How do you divide $\frac{8 y^{2} - 6 y - 3}{2 y - 1}$ $\frac { 8y ^ { 2} - 6y - 3} { 2y - 1}$ ?

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How do you divide $\frac{8 y^{2} - 6 y - 3}{2 y - 1}$ $\frac { 8y ^ { 2} - 6y - 3} { 2y - 1}$ ?

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Answer 1 · 2017-07-22T09:39:54

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

Explanation:

I had to use algebraic division for this one: enter image source here

Answer 2 · 2017-07-22T11:18:38

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

Explanation:

$one way is to use the divisor as a factor in the numerator$ $"one way is to use the divisor as a factor in the numerator"$

$consider the numerator$ $"consider the numerator"$

$4 y (2 y - 1) + 4 y - 6 y - 3$ $color(red)(4y)(2y-1)color(magenta)(+4y)-6y-3$

$= 4 y (2 y - 1) - 1 (2 y - 1) - 1 - 3$ $=color(red)(4y)(2y-1)color(red)(-1)(2y-1)color(magenta)(-1)-3$

$= 4 y (2 y - 1) - 1 (2 y - 1) - 4$ $=color(red)(4y)(2y-1)color(red)(-1)(2y-1)-4$

$quotient = 4 y - 1, remainder = - 4$ $"quotient "=color(red)(4y-1)," remainder "=-4$

$\Rightarrow \frac{8 y^{2} - 6 y - 3}{2 y - 1} = 4 y - 1 - \frac{4}{2 y - 1}$ $rArr(8y^2-6y-3)/(2y-1)=4y-1-4/(2y-1)$

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How do you divide $\frac{8 y^{2} - 6 y - 3}{2 y - 1}$ $\frac { 8y ^ { 2} - 6y - 3} { 2y - 1}$ ?

2 Answers

Explanation:

Explanation:

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