Solve the following quadratic equations by factorization:

$\frac{4}{x}-3=\frac{5}{2 x+3}$

$\Rightarrow \frac{4-3 x}{x}=\frac{5}{2 x+3}$

$\Rightarrow(4-3 x)(2 x+3)=5 x$

$\Rightarrow 8 x+12-6 x^{2}-9 x=5 x$

$\Rightarrow-6 x^{2}-6 x+12=0$

$\Rightarrow x^{2}+x-2=0$

$\Rightarrow x^{2}+2 x-x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow(x-1)(x+2)=0$

$\Rightarrow x-1=0$ or $x+2=0$

$\Rightarrow x=1$ or $x=-2$

Hence, the factors are 1 and −2.