How do you long divide $\frac{16 y^{2} - 8 y - 8}{4 y - 2}$ $(16y^(2)-8y-8) /( 4y-2)$ ?

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Answer 1 · 2017-06-19T11:48:58

$\frac{16 y^{2} - 8 y - 8}{4 y - 2} = 4 y - \frac{4}{2 y - 1}$ $(16y^2-8y-8)/(4y-2)=4y-4/(2y-1)$

$\frac{16 y^{2} - 8 y - 8}{4 y - 2}$ $(16y^2-8y-8)/(4y-2)$

$= \frac{4 y (4 y - 2) - 8}{4 y - 2}$ $=(4y(4y-2)-8)/(4y-2)$

$= \frac{4 y (4 y - 2)}{4 y - 2} - \frac{8}{2 (2 y - 1)}$ $=(4y(4y-2))/(4y-2) -8/(2(2y-1))$

$= 4 y - \frac{4}{2 y - 1}$ $=4y-4/(2y-1)$

How do you long divide (16y^(2)-8y-8) /( 4y-2)16y2−8y−84y−2(16y^(2)-8y-8) /( 4y-2)?