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

Answer 1

First note that there are no terms with powers 2, 1 and 0 --> a³ is a factor

So we get:
6a³ (2a² - 3a - 1)

Then we find the roots of the second factor:
D = 9 + 8 = 17
a = (3 +/- sqrt(17)) / 4

6a³ (a-(3+sqrt(17))/4)(a-(3-sqrt(17))/4)

This makes me think there is an error in your question (could be right though but in school math problems the answers tend to be easy)

if it is +6a³ then
D = 1

a = 1 and a = 1/2

So we get
6a³ (a-1) (2a-1)

Answer 2

Your first step is to pull out the biggest factor you can, out of all of these. The first indicator of knowing to do this is realizing each term has an "a". Given that they all have "a" terms, and given that each constant is divisible by 6, the best term to pull out would be 6a^3.

6a^3 [2a^2 - 3a - 1]

This doesn't factor any further, since the discriminant (b^2 - 4ac, equal to 17) isn't a square number.

Answer 3

12a^5-18a^4-6a³ =
a³(12a²-18a-6) =
3a³(4a²-6a-2) =
6a³(2a²-3a-1) =

₢

Answer 4

Take 6a^3 outside(in otherwords, divide and multiply thropughout by a^3)
You have
6a^3(2a^2-3a-1)
If you want to further simplify it, u will get sqrts and make it more complex

Answer 5

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

Answer 6

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

Answer 7

that's impossible. you cannot subtract different exponents from each other. this is the simplest form you can get to.