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

2 Answers
George C.
Jan 21, 2016

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)#

A08
Jan 21, 2016

#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)#
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