How do you factor $2 x^{6} - 3 x^{4}$ $2x^6-3x^4$ ?

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How do you factor $2 x^{6} - 3 x^{4}$ $2x^6-3x^4$ ?

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Answer 1 · 2015-06-15T04:03:16

Factor out the greatest common factor $x^{4}$ $x^4$ to get $x^{4} (2 x^{2} - 3)$ $x^4(2x^2-3)$ .

Explanation:

Factor out the GCF $x^{4}$ $x^4$ .

$2 x^{6} - 3 x^{4}$ $2x^6-3x^4$ =

$x^{4} (2 x^{2} - 3)$ $x^4(2x^2-3)$

Answer 2 · 2015-06-15T06:46:44

$2 x^{6} - 3 x^{4} = x^{4} (2 x^{2} - 3) = x^{4} (\sqrt{2} x - \sqrt{3}) (\sqrt{2} x + \sqrt{3})$ $2x^6-3x^4 = x^4(2x^2-3) = x^4(sqrt(2)x-sqrt(3))(sqrt(2)x+sqrt(3))$

Explanation:

First separate out the common factor $x^{4}$ $x^4$ to get:

$2 x^{6} - 3 x^{4} = x^{4} (2 x^{2} - 3)$ $2x^6-3x^4 = x^4(2x^2-3)$

The remaining quadratic factor can be treated as a difference of squares with irrational coefficients:

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

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

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

using the difference of squares identity:

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

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How do you factor $2 x^{6} - 3 x^{4}$ $2x^6-3x^4$ ?

2 Answers

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