What is the standard form of $y = (4 x - 4) (6 x - 2) (\frac{1}{8} x - \frac{1}{4})$ $y= (4x-4)(6x-2)(1/8x-1/4)$ ?

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Answer 1 · 2017-04-18T12:49:17

$3 x^{3} - 10 x^{2} + 9 x - 2$ $3x^3-10x^2+9x-2$

When you multiply the first two parts of the equation you will get:

$y = (24 x^{2} - 8 x - 24 x + 8) ((\frac{x}{8}) - (\frac{1}{4}))$ $y=(24x^2-8x-24x+8)((x/8)-(1/4))$

Now you can arrange it:

$y = (24 x^{2} - 32 x + 8) ((\frac{x}{8}) - (\frac{1}{4}))$ $y=(24x^2-32x+8)((x/8)-(1/4))$

Multiply terms again:

$y = 3 x^{3} - 6 x^{2} - 4 x^{2} + 8 x + x - 2$ $y=3x^3-6x^2-4x^2+8x+x-2$

Arrange the above equation:

$y = 3 x^{3} - 10 x^{2} + 9 x - 2$ $y=3x^3-10x^2+9x-2$

This is the answer.

What is the standard form of y= (4x-4)(6x-2)(1/8x-1/4)y=(4x−4)(6x−2)(18x−14) y= (4x-4)(6x-2)(1/8x-1/4)?