Question

Reduce to its simplest form:$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$.

Answer 1

Now,

$\begin{array}{c}\left(x^2+1\right)\left(x^2+2\right)\\ Leta=x^2,b=1c=x^2,d=2\\ =x^2{x}^2+x^2.2+1.x^2+1.2\\ =x^4+3x^2+2\\ =\left(x^2+3\right)\left(x^2+4\right)\\ weknowthat,\\ \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd\\ =x^2{x}^2+x^2.4+3x^2+3.4\\ =x^4+7x^2+12\\ =\frac{x^4+3x^2+2}{x^4+7x^2+12}\\ Divide:\frac{x^4+3x^2+2}{x^4+7x^2+12}=1+\frac{-4x^2-10}{x^4+7x^2+12}\\ =1+\frac{-4x^2-10}{x^4+7x^2+12}\end{array}$