Question

show your solution
B. perform the indicated operation.
$\frac{2x - 1}{ {2x}^{2} + 5 \times + 3 } + \frac{2}{3x + 3}$
​

Answer 1

$\frac{10x + 3}{3 (x + 1)(2x + 3)}$

Question :

$\frac{2x - 1}{2 {x}^{2} + 5x + 3 } + \frac{2}{3x + 3}$

Solution :

$= \frac{2x - 1}{2 {x}^{2} + 5x + 3} + \frac{2}{3x + 3}$

Let's find the roots of quadratic equation.

$= \frac{2x - 1}{2 {x}^{2} + 2x + 3x + 6 } + \frac{2}{3x + 3}$

$= \frac{2x - 1}{2x(x + 1) + 3(x + 1)} + \frac{2}{3x + 3}$

$= \frac{2x - 1}{(2x + 3)(x + 1)} + \frac{2}{3x + 3}$

Cross Multiplying the terms

$= \frac{(3x + 3)(2x - 1) + 2(2x + 3)(x + 1)}{(2x + 3)(x + 1)(3x + 3)}$

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

$= \frac{6 {x}^{2} + 3x - 3 + 4 {x}^{2} + 4x + 6x + 6}{(2x + 3)(x + 1)(3x + 3)}$

$= \frac{6 {x}^{2} + 4 {x}^{2} + 3x + 4x + 6x + 6 - 3}{(2x + 3)(x + 1)(3x + 3)}$

$= \frac{10 {x}^{2} + 13x + 3 }{(2x + 3)(x + 1)(3x + 3)}$

Let's find the roots of quadratic equation,

$= \frac{10 {x}^{2} + 10x + 3x + 3 }{(2x + 3)(x + 1)(3x + 1)}$

$= \frac{10x(x + 1) + 3(x + 1)}{(2x + 3)(x + 1)(3x + 1)}$

$= \frac{(10x + 3)(x + 1)}{(2x + 3)(x + 1)(3x + 1)}$

$= \frac{(10x + 3)}{(2x + 3)(3x + 3)}$

$= \frac{10x + 3}{3 (x + 1)(2x + 3)}$

Answer 2

Answer:

$Question :</p><p></p><p>\frac{2x - 1}{2 {x}^{2} + 5x + 3 } + \frac{2}{3x + 3}2x2+5x+32x−1+3x+32</p><p></p><p>Solution :</p><p></p><p>= \frac{2x - 1}{2 {x}^{2} + 5x + 3} + \frac{2}{3x + 3}=2x2+5x+32x−1+3x+32</p><p></p><p>Let's find the roots of quadratic equation.</p><p></p><p>= \frac{2x - 1}{2 {x}^{2} + 2x + 3x + 6 } + \frac{2}{3x + 3}=2x2+2x+3x+62x−1+3x+32</p><p></p><p>= \frac{2x - 1}{2x(x + 1) + 3(x + 1)} + \frac{2}{3x + 3}=2x(x+1)+3(x+1)2x−1+3x+32</p><p></p><p>= \frac{2x - 1}{(2x + 3)(x + 1)} + \frac{2}{3x + 3}=(2x+3)(x+1)2x−1+3x+32</p><p></p><p>Cross Multiplying the terms</p><p></p><p>= \frac{(3x + 3)(2x - 1) + 2(2x + 3)(x + 1)}{(2x + 3)(x + 1)(3x + 3)}=(2x+3)(x+1)(3x+3)(3x+3)(2x−1)+2(2x+3)(x+1)</p><p></p><p>= \frac{(6 {x}^{2} - 3x + 6x - 3) + 2(2 {x}^{2}+ 2x + 3x + 3 )}{(2x + 3)(x + 1)(3x + 3)}=(2x+3)(x+1)(3x+3)(6x2−3x+6x−3)+2(2x2+2x+3x+3)</p><p></p><p>= \frac{6 {x}^{2} + 3x - 3 + 4 {x}^{2} + 4x + 6x + 6}{(2x + 3)(x + 1)(3x + 3)}=(2x+3)(x+1)(3x+3)6x2+3x−3+4x2+4x+6x+6</p><p></p><p>= \frac{6 {x}^{2} + 4 {x}^{2} + 3x + 4x + 6x + 6 - 3}{(2x + 3)(x + 1)(3x + 3)}=(2x+3)(x+1)(3x+3)6x2+4x2+3x+4x+6x+6−3</p><p></p><p>= \frac{10 {x}^{2} + 13x + 3 }{(2x + 3)(x + 1)(3x + 3)}=(2x+3)(x+1)(3x+3)10x2+13x+3</p><p></p><p>Let's find the roots of quadratic equation,</p><p></p><p>= \frac{10 {x}^{2} + 10x + 3x + 3 }{(2x + 3)(x + 1)(3x + 1)}=(2x+3)(x+1)(3x+1)10x2+10x+3x+3</p><p></p><p>= \frac{10x(x + 1) + 3(x + 1)}{(2x + 3)(x + 1)(3x + 1)}=(2x+3)(x+1)(3x+1)10x(x+1)+3(x+1)</p><p></p><p>= \frac{(10x + 3)(x + 1)}{(2x + 3)(x + 1)(3x + 1)}=(2x+3)(x+1)(3x+1)(10x+3)(x+1)</p><p></p><p>= \frac{(10x + 3)}{(2x + 3)(3x + 3)}=(2x+3)(3x+3)(10x+3)</p><p></p><p>= \frac{10x + 3}{3 (x + 1)(2x + 3)}=3(x+1)(2x+3)10x+3</p><p></p><p>$