Chapter 6.1, Problem 48WE

To determine

To calculate: The simplified form of the expression $\frac{x^4-10x^2+9}{3-2x-x^2}$ and state restrictions on variable.

Answer to Problem 48WE

The simplified value of the expression $\frac{x^4-10x^2+9}{3-2x-x^2}$ is $-\left(x-1\right)\left(x+3\right)$ and restriction is $x\ne 1,x\ne -3$.

Explanation of Solution

Given information:

The expression $\frac{x^4-10x^2+9}{3-2x-x^2}$.

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 $\frac{x^4-10x^2+9}{3-2x-x^2}$.

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 both numerator and denominator by using middle term splitting method and simplify it as,

  $\begin{array}{c}\frac{x^4-10x^2+9}{3-2x-x^2}=\frac{x^4-9x^2-x^2+9}{-1\left(x^2+2x-3\right)}\\ =-\frac{x^2\left(x^2-9\right)-1\left(x^2-9\right)}{x^2+3x-x-3}\\ =-\frac{\left(x^2-1\right)\left(x^2-9\right)}{x\left(x+3\right)-1\left(x+3\right)}\\ =-\frac{\left(x^2-1\right)\left(x^2-9\right)}{\left(x-1\right)\left(x+3\right)}\end{array}$

Now, apply $X^2-Y^2=\left(X+Y\right)\left(X-Y\right)$ in numerator and then strike out common terms and simplify it further as,

  $\begin{array}{c}\frac{x^4-10x^2+9}{3-2x-x^2}=-\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\\ =-\left(x+1\right)\left(x-3\right)\end{array}$

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

So, denominator will be zero when,

  $\begin{array}{c}-1\left(x-1\right)\left(x+3\right)=0\\ x-1=0,x+3=0\\ x=1,x=-3\end{array}$

For denominator not to be zero, $x\ne 1,x\ne -3$ . 3

Thus, the simplified value of the expression $\frac{x^4-10x^2+9}{3-2x-x^2}$ is $-\left(x+1\right)\left(x-3\right)$ and restriction is $x\ne 1,x\ne -3$.