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Factorize x^6 – 64.  

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xetaalpha | Student | Honors

Posted November 24, 2010 at 9:33 PM via web

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Factorize x^6 – 64.

 

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 24, 2010 at 9:33 PM (Answer #1)

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We have to factorize x^6 – 64.

Now x^6 = (x^3) ^2 and 64 = 8^2

x^6 – 64

=> (x^3) ^2 – 8^2

We use x^2 – y^2 = (x - y) (x + y)

=> (x^3 – 8) (x^3 + 8)

=> (x^3 – 2^3) (x^2 + 2^3)

Now use the expansion for the sum of cubes x^3 + y^3 =(x + y) (x^2 - xy + y^2) and the difference of cubes x^3 – y^3 = (x – y) (x^2 + xy +y^2)

=> (x – 2) (x^2 + 2x + 4) (x + 2) (x^2 – 2x + 4)

=> (x – 2) (x+2) (x^2 + 2x + 4) (x^2 – 2x + 4)

Therefore x^6 – 64 can be factorized as (x – 2) (x+2) (x^2 + 2x + 4) (x^2 – 2x + 4)

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giorgiana1976 | College Teacher | Valedictorian

Posted November 24, 2010 at 9:41 PM (Answer #2)

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We could write 64 as a power of 2:

64 = 2^6

Now, we can apply the formula:

x^n - a^n = (x-a)(x^(n-1) + x^(n-2)*a + .... + a^(n-1))

We'll put n = 6 and a = 2:

x^6 - 2^6 = (x-2)(x^5 + 2x^4 + 4x^3 + 8x^2 + 16x + 32)

We also could write:

(x^2)^3 - (2^2)^3

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

We'll put a = x^2 and b = 2^2

(x^2)^3 - (2^2)^3 = (x^2 - 4)(x^4 + 4x^2 + 16)

But x^2 - 4 is a difference of squares:

x^2 - 4 = (x-2)(x+2)

(x^2)^3 - (2^2)^3 = (x-2)(x+2)(x^4 + 4x^2 + 16)

x^6 - 2^6 = (x-2)(x+2)(x^4 + 4x^2 + 16)

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neela | High School Teacher | Valedictorian

Posted November 24, 2010 at 9:44 PM (Answer #3)

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To factorise x^6 -64.

We know that a^2-b^2 = (a+b)(a-b).

Therefore, x^6-8^2 = (x^3)^2- 8^2 = (x^3+8)(x^3-8)

(x^3+8)(x^2-8) = (x^3+2^3)(x^3-2^3).....(1)

We  use a^3+b^3 = (a+b)(a^2-ab+b^2) and a^3-b^3 = (a-b)(a^2+ab+b^2).

Therefore x^3+2^3 = (x+2)(x^2+2x+4) and (x^3-2^3) = (x-2)(x^2+2x+4). We use this result in (1).

(x^3+8)(x^3-8) = (x+2)(x^2-2x+4)(x-2)(x^2+2x+4).

Therefore x^6-64 = (x-2)(x+2)(x^2-2x+4)(x^2+2x+4).

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