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.
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.
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)
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.
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Verified answer
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)
(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.
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)
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
8a*3=27. a*3=27/8. a= 3/2.
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.
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.
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)
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.