Question

How do you factor $\dfrac{1}{4}{x^2} - 4$ ?

Answer 1

Hint: In this question, we have been asked how do we factorise the given expression. If you observe carefully, the given expression resembles an identity. Write down the identity and expand the given equation using that identity. You will get your answer.

Formula used: ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$

Complete step-by-step solution:
We have been asked to factorise $\dfrac{1}{4}{x^2} - 4$.
This expression resembles the identity ${a^2} - {b^2}$.
We can write the given expression as ${\left( {\dfrac{x}{2}} \right)^2} - {2^2}$. Let ${a^2} = \dfrac{{{x^2}}}{4}$ and ${b^2} = 4$. Therefore, we can say that $a = \dfrac{x}{2}$ and $b = 2$.
We know that ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. Using this identity, we will get,
$ \Rightarrow \dfrac{1}{4}{x^2} - 4 = \left( {\dfrac{x}{2} + 2} \right)\left( {\dfrac{x}{2} - 2} \right)$

Therefore, the factors of $\dfrac{1}{4}{x^2} - 4$ are $\left( {\dfrac{x}{2} + 2} \right)\left( {\dfrac{x}{2} - 2} \right)$.

Note: We can solve this question in the following way also:
Instead of finding the factors at an early stage, we will find the factors later.
We are given an equation - $\dfrac{1}{4}{x^2} - 4$
In the first step, let us find the LCM and make the denominators equal.
The LCM will be $4$. In order to make the denominators equal, we will multiply and divide the second term by $4$.
$ \Rightarrow \dfrac{1}{4}{x^2} - \dfrac{{4 \times 4}}{4}$
On simplifying, we get,
$ \Rightarrow \dfrac{{{x^2}}}{4} - \dfrac{{16}}{4}$
$ \Rightarrow \dfrac{{{x^2} - 16}}{4}$
Now, we will use the identity ${a^2} - {b^2}$ in the numerator.
The identity says that ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. In our case, $a = x$, $b = 4$.
Using the identity in the numerator, we get,
$ \Rightarrow {x^2} - {4^2} = \left( {x + 4} \right)\left( {x - 4} \right)$
These are the factors of the numerator. Putting them back in the question,
$ \Rightarrow \dfrac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{4}$
Now, we will take this equation equal to $0$.
$ \Rightarrow \dfrac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{4} = 0$
Sending the denominator of LHS to the other side
$ \Rightarrow \left( {x + 4} \right)\left( {x - 4} \right) = 0 \times 4 = 0$
Now, we will keep each factor equal to $0$ and find the value of x.
$ \Rightarrow x + 4 = 0,x - 4 = 0$
Shifting to find the required values,
$ \Rightarrow x = 4, - 4$
Hence, we can solve the given question in this way also.