Fan And Medal!
Factor Completely: x2 + 25
(x + 5)(x + 5)
(x + 5)(x − 5)
(x − 5)(x − 5)
Prime
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OpenStudy (anonymous):
(x + 5)(x + 5)
OpenStudy (anonymous):
@SkydexKiwi that's wrong
OpenStudy (anonymous):
You are missing one term. This expression is not factorable
OpenStudy (anonymous):
what!?
OpenStudy (anonymous):
You need to use complex solutions.
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OpenStudy (anonymous):
So its a prime?
OpenStudy (anonymous):
@robz8 please explain?
OpenStudy (anonymous):
oh wait it'
(sx + 5)(x − 5)
isn't it?
OpenStudy (psymon):
How is it not prime?
OpenStudy (anonymous):
(x + 5)(x − 5)
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OpenStudy (anonymous):
it is prime
OpenStudy (psymon):
@SkydexKiwi Thats the factored form of x^2 - 25
OpenStudy (anonymous):
haha i suck then been a while for this
OpenStudy (anonymous):
You could do (a +b)^2 -2ab)
OpenStudy (anonymous):
|dw:1375992841262:dw|
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OpenStudy (mathstudent55):
The difference of squares is factorable.
The sum of squares is not factorable.
The difference of cubes is factorabnle.
The sum of cubes is factorable.
This is a sum of squares, so it's not factorable.
OpenStudy (anonymous):
x2 + 25? If x = 2 for example then x2 + 25 = 4 + 25 = 29 which is a prime number. Thus the only factors of 29 are 29 and 1 and hence the only factorization of 29 is 29 × 1. Similarly if x = 4 then x2 + 25 = 41 which is again prime. Hence the only factorization of x2 + 25 which is valid for every number x is
OpenStudy (anonymous):
thats wha t i got from another site so i think im going with robs
OpenStudy (anonymous):
You could also do (a+bi)(a-bi)... which is not a valid factorization... I knew it!! @robz8 ...well ?