Mathematics
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OpenStudy (samigupta8):
if x^2+px+1 is a factor of the expression ax^3+bx+c
a.a^2+c^2=-ab
b.a^2-c^2=-ab
a^2-c^2=ab
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OpenStudy (samigupta8):
@RadEn pls..hlp
OpenStudy (raden):
what's the question here ?
OpenStudy (samigupta8):
this is the question
OpenStudy (anonymous):
What is "the factor of the expression"? Sorry, my English is not so good.
OpenStudy (samigupta8):
ax^3+bx+c is any expression and x^2+px+1 is factor of it that's what ques says
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OpenStudy (anonymous):
Thank you, I get it.
hartnn (hartnn):
lets try
whats sum , product of roots for x^2+px+1 ?
and whats for that cubic expression ?
OpenStudy (samigupta8):
sum is -p and product is 1
OpenStudy (samigupta8):
sum for cubic expression is 0 and product is c
hartnn (hartnn):
product is ? c? no...
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OpenStudy (samigupta8):
bt why
hartnn (hartnn):
ok, so let 3 roots of cubic exp be A,B,C
product of roots for cubic expression ax^3+bx^2+cx+d is -d/a , right ?
hartnn (hartnn):
so, here. won't it be -c/a ?
OpenStudy (samigupta8):
k...i undrstand it
hartnn (hartnn):
so,
since x^2+px+1 is a factor of the expression ax^3+bx+c,
two from A,B,C are also roots of x^2+px+1
got this ? this is important step
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hartnn (hartnn):
let that 2 roots be A,B
OpenStudy (samigupta8):
kk...
hartnn (hartnn):
so, now we have
AB =1
A+B=-p
ABC=-c/a
A+B+C =0
what about AB+BC+CA =...?
OpenStudy (samigupta8):
that's simply b
hartnn (hartnn):
not b/a ?
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OpenStudy (samigupta8):
oh sorry i m considering a=1
OpenStudy (samigupta8):
then wt shud we do now
hartnn (hartnn):
don't.
so,
AB =1
A+B=-p
ABC=-c/a
A+B+C =0
AB+BC+CA =b/a
everything clear till here ?
see if you get C=p or not.....
OpenStudy (samigupta8):
yaa c=p
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hartnn (hartnn):
also fro ABC = -c/a
since AB=1
C=p=-c/a
ok?
OpenStudy (samigupta8):
yaa i got it uptill here
hartnn (hartnn):
good.
AB+BC+CA =b/a
1 + C (A+B) = b/a
got this ? as AB=1
OpenStudy (samigupta8):
yaaaa
hartnn (hartnn):
now just plug in values!
C = p = -b/a
A+B = -p = b/a
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OpenStudy (samigupta8):
c=p=-c/a then hw it came c=p=-b/a
hartnn (hartnn):
thats why i took capital c ---->C to avoid confusion
C = p
not c=p
OpenStudy (samigupta8):
no no i m saying dat C=p=-c/a and u r getting it as C=p=-b/a
hartnn (hartnn):
and my bad, typing mistake
C = p = -c/a
hartnn (hartnn):
sorry
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OpenStudy (samigupta8):
it's jst ok...
hartnn (hartnn):
C = p = -c/a
A+B = -p = c/a
hartnn (hartnn):
1-c^2/a^2 = b/a
....