Question

Factorise the following: $1 - {y^2}$

Answer 1

Hint: In this question, we are given an expression and we have been asked to factorize it. Observe the expression carefully and you will find an identity in the question. Using that identity, expand the given expression and you will have your answer.

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

Complete step-by-step solution:
We are given an expression, $1 - {y^2}$ and we have to factorize the given expression. Let us see below how it can be done.
$ \Rightarrow 1 - {y^2}$ …. (given)
We can also write it as –
$ \Rightarrow {1^2} - {y^2}$
Now, if we see, both the terms are in complete squares. It reminds us of the formula ${a^2} - {b^2}$.
Let us take $a = 1$ and $b = y$.
Our formula says - ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$.
Using the formula, we will get,
$ \Rightarrow {1^2} - {y^2} = \left( {1 + y} \right)\left( {1 - y} \right)$

Hence, the factors of ${1^2} - {y^2}$ are $\left( {1 + y} \right)\left( {1 - y} \right)$

Note: 1) Here we used the word “expression” and not “equation”. This is because an equation should necessarily have the sign of “equals” (=). But here, in our case, our expression does not have the required sign. Hence, we call it “expression” and not “equation”.
2) Using the factors, we can find the value of $y$ as well. Let us expand our question and find the values.
We have already found the factors of ${1^2} - {y^2}$. They are$\left( {1 + y} \right)\left( {1 - y} \right)$. Now, we will put each factor equal to 1.
$ \Rightarrow 1 + y = 0,1 - y = 0$
Shifting the terms to the other side,
$ \Rightarrow y = - 1,y = 1$
Now, we have the values of $y$ also.
3) We can also find the values of $y$ without finding the factors. Let us see how that can be done.
The expression given to us is - $1 - {y^2}$.
Let us keep the expression equal to $0$.
$ \Rightarrow 1 - {y^2} = 0$
We will shift the variable to the other side,
$ \Rightarrow 1 = {y^2}$
Square rooting both the sides,
$ \Rightarrow \sqrt 1 = \sqrt {{y^2}} $
Therefore, $y = \pm 1$.