What is the standard form of $y = (\frac{1}{5} x^{2} - \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $y= (1/5x^2-1/12)(1/3x+3/8)$ ?

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What is the standard form of $y = (\frac{1}{5} x^{2} - \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $y= (1/5x^2-1/12)(1/3x+3/8)$ ?

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Answer 1 · 2016-03-12T10:16:59

$y = \frac{1}{15} x^{3} + \frac{3}{40} x^{2} - \frac{1}{36} x - \frac{3}{96}$ $y=1/15x^3+3/40x^2-1/36x-3/96$

Explanation:

Given: $y = (\frac{1}{5} x^{2} - \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $color(brown)(y=color(blue)((1/5x^2-1/12)) (1/3x+3/8)$

$y = \frac{1}{5} x^{2} (\frac{1}{3} x + \frac{3}{8}) + (- \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $color(brown)(y=color(blue)(1/5x^2) (1/3x+3/8)+color(blue)((-1/12))(1/3x+3/8))$

$y = (\frac{1}{15} x^{3} + \frac{3}{40} x^{2}) + (- \frac{1}{36} x - \frac{3}{96})$ $y=(1/15x^3 +3/40 x^2)+(-1/36x-3/96)$

$y = \frac{1}{15} x^{3} + \frac{3}{40} x^{2} - \frac{1}{36} x - \frac{3}{96}$ $y=1/15x^3+3/40x^2-1/36x-3/96$

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What is the standard form of $y = (\frac{1}{5} x^{2} - \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $y= (1/5x^2-1/12)(1/3x+3/8)$ ?

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What is the standard form of $y = (\frac{1}{5} x^{2} - \frac{1}{12}) (\frac{1}{3} x + \frac{3}{8})$ $y= (1/5x^2-1/12)(1/3x+3/8)$ ?