What is the standard form of $y = (x - 5) (2 x - 2) (3 x - 1)$ $y= (x-5)(2x-2)(3x-1)$ ?

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Answer 1 · 2017-10-26T00:47:52

Seems to me the standard form follows this pattern: $A x^{3} + B x^{2} + C x + D = 0$ $Ax^3+Bx^2 +Cx +D = 0$

So, let's start multiplying out the factors in parentheses:

$y = (x - 5) \cdot (2 \cdot x - 2) \cdot (3 x - 1)$ $y = (x-5)*(2*x-2)*(3x-1)$ .

FOIL the first two parentheses and we get:

$y = (2 x^{2} - 2 x - 10 x + 10) \cdot (3 x - 1)$ $y = (2x^2-2x-10x+10)*(3x-1)$

OR

$y = (2 x^{2} - 12 x + 10) \cdot (3 x - 1)$ $y = (2x^2-12x+10) * (3x-1)$

FOIL these parentheses:

$y = 6 x^{3} - 38 x^{2} + 42 x - 10$ $y = 6x^3-38x^2+42x-10$

OR

$6 x^{3} - 38 x^{2} + 42 x - 10 = 0$ $6x^3-38x^2+42x-10 = 0$ .

What is the standard form of y=(x−5)(2x−2)(3x−1) y= (x-5)(2x-2)(3x-1)?