12a^5-18a^4-6a^3?

Answer 1

=6a^3(2a^2-3a-1)

Answer 2

I think you want it to be factored
12a^5-18a^4-6a^3

take 6a^3 common

= 6a^3(2a^2-3a-1)
now term in bracket cannot be factored further to rational

Answer 3

12a^5-18a^4-6a^3 = 6a^3(2a^2 - 3a - 1)

What were you looking for?

Answer 4

It factors down to 6a^3(2a^2-3a-1)

Answer 5

i imagine you want to examine your question back, because it does not look to grant adequate files. If i assume you're talking about 6A3 the position A=3 ought to provide 633, then there's no thanks to operate 6A3+2B5 mutually to get a style gently divisible with techniques from 5, so the question is moot. If i assume you advise 6*A*3 and 2*B*5, then as a fashion to have a style divisible with techniques from 5 even as both are extra, A must be a distinct of 5, if we extra assume A and B are entire numbers lower than 10, then A must be 0 or 5. B must be any style, the biggest of that may be 9. So we hae 5+9=14. yet I doubt it is the answer, as i imagine the question is misstated

Answer 6

Your first step is to factor out the GCF of all those terms.
In this case, it would be 6a^3.

6a^3 (2a^2 - 3a - 1)

This doesn't factor any further.

Answer 7

6a^3(2a^2 - 3a - 1)
[to factorise use formula (-b +- sqrt(b^2-4ac))/2a]

a = 0 or a = (3+sqrt17)/4 or a = (3-sqrt17)/4

Answer 8

12a^5-18a^4-6a^3
=6a^3(2a^2-3a-1)
=6a^3(2a^2-2a-a-1)
=6a^3*2(a^2-3/2a-1/2)
=12a^3((a-3/4)^2+(1/4)^2)

Answer 9

What do you want to do with this expression?
If you thought you want to solve for a, you have to put an = sign somewhere.