How do you write $4 y^{3} - 4 y^{2} + 3 - y$ $4y^3-4y^2+3-y$ in standard form?

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Answer 1 · 2018-05-23T00:49:16

$4 y^{3} - 4 y^{2} - y + 3$ $4y^3-4y^2-y+3$

$4 y^{3} - 4 y^{2} + 3 - y$ $4y^3-4y^2+3-y$

Standard form is descending exponents:

$4 y^{3} - 4 y^{2} - y + 3$ $4y^3-4y^2-y+3$

How do you write 4y3−4y2+3−y4y^3-4y^2+3-y in standard form?