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## Factor the differences

Factor the differences of two squares. Be sure to factor completely.

X6-64

There are two ways at least you can go to factor X6-64.

Method 1. Treat X6-64 as the difference of two squares first.

X6-64

= (x3)2 - 82

= (x3 + 8)(x3 - 8)

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

Method 2. Treat X6-64 as the difference of two cubes first.

x6-64

= (x2)3 - (4)3

= (x2 - 4)(x4 + 4x2 + 16)

= (x+2)(x-2)[(x2+4)2 - (2x)2], by completing the square

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

x6-64= (x3)- (8)

= (x3+8) (x3-8)

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

Kayla, for any polynomial, [f(x)]^2 - [g(x)]^2, will factor as [f(x)-g(x)] times [f(x)+g(x)], so if you set f(x) = x^3 and g(x) = 8, you can rewrite the equation as [x^3]^2 - [8]^2, which you can factor as the difference of two squares.  The factor "completely" part of the question is important because once you get your two terms, you will find that they are both a sum and difference of cubes, which also have a special factoring process.  You can find a good illustration of that process (assuming your text does not provide illustrations) at http://www.cliffsnotes.com/study_guide/Sum-or-Difference-of-Cubes.topicArticleId-257309,articleId-257149.html

I'll leave the rest to you, but feel free to email or ask any follow-up questions if this does not make sense.  John