
How do you find the restricted values for for the rational expression $\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$ ?
Answer
543.6k+ views
Hint:First of all, simplify the equation given in the question by taking $x$ common from both numerator and denominator. Recall the fact that any fraction cannot have a denominator equal to 0 as it gives infinity.
Complete Step by Step Solution:
The equation given in the question is –
$\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$
In the above equation, we have to find the restricted values for $x$ which means that the value of $x$ in the given equation is not possible. So, we cannot directly find it, first of all, we should simplify the equation which will make it easy to find the restricted values for $x$.
Now, simplifying the given equation in the question –
$\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$
Taking $x$ common from numerator and denominator from the above given equation, we get –
$ \Rightarrow \dfrac{{x\left( {{x^2} - 2x - 8} \right)}}{{x\left( {x - 4} \right)}}$
The above equation after factorization can also be written as –
$ \Rightarrow \dfrac{{x\left( {x - 4} \right)\left( {x + 2} \right)}}{{x\left( {x - 4} \right)}}$
If in the fraction, the denominator becomes equal to 0 then, the answer we get is infinite and if the numerator and denominator is equal to 0 then, the value we get is also infinite. So, the denominator cannot be equal to zero. This one of the restrictions for $x$. To find other restrictions set each polynomial or term in the denominator not equal to 0 and then solve it for $x$.
Finding the restrictions for $x$ -
$x \ne 0$
$
x - 4 \ne 0 \\
x \ne 4 \\
$
Therefore, the values of $x$ cannot be equal to 4 and 0 which are the restrictions for $x$ in the equation $\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$.
Note: Always simplify the equation as we can because it will make the calculations to be easy and the answer will be correct while if it is not simplified it will become hard to find the answer and there are higher chances for the answer to be wrong.
Complete Step by Step Solution:
The equation given in the question is –
$\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$
In the above equation, we have to find the restricted values for $x$ which means that the value of $x$ in the given equation is not possible. So, we cannot directly find it, first of all, we should simplify the equation which will make it easy to find the restricted values for $x$.
Now, simplifying the given equation in the question –
$\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$
Taking $x$ common from numerator and denominator from the above given equation, we get –
$ \Rightarrow \dfrac{{x\left( {{x^2} - 2x - 8} \right)}}{{x\left( {x - 4} \right)}}$
The above equation after factorization can also be written as –
$ \Rightarrow \dfrac{{x\left( {x - 4} \right)\left( {x + 2} \right)}}{{x\left( {x - 4} \right)}}$
If in the fraction, the denominator becomes equal to 0 then, the answer we get is infinite and if the numerator and denominator is equal to 0 then, the value we get is also infinite. So, the denominator cannot be equal to zero. This one of the restrictions for $x$. To find other restrictions set each polynomial or term in the denominator not equal to 0 and then solve it for $x$.
Finding the restrictions for $x$ -
$x \ne 0$
$
x - 4 \ne 0 \\
x \ne 4 \\
$
Therefore, the values of $x$ cannot be equal to 4 and 0 which are the restrictions for $x$ in the equation $\dfrac{{{x^3} - 2{x^2} - 8x}}{{{x^2} - 4x}}$.
Note: Always simplify the equation as we can because it will make the calculations to be easy and the answer will be correct while if it is not simplified it will become hard to find the answer and there are higher chances for the answer to be wrong.
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