Question

$\frac{ {x}^{3} - 1 }{ {x}^{2} + 2 }$
मैं क्या जोड़ा जाए कि योगफल$\frac{x {}^{3} - 2x {}^{2} - 5}{ {x}^{2} + 2}$ हो जाए

Answer 1

$\bold{\huge{Hay!!}}$


$\bold{Dear\:user!!}$



$\bold{\underline{Question-}}$


$\frac{ {x}^{3} - 1 }{ {x}^{2} + 2 }$
मैं क्या जोड़ा जाए कि योगफल$\frac{x {}^{3} - 2x {}^{2} - 5}{ {x}^{2} + 2}$ हो जाए


$\bold{\underline{Answer-}}$

हल- माना अभीष्ट परिमेय व्यंजक = p(x)/q(x)


तब,
$\frac{x {}^{3} - 1}{ {x}^{2} + 2 } + \frac{p(x)}{q(x)} = \frac{x {}^{3} - {2x}^{2} - 5}{ {x}^{2} + 2} \\ \\ \\ \\ = > \frac{p(x)}{q(x)} = \frac{ {x}^{3} - {2x}^{2} - 5}{ {x}^{2} + 2 } - \frac{ {x}^{ 3} + 1}{ {x}^{2} + 2 }$



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

Answer 2

Heya....

===== Answer ====

Solution :-

x^3-1/x^2+2 = P(x)/q(x) = x^3-2x-5 / x^2+2

= P(x)/q(x) = x^3-2x-5/x^2+2 - x^2+1/x^2+2

= x^3-2x^2-5x-x^3+1 / x^2+2

= -2x^2-4/x^2+2 = -2x(x^2+2)/12x^2+2

= -2

So.. in the given eq u must add -2 to get the given one...

Thank you