How do you factor completely $64 x^{4} - 9$ $64x^4-9$ ?

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How do you factor completely $64 x^{4} - 9$ $64x^4-9$ ?

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Answer 1 · 2016-08-02T06:16:47

$64 x^{4} - 9$ $64x^4-9$

$= (8 x^{2} - 3) (8 x^{2} + 3)$ $=(8x^2-3)(8x^2+3)$

$= (2 \sqrt{2} x - \sqrt{3}) (2 \sqrt{2} x + \sqrt{3}) (8 x^{2} + 3)$ $=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(8x^2+3)$

$= (2 \sqrt{2} x - \sqrt{3}) (2 \sqrt{2} x + \sqrt{3}) (2 \sqrt{2} x - \sqrt{3} i) (2 \sqrt{2} x + \sqrt{3} i)$ $=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(2sqrt(2)x-sqrt(3)i)(2sqrt(2)x+sqrt(3)i)$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We will use this three times below.

$64 x^{4} - 9$ $64x^4-9$

$= {(8 x)}^{2} - 3^{2}$ $=(8x)^2-3^2$

$= (8 x^{2} - 3) (8 x^{2} + 3)$ $=(8x^2-3)(8x^2+3)$

If we allow irrational coefficients then we can factor this further:

$= ({(2 \sqrt{2} x)}^{2} - {(\sqrt{3})}^{2}) (8 x^{2} + 3)$ $=((2sqrt(2)x)^2-(sqrt(3))^2)(8x^2+3)$

$= (2 \sqrt{2} x - \sqrt{3}) (2 \sqrt{2} x + \sqrt{3}) (8 x^{2} + 3)$ $=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(8x^2+3)$

That is as far as we can go with Real coefficients, since $8 x^{2} + 3 > 0$ $8x^2+3 > 0$ for all Real values of $x$ $x$ . If we allow Complex coefficients then we can factor this further:

$= (2 \sqrt{2} x - \sqrt{3}) (2 \sqrt{2} x + \sqrt{3}) ({(2 \sqrt{2} x)}^{2} - {(\sqrt{3} i)}^{2})$ $=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))((2sqrt(2)x)^2-(sqrt(3)i)^2)$

$= (2 \sqrt{2} x - \sqrt{3}) (2 \sqrt{2} x + \sqrt{3}) (2 \sqrt{2} x - \sqrt{3} i) (2 \sqrt{2} x + \sqrt{3} i)$ $=(2sqrt(2)x-sqrt(3))(2sqrt(2)x+sqrt(3))(2sqrt(2)x-sqrt(3)i)(2sqrt(2)x+sqrt(3)i)$

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How do you factor completely $64 x^{4} - 9$ $64x^4-9$ ?

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