Chapter 6.1, Problem 5OE

To determine

To calculate: The simplified form of the expression $\frac{x^2-2x+1}{x-1}$ and state restrictions on variable.

Answer to Problem 5OE

The simplified value of the expression $\frac{x^2-2x+1}{x-1}$ is $x-1$ and restriction is $x\ne 1$.

Explanation of Solution

Given information:

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

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

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

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, factorize the numerator by using the formula of perfect square ${\left(X-Y\right)}^2=X^2-2XY+Y^2$ and then strike out common terms and simplify it as,

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

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

For denominator not to be zero, $x\ne 1$.

Thus, the simplified value of the expression $\frac{x^2-2x+1}{x-1}$ is $x-1$ and restriction is $x\ne 1$.