Question

How do you factor by grouping: $1 - a + ab - b$?

Answer 1

Hint: In this question, we will find the factor of the given term. Then we will group the similar terms and rearrange the signs to get the final factored answer.

Complete step-by-step solution:
We have the given question as:
$ \Rightarrow 1 - a + ab - b$
Since the given equation is not directly factorable, we will rearrange the equations to get the similar terms, it can be written as:
$ \Rightarrow ab - a + 1 - b$
Now let’s split the terms into two separate equations,
We can make $ab - a \to (1)$ and $1 - b \to (2)$, therefore the original equation will be $(1) + (2)$.
Now consider $(1)$
$ \Rightarrow ab - a$
Now since $a$ is common in both the terms, we can take it out as common and write it as:
$ \Rightarrow a(b - 1)$
Now consider $(2)$
$ \Rightarrow 1 - b$
Now on taking $1$ as common we get:
$ \Rightarrow 1(1 - b)$
Since it is not similar to the term in equation $(1)$, we can rewrite the above term as:
$ \Rightarrow - 1(b - 1)$
Now since both the equations are simplified, we can join equation $(1)$ and $(2)$ write it as:
$ \Rightarrow a(b - 1) - 1(b - 1)$
Now since the term $(b - 1)$ is common in both the terms we can take it out as common and write it as:
$ \Rightarrow (a - 1)(b - 1)$

$(a-1)(b-1)$ is the final factorized way of writing the given equation.

Note: To check whether the final factored equation is correct we can multiply the factors, and if we get the given equation back, the factors are correct.
We can multiply the factors as:
$ \Rightarrow a \times b - a \times - 1 + - 1 \times b + - 1 \times - 1$
On simplifying we get:
$ \Rightarrow ab - a - b + 1$
On rearranging we get:
$ \Rightarrow 1 - a + ab - b$, which is the original equation, therefore the factors are correct.
It is to be remembered that factors are the digits which make up a number, a factor should be indivisible by any number except $1$ and it should not be in the form of a number which is raised to a power.