i've been working on this one for a long time and just can't find an answer. if you could tell me step by step how it can be done, that would be great.
2007-12-11 05:09:58 · 9 answers · asked by mr j 4 in Science & Mathematics ➔ Mathematics
8a³-27
({2a}^3-{3}^3)
this is of the form form a^3-b^3=(a-b)(a^2+ab+b^2)
where a=2a, b=3,
now substuttie in this formula
(2a-3)(4a^2+2a*3+9)
(2a-3)(4a^2+6a+9)
2007-12-11 05:13:57 · answer #1 · answered by Siva 5 · 2⤊ 0⤋
(2a-3)(4a^2+6a+9)
There's no real "steps" to find the (2a-3). Just know that the coefficient of "a" needs to be a factor of 8 (1,2,4, or 8) and the constant needs to be a factor of 27 (1, 3, 9, or 27).
Because we have 8 and 27 in the polynomial and both are cubes, this is probably not a coincidence. Thus, try a polynomial that uses 2 and 3. Then, notice that it is 8a^3 MINUS 27, so try 2a MINUS 3 just to see.
Then do polynomial long division to get (8a^3+0a^2+0a-27) divided by (2a-3) equals (4a^2+6a+9). Then it's a matter of being able to see that this 4a^2+6a+9 is not factorable.
You're done.
2007-12-11 05:19:06 · answer #2 · answered by Kyle G 1 · 0⤊ 1⤋
difference of two cubes
2a cubed is 8a cubed and -3 cubed is -27
so
factor is
(2a-3)(4a^2+6a+9)
for second factor you square the first term (4a^2), take opposite sign of the negative and multiply the two monomials in the first factor (+6a) then square the last term (9)
a^3+64
would factor to
(a+4)(a^2-4a+16)
2007-12-11 05:21:39 · answer #3 · answered by mauler012001 1 · 1⤊ 0⤋
Use the factor theorem/remainder theorem.
(f)x= 8a^3 - 27 = 0 (because factors make the sum = 0, think graphically)
therefore
8 a^3 = 27
a^3 = 27 / 8
a= (27/8)^ 1/3 (or 27/3 cube root)
which = 1.5
2007-12-11 05:14:55 · answer #4 · answered by Anonymous · 0⤊ 2⤋
8a*3=27. a*3=27/8. a= 3/2.
2007-12-11 05:39:37 · answer #5 · answered by Sasi Kumar 4 · 1⤊ 1⤋
As I recall the rules, there is no common factor you can divide out. ( 2a-3)cubed seems like it might work, for you would have 8a cubed and -27 but you would also have a bunch of other stuff in a squared, and a.
2007-12-11 05:16:08 · answer #6 · answered by hasse_john 7 · 0⤊ 2⤋
OK
8a^3 - 27
(2a-3)(4a^2 +6a + 9)
Check
(2a-3)(4a^2 +6a + 9)
8a^3 (- 12a^2 +12a^2)( -18a +18a) - 27
8a^3 -27
Hope that helps.
2007-12-11 05:18:44 · answer #7 · answered by pyz01 7 · 0⤊ 0⤋
8a^3 - 27
=>(2a)^3 - (3)^3
now recall that a^3 - b^3 = (a-b)(a^2 + ab + b^2)
here a = 2a and b = 3
so (2a)^3 - (3)^3 = (2a - 3)(4a^2 + 6a + 9)
2007-12-11 05:14:39 · answer #8 · answered by mohanrao d 7 · 2⤊ 0⤋
what does this 'a' be relating to?
If 'a' be a constant then I can be telling you there is somethin
No one can know.
If it be a variable Any one can answer it.
2007-12-11 05:16:18 · answer #9 · answered by Anonymous · 0⤊ 2⤋