Factoring polymonial with complex roots

[ACLerok]

This may be a bit silly but i forget how to factor this into complex factors:

s^2 + 6s + 25

i know the answer is (s +3 - i4)(s +3 - j4)

but how do i get that?

 

[lurflurf]

This may be a bit silly but i forget how to factor this into complex factors:

s^2 + 6s + 25

i know the answer is (s +3 - i4)(s +3 - j4)

but how do i get that?

You could use the quadratic formula.

 

[Nylex]

Yes, use the quadratic formula to find the roots of $s^2 + 6s + 25 = 0$ and then use the factor theorem: if f(a) = 0, then (x - a) is a factor of f(x). Your a here will be the complex number you get.

Edit: no doubt dexter or someone will tell me this is wrong .

 

[George Jones]

This may be a bit silly but i forget how to factor this into complex factors:

s^2 + 6s + 25

i know the answer is (s +3 - i4)(s +3 - j4)

but how do i get that?

Set your expression equal to zero and the roots, i.e. find s = a and s = b such that

0 = s^2 +6s +25.

Then,

0 = s^2 +6s +25
= (s - a)(s - b).

You could use the quadratic formula, but I think completing the square offers more insight.

Write

0 = s^ + 6s + c^2 - c^2 +25.

Now find c such that

s^ + 6s +c^2 = (s + c)^2.

This means that 2c = 6 and c = 3. Therefore,

0 = s^2 + 6s + 9 - 9 +25
= (s+3)^2 +16

So,

(s + 3)^2 = -16.

Regards,
George