Question

How do you factor `x^2 + 25` completely?

Answer 1

This can only be factored using non-Real Complex coefficients:

`x^2+25 = (x-5i)(x+5i)`

Explanation:

Notice that if `x` is Real then `x^2 >= 0`, so `x^2+25` has no linear factors with Real coefficients.

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We can write `x^2+25` as a difference of squares and factor it as follows:

`x^2+25 = x^2-(5i)^2 = (x-5i)(x+5i)`

Answer 2

`x^2+25=(x+5iota)(x-5iota)`
where `iota-=sqrt (-1)`

Explanation:

There are two methods.
1. By using the general expression to find the roots of a quadratic expressions. If `x` are the roots of quadratic expression
`ax^2+bx+c=0`
Then `x=(-b+-sqrt(b^2-4ac))/(2a)`

In the given expression `x^2+25` we note that the discriminant `sqrt(b^2-4ac)` is a negative quantity.

As such the quadratic has only imaginary roots. These can be calculated and factors found as

`(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))`

1. By inspection. For the given problem
   `x^2+25` can be written as
   `x^2-(5iota)^2` where `iota-=sqrt (-1)`
   To find the two factors use `x^2-y^2=(x+y)(x-y)`
   Hence, `x^2-(5iota)^2=(x+5iota)(x-5iota)`