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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    BG: Do you want to simplify that?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    another way to do it.

  5. anonymous
    • 5 years ago
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    Can you clarify what you need?

  6. anonymous
    • 5 years ago
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    27z3-18z2+3z =( )( )( )

  7. anonymous
    • 5 years ago
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    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)

  8. anonymous
    • 5 years ago
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    thanks and i am still brute

  9. anonymous
    • 5 years ago
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    and this one \[6x ^{2}+11x+4\]

  10. anonymous
    • 5 years ago
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    (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.

  11. anonymous
    • 5 years ago
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    what method did you use

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