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What are the factors of the given expressions?

1).

27x^4 - 54x^3 + 36x^2 - 10x + 1

2).

27x^4 - 27x^3 - x + 1

4 Answers

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  • Sepia
    Lv 7
    9 years ago
    Favorite Answer

    1).

    27x^4 - 54x^3 + 36x^2 - 10x + 1

    = (3x^2 - 4x + 1)(9x^2 - 6x + 1)

    = (3x - 1)(x - 1)(3x - 1)^2

    = (x - 1)(3x - 1)^3

    2).

    27x^4 - 27x^3 - x + 1

    = (3x^2 - 4x + 1)(9x^2 + 3x + 1)

    = (3x - 1)(x - 1)(9x^2 + 3x + 1)

  • Raoul
    Lv 4
    9 years ago

    1. The expression is difficult to factor.So we use the following procedure:

    x-a is a factor if, when a is substituted into x in the given expression, the answer is 0.

    Possible values of a are in the form: (factor of the constant) / (factor of the leading coefficient).

    In this case, the constant is 1 and the leading coefficient is 27.

    Possible values of a are : +/- 1, 1/3, 1/9, 1/27.

    Let's try 1.

    27x^4 - 54x^3 + 36x^2 - 10x + 1 when x=1 gives 0.

    Thus, (x-1) is a factor.

    We then divide 27x^4 - 54x^3 + 36x^2 - 10x + 1 by (x-1) for ease in looking for the other factors:

    27x^4 - 54x^3 + 36x^2 - 10x + 1 / (x-1) = 27x^3 - 27x^2 + 9x - 1

    Now, 27x^3 - 27x^2 + 9x - 1 is easier to factor:

    = (27x^3 -1) - (27x^2 - 9x)

    = (3x-1)(9x^2 + 3x + 1) - 9x(3x-1) <-----difference of 2 cubes!

    = (3x-1)(9x^2 + 3x + 1 - 9x)

    = (3x-1)(9x^2 - 6x + 1)

    = (3x-1)(3x-1)^2

    = (3x-1)^3

    The factors of 27x^4 - 54x^3 + 36x^2 - 10x + 1 are (x-1)(3x-1)^3

    2. 27x^4 - 27x^3 - x + 1

    This time, we can factor more easily.

    = 27x^3(x-1) - (x-1)

    = (27x^3 - 1)(x-1)

    =(3x-1)(9x^2 + 3x + 1)(x-1) <---- difference of 2 cubes!

  • Colin
    Lv 7
    9 years ago

    1).

    27x^4 - 54x^3 + 36x^2 - 10x + 1

    = (x - 1)*(27*x^3 - 27*x^2 + 9*x - 1)

    = (x - 1)*(3*x - 1)*(9*x^2 - 6*x + 1)

    = (x - 1)*(3*x - 1)*(3*x - 1)^2

    = (x - 1)*(3*x - 1)^3 <<<

    2).

    27x^4 - 27x^3 - x + 1

    = 27*x^3*(x - 1) - (x - 1)

    = (x - 1)*(27^x^3 - 1)

    = (x - 1)*((3^x)^3 - 1)

    = (x - 1)*(3*x - 1)*(9^x^2 + 3*x + 1) <<<

  • Pranil
    Lv 7
    9 years ago

    1).

    27x^4 - 54x^3 + 36x^2 - 10x + 1

    sum of coefficients

    27 – 54 + 36 – 10 + 1 = 0

    so (x – 1) is a factor of it

    find other factor by synthetic division or by long division

    1| 27 – 54 + 36 – 10 + 1

    ------- + 27 – 27 + 9 – 1

    -----------------------------------------

    --- 27 – 27 + 9 – 1 + 0

    so other factor is

    (27x³ – 27x² + 9x – 1)

    = (27x³ – 1 – 27x² + 9x )

    = (27x³ – 1) – (27x² – 9x )

    = (3x – 1)(9x² + 3x + 1) – 9x(3x – 1)

    = (3x – 1)(9x² + 3x + 1 – 9x)

    = (3x – 1)(9x² – 6x + 1)

    = (3x – 1)³

    27x^4 - 54x^3 + 36x^2 - 10x + 1

    = (x – 1)(3x – 1)³

    27x^4 - 27x^3 - x + 1

    = 27x^3(x - 1) - 1(x – 1)

    = (27x^3 - 1) (x – 1)

    = (3x^3 - 1)(9x² + 3x + 1) (x – 1)

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