How do you find the restricted values of x or the rational expression (x^3-2x^2-8x)/(x^2-4x)?

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5
Dec 16, 2015

$x \ne 0 , 4$

Explanation:

Start by simplifying the equation:

$\frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x}$

$= \frac{x \left({x}^{2} - 2 x - 8\right)}{x \left(x - 4\right)}$

$= \frac{x \left(x - 4\right) \left(x + 2\right)}{\textcolor{red}{x} \textcolor{b l u e}{\left(x - 4\right)}}$

Recall that any fraction cannot have a denominator of $0$. To find the restrictions for $x$, set each polynomial or term in the denominator to cannot equal to $0$, and solve for $x$.

Finding the restrictions

$1. \textcolor{red}{x} \ne 0$

$2. \textcolor{b l u e}{x - 4} \ne 0$
$\textcolor{w h i t e}{i \times \times} x \ne 4$

$\therefore$, the restrictions are $x$ are $x \ne 0$ and $x \ne 4$.

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NagrajB Share
Dec 16, 2015

x cannot be zero
x cannot be 4

Explanation:

$\left(x\right) \cdot \frac{{x}^{2} - 2 x - 8}{x \left(x - 4\right)}$
x can be cancelled in both numerator and denominator if and only if x not equal to zero

$\frac{\left(x - 4\right) \left(x + 2\right)}{x - 4}$

x - 4 cannot be cancelled if x = 0

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