What is the standard form of $y=-(x - 1) (x^2 +6)(x/3 + 2)$ ?

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Answer 1 · 2018-02-24T14:36:02

$y=-1/3x^4-5/3x^3-10x+12$

$y=-(x-1)(x^2+6)(x/3+2)$

$y=-(x^3+6x-x^2-6)(1/3)(x+6)$

$y=-1/3(x^3-x^2+6x-6)(x+6)$

$y=-1/3(x^4-x^3+6x^2-6x+6x^3-6x^2+36x-36)$

$y=-1/3(x^4+5x^3+30x-36)$

$y=-1/3x^4-5/3x^3-10x+12$
Which fits the standard form of $y=Ax^4+Bx^3+Cx^2+Dx+E$

What is the standard form of y=-(x - 1) (x^2 +6)(x/3 + 2) y=-(x - 1) (x^2 +6)(x/3 + 2) ?