Chapter 1.3, Problem 128E

To determine

To calculate: The common factor of the expression $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ .

Answer to Problem 128E

The common factor of the expression $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ is $2x^{-1/2}{\left(3x+4\right)}^{-1/2}$ .

Explanation of Solution

Given information:

The expression $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ .

Formula used:

To factor out the common factor from a polynomial, find out the greatest common factor and express the polynomial as a product of the simpler ones.

Calculation:

Consider the given expression $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ .

Recall that to factor out the common factor from a polynomial, find out the greatest common factor and express the polynomial as a product of the simpler ones.

Here, the terms have the common factor $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{-1/2}$ .

So, $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ can be written in simplified form as,

  $\begin{array}{c}\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}=\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{-1/2}\left[{\left(3x+4\right)}^{1/2}\cdot {\left(3x+4\right)}^{1/2}-3\left(x^{1/2}\cdot x^{1/2}\right)\right]\\ =\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{-1/2}\left(3x+4-3x\right)\\ =\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{-1/2}\left(4\right)\\ =2x^{-1/2}{\left(3x+4\right)}^{-1/2}\end{array}$

Thus, the common factor of the expression $\frac{1}{2}{x}^{-1/2}{\left(3x+4\right)}^{1/2}-\frac{3}{2}{x}^{1/2}{\left(3x+4\right)}^{-1/2}$ is $2x^{-1/2}{\left(3x+4\right)}^{-1/2}$ .