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Mathematics 11 Online
OpenStudy (anonymous):

@triciaal

OpenStudy (anonymous):

OpenStudy (triciaal):

x^3 polynomial 3 roots

OpenStudy (anonymous):

There's two answers with 3 roots.

OpenStudy (anonymous):

Tri?

OpenStudy (triciaal):

-1 is not a factor 1 is a factor I substituted the values I am not sure the systematic way to prove D

OpenStudy (anonymous):

So D's the correct answer because -1 isnt a factor?

OpenStudy (triciaal):

I think i have it switched when x = 1 remainder = 0 so x-1 is a factor

OpenStudy (anonymous):

So the correct answer would be C in that case?

OpenStudy (triciaal):

yes

OpenStudy (triciaal):

when we use the conjugate and "a" is positive it becomes a^2 + 1

OpenStudy (anonymous):

It was incorrect :S I got like a 40 on this whole thing, algebra 2 sucks. lol.

OpenStudy (triciaal):

sorry i am getting mixed up

OpenStudy (triciaal):

complex numbers can be tricky

OpenStudy (anonymous):

Its okay, I can retake. But I got these wrong:

OpenStudy (anonymous):

OpenStudy (triciaal):

product is -12 and sum +1 you know the numbers are 3, 4 you know one number is -ve because the product is -ve how do you combine 3, 4 with + and - to get positive 1 ?

OpenStudy (triciaal):

(x+ 4)(x -3) roots -4 and 3 so A

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