How do you divide $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by $2 x - 6$ $2x-6$ ?

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How do you divide $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by $2 x - 6$ $2x-6$ ?

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Answer 1 · 2017-03-04T21:20:29

$\frac{2 x^{2} - 5 x - 3}{2 x - 6} = x + \frac{1}{2}$ $(2x^2-5x-3)/(2x-6) = x+1/2$

Explanation:

We can split the numerator up into parts which are monomial multiples of $(2 x - 6)$ $(2x-6)$ , then combine them like this:

$\frac{2 x^{2} - 5 x - 3}{2 x - 6} = \frac{2 x^{2} - 6 x + x - 3}{2 x - 6}$ $(2x^2-5x-3)/(2x-6) = (2x^2-6x+x-3)/(2x-6)$

$\frac{2 x^{2} - 5 x - 3}{2 x - 6} = \frac{x (2 x - 6) + \frac{1}{2} (2 x - 6)}{2 x - 6}$ $color(white)((2x^2-5x-3)/(2x-6)) = (x(2x-6)+1/2(2x-6))/(2x-6)$

$color(white)((2x^2-5x-3)/(2x-6)) = ((x+1/2)color(red)(cancel(color(black)((2x-6)))))/color(red)(cancel(color(black)((2x-6))))$

$color(white)((2x^2-5x-3)/(2x-6)) = x+1/2$

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How do you divide $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by $2 x - 6$ $2x-6$ ?

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Explanation:

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How do you divide $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by $2 x - 6$ $2x-6$ ?