Question

Factor the expression completely. Begin by factoring out the lowest power of each common factor.

${\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Short answer

The solution${\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{x^2+3}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Step-by-step solution

Step 1. Apply the concept of distributive property.

If A, B and C are real numbers, then

$A\left(B+C\right)=\left(B+C\right)A=AB+AC$

Step 2. Apply the concept of product of sum and difference.

If A and B are real numbers, then

$\left(A+B\right)\left(A-B\right)=A^2-B^2$

Step 3. Analyze the expression.

Given${\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

We have${\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{2}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\mathrm{........}\left[x^{-n}=\frac{1}{x^n}\right]$

Step 4. Factor the polynomial.

Now, using the law of exponents polynomial can be factored as:

$\begin{array}{l}{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{2}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \Rightarrow {\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cdot \frac{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{2}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ \Rightarrow {\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\left(x^2+1\right)+2}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\mathrm{......}\left[x^a\cdot x^b=x^{a+b}\right]\\ \Rightarrow {\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2{\left(x^2+1\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{x^2+3}{{\left(x^2+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$