Question

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

Answer 1

Hint: As we can see that the given expression is a quadratic expression, it will have two roots. Each root will give one factor. So, there will be two factors in total. To find the roots we will first compare it with the standard equation. Then, we will find the determinant and then we will use the formula for finding the root of the quadratic equation.

Complete step by step answer:
Given expression;
\[{x^2} + 1\]
Now we will equate it to zero for finding the roots. So, we have;
\[ \Rightarrow {x^2} + 1 = 0\]
Now we know that the standard form of the quadratic equation is \[a{x^2} + bx + c = 0\]. So, on comparing we get;
\[a = 1,b = 0,c = 1\]
Now we will find the discriminant of the above equation. We know;
\[D = {b^2} - 4ac\]
On putting the values, we get;
\[ \Rightarrow D = {0^2} - 4 \times 1 \times 1\]
Solving we get;
\[ \Rightarrow D = - 4\]

As we can see that the discriminant of the equation is negative. So, we will have imaginary roots.Now we know that;
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
We can also write it as;
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
On putting the values, we get;
\[ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt { - 4} }}{{2 \times 1}}\]
On simplification by writing \[\sqrt { - 1} = i\], we get;
\[ \Rightarrow x = \dfrac{{ \pm 2i}}{2}\]
On dividing we get;
\[ \Rightarrow x = \pm i\]
Since, \[x = \pm i\] are the roots.

Therefore, \[\left( {x + i} \right){\text{ and }}\left( {x - i} \right)\] are the factors of \[{x^2} + 1\].

Note: We can also solve this question simply by just manipulating the terms. We have the given expression as;
\[{x^2} + 1\]
We can also write it as;
\[{x^2} - {\left( {\sqrt { - 1} } \right)^2}\]
Now we will use \[\sqrt { - 1} = i\]. So, on replacing we get;
\[ {x^2} - {\left( i \right)^2}\]
Now we know that \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]. So, using this we get;
\[ = \left( {x + i} \right)\left( {x - i} \right)\]
Hence, \[{x^2} + 1 = \left( {x + i} \right)\left( {x - i} \right)\]. One thing to note here is that the imaginary roots always occur in pairs. As we can see in the above question, if one root is \[i\] then the other root is \[ - i\].