How do you divide $\frac{4 x^{3} - 5 x^{2} + 9 x - 3}{2 x - 3}$ $(4x^3-5x^2+9x-3)/(2x-3)$ ?

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How do you divide $\frac{4 x^{3} - 5 x^{2} + 9 x - 3}{2 x - 3}$ $(4x^3-5x^2+9x-3)/(2x-3)$ ?

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Answer 1 · 2015-12-17T02:45:54

$y = \frac{(x^{2} + 3) (4 x - 5) (3 x - 1)}{2 x - 3}$ $y=((x^2+3)(4x-5)(3x-1))/(2x-3)$

Explanation:

$y = \frac{4 x^{3} - 5 x^{2} + 9 x - 3}{2 x - 3}$ $y=(4x^3-5x^2+9x-3)/(2x-3)$

Divide $2$ $2$ terms at once by a common factor.
$y = \frac{x^{2} (4 x - 5) + 3 (3 x - 1)}{2 x - 3}$ $y=(x^2(4x-5)+3(3x-1))/(2x-3)$

Rewrite the equation.
$y = \frac{(x^{2} + 3) (4 x - 5) (3 x - 1)}{2 x - 3}$ $y=((x^2+3)(4x-5)(3x-1))/(2x-3)$ , $x \neq \frac{3}{2}$ $x!=3/2$

Since the equation cannot be simplified any further, the final answer is $y = \frac{(x^{2} + 3) (4 x - 5) (3 x - 1)}{2 x - 3}$ $y=((x^2+3)(4x-5)(3x-1))/(2x-3)$ where $x \neq \frac{3}{2}$ $x!=3/2$ .

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How do you divide $\frac{4 x^{3} - 5 x^{2} + 9 x - 3}{2 x - 3}$ $(4x^3-5x^2+9x-3)/(2x-3)$ ?

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Explanation:

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How do you divide $\frac{4 x^{3} - 5 x^{2} + 9 x - 3}{2 x - 3}$ $(4x^3-5x^2+9x-3)/(2x-3)$ ?