dj asked in Science & MathematicsMathematics · 8 years ago

# try and factor x^12-64?

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• PaulR2
Lv 7
8 years ago

Let u² = x^12

(u² - 64)

(u + 8)(u - 8)

(x^6 + 8)(x^6 - 8)

Let u³ = x^6

(u³ + 8)(u³ - 8)

(u³ + 8)(u - 2)(u² + 2u - 4)

(u + 2)(u² - 2u + 4)(u - 2)(u² + 2u + 4)

(x² + 2)(x^4 - 2x² + 4)(x² - 2)(x^4 + 2x² + 4)

At this point all our terms are prime and cannot be factored anymore. The proof of this is that after factoring cubic, the trinomial is always prime. You should also take notice that the binomials are prime because their constants are prime. I hope this helps. Have a good day.

• 8 years ago

Factor as the difference of two squares

(x^6 + 8)(x^6 - 8).

Now factor as the sum and difference of two cubes

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

• 8 years ago

x^12 - 64

USE FORMULA a^2 - b^2 = ( a+b) (a-b)

( x^6 + 8) ( x^6 - 8 )

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

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

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