How do you factor: $y = 2 x^{3} - 32 x$ $y= 2x^3 -32x$ ?

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How do you factor: $y = 2 x^{3} - 32 x$ $y= 2x^3 -32x$ ?

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Answer 1 · 2018-04-21T08:05:44

$y = 2 x (x + 4) (x - 4)$ $y=2x(x+4)(x-4)$

Explanation:

$(1)$ $(1)" "$ we take out the $h c f s$ $hcfs$

y=2x^3-32x#

$h c f (2, 32) = 2$ $hcf(2,32)=2$

$y = 2 (x^{3} - 16 x)$ $y=color(red)(2)(x^3-16x)$

$h c f (x, x^{3}) = x$ $hcf(x,x^3)=x$

$y = 2 x (x^{2} - 16)$ $y=2color(red)(x)(x^2-16)$

$(2)$ $(2)" "$ apply difference of squares

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

$y = 2 x (x + 4) (x - 4)$ $y=2x(x+4)(x-4)$

Answer 2 · 2018-04-21T08:05:55

$y = 2 x (x - 4) (x + 4)$ $y=2x(x-4)(x+4)$

Explanation:

$take out a common factor 2 x$ $"take out a "color(blue)"common factor "2x$

$= 2 x (x^{2} - 16)$ $=2x(x^2-16)$

$x^{2} - 16 is a difference of squares$ $x^2-16" is a "color(blue)"difference of squares"$

$which factors in general as$ $"which factors in general as"$

$∙ x a^{2} - b^{2} = (a - b) (a + b)$ $•color(white)(x)a^2-b^2=(a-b)(a+b)$

$here a = x and b = 4$ $"here "a=x" and "b=4$

$\Rightarrow x^{2} - 16 = (x - 4) (x + 4)$ $rArrx^2-16=(x-4)(x+4)$

$\Rightarrow y = 2 x (x - 4) (x + 4)$ $rArry=2x(x-4)(x+4)$

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How do you factor: $y = 2 x^{3} - 32 x$ $y= 2x^3 -32x$ ?

2 Answers

Explanation:

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