What is the standard form of $y = (- 2 x^{2} - 2) (4 x - 1)$ $y= (-2x^2-2)(4x-1)$ ?

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What is the standard form of $y = (- 2 x^{2} - 2) (4 x - 1)$ $y= (-2x^2-2)(4x-1)$ ?

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Answer 1 · 2016-01-21T12:04:50

$- 8 x^{3} + 2 x^{2} - 8 x + 2$ $- 8x^3 + 2x^2 - 8x + 2$

Explanation:

Multiply out brackets (distributive law ) and collect ' like terms'.

$- 2 x^{2} (4 x - 1) - 2 (4 x - 1) = - 8 x^{3} + 2 x^{2} - 8 x + 2$ $-2x^2 (4x - 1 ) - 2 (4x - 1 ) = -8x^3 + 2x^2 - 8x + 2$

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

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What is the standard form of $y = (- 2 x^{2} - 2) (4 x - 1)$ $y= (-2x^2-2)(4x-1)$ ?

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Explanation:

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What is the standard form of y=(−2x2−2)(4x−1)y= (-2x^2-2)(4x-1) ?

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What is the standard form of $y = (- 2 x^{2} - 2) (4 x - 1)$ $y= (-2x^2-2)(4x-1)$ ?