Question

How do you factor \[{{x}^{2}}-1\] ?

Answer 1

Hint: To find the factors of a quadratic equation we first equate that polynomial to zero that it leads to an equation. Now calculate the roots of this equation that is the value of \[x\] which satisfies this equation means putting it will result in zero. In this particular question, we just have to simplify and don't need any manipulations as the coefficient of \[x\] is zero.

Complete step by step solution:
Let equate this polynomial with zero
\[\Rightarrow {{x}^{2}}-1=0\]
\[\Rightarrow {{x}^{2}}=1\]
Now taking square root both sides
\[\Rightarrow \sqrt{{{x}^{2}}}=\pm \sqrt{1}\]
Since square root means the base has exponential power as \[\dfrac{1}{2}\]
And the square root \[1\] is \[1\]
\[\Rightarrow {{({{x}^{2}})}^{\dfrac{1}{2}}}=\pm 1\]
\[\Rightarrow x=\pm 1\]
Thus, we have two roots of this quadratic equation, \[x=1\] and \[x=-1\]
And we also know that if a quadratic equation has roots \[a\] and \[b\] then its factors will be
\[(x-a)\] and \[(x-b)\]
Since in this equation we have roots \[1\] and \[-1\]

Hence, the factors of quadratic equation \[({{x}^{2}}-1)=(x-1)(x+1)\]

Note: To find the roots of quadratic equation always we first need to equate with zero the either we will use completing the perfect square method that we have used but in this particular question we can use the identity \[({{a}^{2}}-{{b}^{2}})=(a-b)(a+b)\] and on comparing \[a\] is replaced by \[x\] and b is replaced by \[1\].