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x^3+px^27x6 can be factorised into three simple factors only when p is equal to ?
 2 years ago
 2 years ago
x^3+px^27x6 can be factorised into three simple factors only when p is equal to ?
 2 years ago
 2 years ago

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moneybirdBest ResponseYou've already chosen the best response.1
what does simple factors mean?
 2 years ago

mkuma36Best ResponseYou've already chosen the best response.0
meaning (x+1) (x+2)(x+4) something like this
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.1
it can't be (2x+1), right?
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
simple had me too, i wanna say integers but simple can mean nothing in math :)
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
(x^37x)+(px^26) x(x^2  7) + p(x^2  6/p) p = 6/7 if i see it right
 2 years ago

mkuma36Best ResponseYou've already chosen the best response.0
let us say p=0 the equation will be x^37x6 = x^2(x+1)x(x+1)6(x+1)=(x3)(x+2)(x+1).
 2 years ago

mkuma36Best ResponseYou've already chosen the best response.0
but this is not concrete solution ..this is hit and trial...As choices were 0,1,2,3
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
(x+a)(x+b)(x+c) (x^2 + 2abx + ab)(x+c) x^3 + 2abx^2 + abx +cx^2 +2abcx + abc  x^3 + p x^2 + 7x  6 if thats workable :)
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.1
possible rational roots are pm 1, pm 2, pm 3, pm 6
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
0 is it according to the trial and error method :)
 2 years ago

dturpin53Best ResponseYou've already chosen the best response.0
\[x^3+px^27x6=x^x(x+p)(7x+6)=(x^21)(x+p) =(x^21)(7x+6)\] hence x+p=7x+6 p=6x+6
 2 years ago

ChlorophyllBest ResponseYou've already chosen the best response.0
when p= 0 > f(x) = x^3  7x  6 = ( x + 1) ( x^2  x  6) => x = 1, x = 2, x = 3
 2 years ago
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