Question

Im having trouble with factoring..

Answer 1

im suppose to factor x^2+4

is this prime?

Answer 2

or (x+2)(x+2)

Answer 3

it will have complex factors.
else its prime.

Answer 4

its NOT (x+2)(x+2)

Answer 5

\[a^2+b^2 = (a+ib)(a-ib)\]

Answer 6

ok wait a second

Answer 7

(x + 2)(x – 2)
(x + 2)(x + 2)
(x + 1)(x + 4)

ao it couldnt be any of these? im so confused with this one question

Answer 8

no, it cannot be any of these.
its prime, if we are considering only real factors.

Answer 9

ok. thankyou. i thought so, just was not sure. :) i appreicaite that.

Answer 10

\[(x+2)(x-2) = x^2 +2x - 2x -4 = x^2 -4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]

Answer 11

That i in there in the (a+ib)(a-ib) does the trick of reversing the sign of the final term:
\[(a+ib)(a-ib) = a^2 + aib - aib -i^2b^2 = a^2 - i^2b^2\]but \[ i^2 = -1\] so \[a^2-i^2b^2=a^2-(-b^2)=a^2+b^2\]

Answer 12

welcome ^_^