Question

How do you factor ${{\left( x-1 \right)}^{2}}-4$ ?

Answer 1

Hint: We are asked to factor ${{\left( x-1 \right)}^{2}}-4$, we will use step by step simplification.
First we take the common multiple which could be common to both terms, after that we see how many terms are there and which identity can be used. We will use $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ to learn about how should be able to factor different kind of terms.

Complete step by step answer:
We are given an expression as ${{\left( x-1 \right)}^{2}}-4$.
We are asked to factor this expression. Factoring an expression means that we have to split this expression into a more simple form.
For example: we take $2x+4$
As we can see that ‘2’ and ‘4’ has ‘2’ as a common factor so we take it out.
So we get –
$2x+4=2\left( x+2 \right)$
Now we see it cannot be split so simplest term is $2x+4=2\left( x+2 \right)$
Consider another example –
${{x}^{2}}-{{4}^{2}}$ , to simplify it –
We use that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Considering ‘a’ as ‘x’ and ‘b’ as ‘y’, we get –
$\left( {{x}^{2}}-{{y}^{2}} \right)$ will be factor as –
${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$
To factor term we will usually follow certain step
Step 1: factor out a greatest common factor first is possible.
In our problem we have ${{\left( x-1 \right)}^{2}}-4$ , we can see that these 2 terms have nothing in common so no term can be taken out as a common factor.
Step 2: count the number of terms is we have 2 terms we usually use ${{a}^{2}}-{{b}^{2}}$ or ${{a}^{3}}-{{b}^{3}}$ or ${{a}^{3}}+{{b}^{3}}$
${{a}^{2}}-{{b}^{2}}$ is defined as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
${{a}^{3}}-{{b}^{3}}$ is defined as $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
Now, in our equation we have, ${{\left( x-1 \right)}^{2}}-4$
We can write $4={{2}^{2}}$ so,
We have ${{\left( x-1 \right)}^{2}}-{{2}^{2}}$
They have two terms and power is ‘2’ so, we use ${{a}^{2}}-{{b}^{2}}$ .
We consider ‘a’ as $x-1$ and ‘b’ as ‘2’. So,
Using ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we get –
${{\left( x-1 \right)}^{2}}-{{2}^{2}}=\left( x-1+2 \right)\left( x-1-2 \right)$
By simplifying we get –
$=\left( x+1 \right)\left( x-3 \right)$
So we get –
${{\left( x-1 \right)}^{2}}-4={{\left( x-1 \right)}^{2}}-{{2}^{2}}=\left( x+1 \right)\left( x-3 \right)$

So factor of ${{\left( x-1 \right)}^{2}}-4$ is $\left( x-3 \right)\left( x+1 \right)$

Note: Remember that here $\left( x+1 \right)$ is considered as a single term then we applied ${{a}^{2}}-{{b}^{2}}$ so here ‘a’ will be $a=x-1$ , so error like considering $a=n$ usually happen so we need to be clear while solving. Also remember that constant is added to constant and variable is added only to variable, we cannot add variable to constant like $x+2=2x$ or $x+3+1=4x$ these are incorrect calculations.