What is the standard form of # 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: #Ax^3+Bx^2 +Cx +D = 0#

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

#y = (x-5)*(2*x-2)*(3x-1)#.

FOIL the first two parentheses and we get:

#y = (2x^2-2x-10x+10)*(3x-1)#

OR

#y = (2x^2-12x+10) * (3x-1)#

FOIL these parentheses:

#y = 6x^3-38x^2+42x-10#

OR

#6x^3-38x^2+42x-10 = 0#.