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

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Answer 1 · 2015-12-17T02:45:54

$y=((x^2+3)(4x-5)(3x-1))/(2x-3)$

$y=(4x^3-5x^2+9x-3)/(2x-3)$

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

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

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

How do you divide (4x^3-5x^2+9x-3)/(2x-3) (4x^3-5x^2+9x-3)/(2x-3) ?