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OpenStudy (anonymous):
Factor completely
3x^3 + 9x^2 + x + 3
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OpenStudy (ajprincess):
Can u factorise this?
3x^3 + 9x^2 @davidkoveleski
OpenStudy (anonymous):
would it be 12x^5?
OpenStudy (anonymous):
@ajprincess
OpenStudy (ajprincess):
Nope:(
\(3x^3=3*x*x*x=3x^2*x\)
\(9x^2=3*3*x*x=3x^2*3\)
Nw can u pls tell me what is common to both \(3x^3\) and \(9x^2\)?
OpenStudy (anonymous):
3 and x?
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OpenStudy (anonymous):
@ajprincess
OpenStudy (ajprincess):
Well actually it is \(3x^2\).
\(3x^3=3x^2*x\) and \(9x^2=3x^2*3\).
getting it @davidkoveleski
OpenStudy (anonymous):
So how should I answer @ajprincess
OpenStudy (ajprincess):
When we take the common one out what remains is x and 3.
So
\(3x^3+9x^2=3x^2(x+3)\)
OpenStudy (ajprincess):
x+1=1(x+3)
can u factorise
\(3x^2(x+3)+1(x+3)\)?
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OpenStudy (ajprincess):
u can assume x+3 as y to make factorising simpler.
So tell me vat u get when u factorise \(3x^2y+y\). @davidkoveleski
OpenStudy (anonymous):
3x^2y^2 @ajprincess
OpenStudy (ajprincess):
hw did u get it?
OpenStudy (anonymous):
Just put the y's together?
OpenStudy (anonymous):
@ajprincess
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OpenStudy (ajprincess):
it can't be done that way.. if it is to do be done there should be * sign between 3x^2y and y. but here u have + sign.
OpenStudy (ajprincess):
|dw:1368799006664:dw|
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