How do you factor: $y = 32 x^{3} - 4$ $y= 32x^3 - 4$ ?

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How do you factor: $y = 32 x^{3} - 4$ $y= 32x^3 - 4$ ?

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Answer 1 · 2017-04-14T10:34:33

$y = 4 (2 x - 1) (4 x^{2} + 2 x + 1)$ $y=4(2x-1)(4x^2+2x+1)$

Explanation:

Recall:

$(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$ $(a^3-b^3)=(a-b)(a^2+ab+b^2)$

Here we go.

$y = 32 x^{3} - 4$ $y=32x^3-4$

By factoring out $4$ $4$ ,

$\Rightarrow y = 4 (8 x^{3} - 1)$ $Rightarrow y=4(8x^3-1)$

By rewriting a bit,

$\Rightarrow y = 4 ({(2 x)}^{3} - 1^{3})$ $Rightarrow y=4((2x)^3-1^3)$

By applying the formula at the top with $a = 2 x$ $a=2x$ and $b = 1$ $b=1$ ,

$\Rightarrow y = 4 (2 x - 1) ({(2 x)}^{2} + 2 x \cdot 1 + 1^{2})$ $Rightarrow y=4(2x-1)((2x)^2+2x cdot 1+1^2)$

By cleaning up a bit,

$y = 4 (2 x - 1) (4 x^{2} + 2 x + 1)$ $y=4(2x-1)(4x^2+2x+1)$

I hope that this was clear.

Answer 2 · 2017-04-14T10:55:31

$y = 4 (2 x - 1) (4 x^{2} + 2 x + 1)$ $y=4(2x-1)(4x^2+2x+1)$

Explanation:

The first step is to factor out $common factor$ $color(blue)"common factor"$ of 4

$\Rightarrow y = 4 (8 x^{3} - 1) \to (1)$ $rArry=4(8x^3-1)to(1)$

now $8 x^{3} - 1$ $8x^3-1$ is a $difference of cubes$ $color(blue)"difference of cubes"$ and factorises in general as.

$color(red)(bar(ul(|color(white)(2/2)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(2/2)|)))$

$8x^3=(2x)^3" and " 1^3=1$

$"using " a=2x" and " b=1$

$rArr8x^3-1=(2x-1)(4x^2+2x+1)$

$"going back to " (1)$

$rArry=4(2x-1)(4x^2+2x+1)$

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How do you factor: $y = 32 x^{3} - 4$ $y= 32x^3 - 4$ ?

2 Answers

Explanation:

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How do you factor: y= 32x^3 - 4 y=32x3−4y= 32x^3 - 4 ?

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How do you factor: $y = 32 x^{3} - 4$ $y= 32x^3 - 4$ ?