How do you factor completely $22 y^{4} - 33 y^{3} + 11 y^{2}$ $22y^4-33y^3+11y^2$ ?

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How do you factor completely $22 y^{4} - 33 y^{3} + 11 y^{2}$ $22y^4-33y^3+11y^2$ ?

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Answer 1 · 2016-04-30T17:18:10

It is

$22 y^{4} - 33 y^{3} + 11 y^{2} = 22 y^{4} - 22 y^{3} - 11 y^{3} + 11 y^{2} = 22 y^{3} (y - 1) - 11 y^{2} (y - 1) = (22 y^{3} - 11 y^{2}) \cdot (y - 1) = 11 y^{2} (2 y - 1) \cdot (y - 1)$ $22y^4-33y^3+11y^2=22y^4-22y^3-11y^3+11y^2= 22y^3(y-1)-11y^2(y-1)=(22y^3-11y^2)*(y-1)= 11y^2(2y-1)*(y-1)$

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How do you factor completely $22 y^{4} - 33 y^{3} + 11 y^{2}$ $22y^4-33y^3+11y^2$ ?

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How do you factor completely $22 y^{4} - 33 y^{3} + 11 y^{2}$ $22y^4-33y^3+11y^2$ ?