Question

How do you simplify and find the excluded values of \[\dfrac{16{{x}^{2}}-4x}{3-x}\]?

Answer 1

Hint: This type of problem is based on the concept of polynomial functions. First, we have to consider the whole function. Convert the first term of the numerator as \[{{\left( 4x \right)}^{2}}\]. Take 4x common from both the terms of the numerator of the function. Then, we get a function \[\dfrac{4x\left( 4x-1 \right)}{3-x}\] which is the simplified form of the function. To find the excluded values of the function, equate the denominator to zero. Solve the denominator and find the value of x for which the denominator is equal to zero which is the excluded value of the function.

Complete step by step answer:
According to the question, we are asked to simplify and find the excluded value of \[\dfrac{16{{x}^{2}}-4x}{3-x}\].
We have been given the equation is \[\dfrac{16{{x}^{2}}-4x}{3-x}\]. -----(1)
We first have to first simplify the given equation.
We know that 16 is the square of 4 and \[{{x}^{2}}\] is the square of x.
Therefore, we can express the function as
\[\dfrac{{{4}^{2}}{{x}^{2}}-4x}{3-x}\]
We know that \[{{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}\]. Using this property of powers, we get
\[\dfrac{16{{x}^{2}}-4x}{3-x}=\dfrac{{{\left( 4x \right)}^{2}}-4x}{3-x}\]
We find that 4x are common in the numerator of the function. Let us now take 4x common from both the terms of the numerator.
\[\Rightarrow \dfrac{16{{x}^{2}}-4x}{3-x}=\dfrac{4x\left( 4x-1 \right)}{3-x}\]
Now, look for any common terms to cancel out. We find that there are no more terms to cancel out or take common.
Therefore, the simplified form of the function is \[\dfrac{4x\left( 4x-1 \right)}{3-x}\].
We now have to find the excluded values of the function (1).
We know that a function is not defined when the denominator is equal to 0. To find the excluded values of the function, we have to equate the denominator equal to 0.
\[\Rightarrow 3-x=0\]
To find the value of x, we have to subtract 3 from the expression.
\[\Rightarrow 3-x-3=0-3\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 3, we get
-x=0-3
On further simplifications, we get
-x=-3.
But, we need the value of x. let us divide the whole expression by -1.
\[\Rightarrow \dfrac{-x}{-1}=\dfrac{-3}{-1}\]
We find that -1 are common in both the numerator and denominator of the LHS and RHS. On cancelling -1, we get
x=3
Therefore, the excluded value of x in the function is 3.
Hence, the simplified form of the function \[\dfrac{16{{x}^{2}}-4x}{3-x}\] is \[\dfrac{4x\left( 4x-1 \right)}{3-x}\] and the excluded value of x is 3.

Note:
Whenever we get such types of questions, we have to equate the denominator of the function equal to 0 to find the excluded values. We should not equate the denominator of the simplified form of the function. We might not get all the excluded values. In this case, we only have one excluded value.