Question

how can i do this 27z3-18z2+3z

Answer 1

BG: Do you want to simplify that?

Answer 2

yes

Answer 3

Ok pull out any common terms. In your case, there are both coefficients and variables that are common.
27, 18, and 3 are all divisible by 3.
Thus, 3(9z^3-6z^2+z)
Also, z^3, z^2, and z^1 all have z^1 in common.
Thus, 3z(9z^2-6z+1) is the most simplified form

Answer 4

another way to do it.

Answer 5

Can you clarify what you need?

Answer 6

27z3-18z2+3z =( )( )( )

Answer 7

Ok
So you want to factor the equation into three quantities.
3z * (3z-1)^2 = 27z3-18z2+3z
Thus, (3z)(3z-1)(3z-1) is your answer.
An easier way to do this by hand (i factored it on my calculator) would be to follow my first method arriving at this answer:
3z(9z^2-6z+1)
Then, you can more easily factor (9z^2-6z+1) into (3z-1)^2
So then you have (3z)(3z-1)(3z-1)

Answer 8

thanks and i am still brute

Answer 9

and this one \[6x ^{2}+11x+4\]

Answer 10

(2x+1)(3x+4)
Take the coefficient from 6x^2 (6) and break it into factors.
The factors of 6 are 6,3,2,1
I normally pick the middle two.
3 and 2.
Then make ( _x + _ ) ( _x + _ ) this setup
Plug in ( 3x + _ ) ( 2x + _ )
then use guess and check for the two blanks that are factors of 4 and when expanded add to 11.
This gives you (2x+1)(3x+4)
Sometimes these problems can take a /long/ time.

Answer 11

what method did you use