Question

How do you write the rational expression in lowest terms \[\dfrac{4p\left( p+4 \right)}{20{{p}^{2}}\left( p-4 \right)}\]?

Answer 1

Hint: In the above question, we have been given a rational expression and are asked to write in the lowest terms. Writing a fraction in the lowest terms means to cancel the highest common factor from the numerator and the denominator. Therefore, in the given fraction \[\dfrac{4p\left( p+4 \right)}{20{{p}^{2}}\left( p-4 \right)}\] we need to factorise both of the numerator and the denominator. The numerator, which is equal to $4p\left( p+4 \right)$, can be factored as $2\cdot 2\cdot p\cdot \left( p+4 \right)$. And the denominator, which is equal to $20{{p}^{2}}\left( p-4 \right)$, can be factored as $2\cdot 2\cdot 5\cdot p\cdot p\cdot \left( p-4 \right)$. Finally, on cancelling all the factors common to the numerator and the denominator, we will obtain the given expression in the lowest terms.

Complete step by step solution:
Let us consider the rational expression given in the above question as
$\Rightarrow E=\dfrac{4p\left( p+4 \right)}{20{{p}^{2}}\left( p-4 \right)}$
According to the question we need to write the above expression in the lowest terms. For this, we need to cancel out the highest common factor from the numerator and the denominator. Therefore, we need to factorise both the numerator and the denominator. Therefore, we first consider the numerator, which is
\[\Rightarrow N=4p\left( p+4 \right)\]
The polynomial in the numerator is already factored. But we can factorise $4$ as $2\cdot 2$ in the above expression to get
\[\begin{align}
  & \Rightarrow N=2\cdot 2p\left( p+4 \right) \\
 & \Rightarrow N=2\cdot 2\cdot p\cdot \left( p+4 \right) \\
\end{align}\]
Now, we consider the denominator which is
$\Rightarrow D=20{{p}^{2}}\left( p-4 \right)$
The polynomial \[{{p}^{2}}\left( p-4 \right)\] can be factored as \[p\cdot p\cdot \left( p-4 \right)\] so that we can write
$\Rightarrow D=20\cdot p\cdot p\cdot \left( p-4 \right)$
Now, \[20\] can be factored as \[2\cdot 2\cdot 5\] so that we can write
$\Rightarrow D=2\cdot 2\cdot 5\cdot p\cdot p\cdot \left( p-4 \right)$
Therefore, the rational expression now becomes
$\Rightarrow E=\dfrac{2\cdot 2\cdot p\cdot \left( p+4 \right)}{2\cdot 2\cdot 5\cdot p\cdot p\left( p-4 \right)}$
Cancelling p, we get
$\Rightarrow E=\dfrac{2\cdot 2\cdot \left( p+4 \right)}{2\cdot 2\cdot 5\cdot p\left( p-4 \right)}$
Finally, cancelling out $2\cdot 2$ we get
\[\begin{align}
  & \Rightarrow E=\dfrac{\left( p+4 \right)}{5\cdot p\left( p-4 \right)} \\
 & \Rightarrow E=\dfrac{\left( p+4 \right)}{5p\left( p-4 \right)} \\
\end{align}\]

Hence, the given rational expression is finally obtained in the lowest term as \[\dfrac{\left( p+4 \right)}{5p\left( p-4 \right)}\].

Note: We must note that not only the polynomials, but also the numbers in the numerator and the denominator have to be factorized. Also, all the factors of the numbers have to be prime only. Then only we can ensure that the highest common factor between the numerator and the denominator has been cancelled.