Chapter 1.3, Problem 91E

To determine

To calculate: The factor of the expression $x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}$ .

Answer to Problem 91E

The factor of the expression $x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}$ is $\frac{1}{\sqrt{x^3}}{\left(x+1\right)}^2$ .

Explanation of Solution

Given information:

The expression $x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}$ .

Formula used:

To factor an expression, split the terms in the expression into multiplication of simpler expressions, then take the common power out and group the expressions together.

To find the factor of the trinomial of the form $x^2+bx+c$ , find two numbers $r$ and $s$ such that sum of the numbers is equal to coefficient of $x$ $\left(r+s=b\right)$ and product of two numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(x+r\right)$ and $\left(x+s\right)$ are the factors of $x^2+bx+c$ .

  $\begin{array}{c}{x}^2+bx+c=x^2+\left(r+s\right)x+rs\\ =x^2+rx+sx+rs\\ =x\left(x+r\right)+s\left(x+r\right)\\ =\left(x+r\right)\left(x+s\right)\end{array}$

Calculation:

Consider the given expression $x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}$ .

Recall that to factor an expression, split the terms in the expression into multiplication of simpler expressions, then take the common power out and group the expressions together.

The greatest common factor of these terms is $x^{-\frac{3}{2}}$ .

Apply it,

  $\begin{array}{c}{x}^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}=x^{-\frac{3}{2}}\left(1+2\frac{x^{-\frac{1}{2}}}{x^{-\frac{3}{2}}}+\frac{x^{\frac{1}{2}}}{x^{-\frac{3}{2}}}\right)\\ =x^{-\frac{3}{2}}\left(1+2x^{-\frac{1}{2}-\left(-\frac{3}{2}\right)}+x^{\frac{1}{2}-\left(-\frac{3}{2}\right)}\right)\\ =\frac{1}{\sqrt{x^3}}\left(1+2x+x^2\right)\\ =\frac{1}{\sqrt{x^3}}\left(x^2+2x+1\right)\end{array}$

Recall to find the factor of the trinomial of the form $x^2+bx+c$ , find two numbers $r$ and $s$ such that sum of the numbers is equal to coefficient of $x$ $\left(r+s=b\right)$ and product of two numbers is equal to constant term $\left(r\cdot s=c\right)$ , such that $\left(x+r\right)$ and $\left(x+s\right)$ are the factors of $x^2+bx+c$ .

  $\begin{array}{c}{x}^2+bx+c=x^2+\left(r+s\right)x+rs\\ =x^2+rx+sx+rs\\ =x\left(x+r\right)+s\left(x+r\right)\\ =\left(x+r\right)\left(x+s\right)\end{array}$

So, $\frac{1}{\sqrt{x^3}}\left(x^2+2x+1\right)$ will be further simplified as,

  $\begin{array}{c}{x}^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}=\frac{1}{\sqrt{x^3}}\left(x^2+2x+1\right)\\ =\frac{1}{\sqrt{x^3}}\left(x^2+x+x+1\right)\\ =\frac{1}{\sqrt{x^3}}\left[x\left(x+1\right)+1\left(x+1\right)\right]\\ =\frac{1}{\sqrt{x^3}}\left(x+1\right)\left(x+1\right)\end{array}$

Simplify it further as,

  $\begin{array}{c}{x}^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}=\frac{1}{\sqrt{x^3}}\left(x+1\right)\left(x+1\right)\\ =\frac{1}{\sqrt{x^3}}{\left(x+1\right)}^2\end{array}$

Thus, the factor of expression $x^{-\frac{3}{2}}+2x^{-\frac{1}{2}}+x^{\frac{1}{2}}$ is $\frac{1}{\sqrt{x^3}}{\left(x+1\right)}^2$ .