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# Is this correctly done?

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Hello I have a new question!

The question is (originally written in swedish):

Divide Factors: 64x^4-4x^2

Mozziee  Jul 27, 2014

#2
+92254
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Thanks Alan, I just explain a little further,

$$\\(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2\\\ so \boxed{a^2-b^2=(a-b)(a+b)}\\\\ \mbox{This is referred to as a difference of two squares and you need to memorize it.}\\\\ \begin{array}{rll} 16x^2-1&=&(4x)^2-1^2\\\\ &=&(4x-1)(4x+1) \end{array}$$

Melody  Jul 27, 2014
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#1
+26642
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Yes, though you can take it a little further to get:  4x^2(4x+1)(4x-1)

So the factors are: 4, x^2, 4x+1 and 4x-1 (and various combinations of these).

Alan  Jul 27, 2014
#2
+92254
+10

Thanks Alan, I just explain a little further,

$$\\(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2\\\ so \boxed{a^2-b^2=(a-b)(a+b)}\\\\ \mbox{This is referred to as a difference of two squares and you need to memorize it.}\\\\ \begin{array}{rll} 16x^2-1&=&(4x)^2-1^2\\\\ &=&(4x-1)(4x+1) \end{array}$$

Melody  Jul 27, 2014

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