How do you factor x2+25 completely?

2 Answers
George C.
Jan 21, 2016

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

x2+25=(x−5i)(x+5i)

Explanation:

Notice that if x is Real then x2≥0, so x2+25 has no linear factors with Real coefficients.

The difference of squares identity can be written:

a2−b2=(a−b)(a+b)

We can write x2+25 as a difference of squares and factor it as follows:

x2+25=x2−(5i)2=(x−5i)(x+5i)

A08
Jan 21, 2016

x2+25=(x+5ι)(x−5ι)
where ι≡√−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
ax2+bx+c=0
Then x=−b±√b2−4ac2a

In the given expression x2+25 we note that the discriminant √b2−4ac is a negative quantity.

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

(x−−b+√b2−4ac2a)(x−−b−√b2−4ac2a)

  1. By inspection. For the given problem
    x2+25 can be written as
    x2−(5ι)2 where ι≡√−1
    To find the two factors use x2−y2=(x+y)(x−y)
    Hence, x2−(5ι)2=(x+5ι)(x−5ι)
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