how to put x^2 + 5 into the form a(x-r(1))(x-r(2))?

Answer 1

a(x - r1)(x - r2)

Expand :
(1) ax^2 - ax(r1 + r2) + a*r1*r2

Your equation is :
x^2 + 5
which can also be written as :
(2) x^2 - x*0 + 5

Equating coefficients of (1) with those of (2) gives:

(i) ax^2 = x^2, therefore, a = 1.
(ii) r1 + r2 = 0, therefore, r2 = -r1.
(iii) r1*r2 = 5

Substituting result from (ii) into (iii) gives :
r1*(-r1) = 5, or (r1)^2 = -5, or r1 = ± i*sqrt(5)

Because r1 and r2 are symmetrical, then one of these, say r1, equals + i*sqrt(5) and the other one, r2, equals - i*sqrt(5).

Plugging these into the original equation gives :
a(x - r1)(x - r2) = 1[x - i*sqrt(5)][x - (-i*sqrt(5))]
as the form you require for x^2 + 5.

Answer 2

x^2+5 = 0
x = +/- sqrt(-5) = +/- i*sqrt(5)
So (x-i*sqrt 5)(x -(-i*sqrt(5)) where a = 1

Answer 3

(x - i sq rt5)(x + i sq rt 5)