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- anonymous

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

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- anonymous

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

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- anonymous

BG: Do you want to simplify that?

- anonymous

yes

- anonymous

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

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- anonymous

another way to do it.

- anonymous

Can you clarify what you need?

- anonymous

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

- anonymous

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)

- anonymous

thanks and i am still brute

- anonymous

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

- anonymous

(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.

- anonymous

what method did you use

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