only equations.
i dont know......
2006-10-26 03:46:59 · answer #6 · answered by bubbly 2 · 0⤊ 0⤋
12x³-16x²+3x-4=?
is that the question?
2006-10-26 03:29:43 · answer #7 · answered by Anonymous · 0⤊ 0⤋
(4x'2+1) (3x-4)
(2a'2)'3 + (3c)'3
2006-10-26 03:31:11 · answer #8 · answered by bogus 3 · 0⤊ 0⤋
WHY ARE YOU PEOPLE DOING THESE KIDS' HOMEWORK FOR THEM???
2006-10-26 03:52:55 · answer #9 · answered by Anonymous · 0⤊ 0⤋
it is so easy
12*3/16*8bbbbbbbbbb+''''''''''''''''''''/hhhhhhhhh=111000000000000000000000000000000000000000000000000000
are you see?
2006-10-26 03:25:09 · answer #10 · answered by hany s 2 · 0⤊ 1⤋