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x^3+px^2-7x-6 can be factorised into three simple factors only when p is equal to ?

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what does simple factors mean?
meaning (x+1) (x+2)(x+4) something like this
it can't be (2x+1), right?

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Other answers:

simple had me too, i wanna say integers but simple can mean nothing in math :)
(x^3-7x)+(px^2-6) x(x^2 - 7) + p(x^2 - 6/p) p = 6/7 if i see it right
let us say p=0 the equation will be x^3-7x-6 = x^2(x+1)-x(x+1)-6(x+1)=(x-3)(x+2)(x+1).
but this is not concrete solution ..this is hit and trial...As choices were 0,1,2,3
try vieta formula
(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 :)
possible rational roots are pm 1, pm 2, pm 3, pm 6
0 is it according to the trial and error method :)
\[x^3+px^2-7x-6=x^x(x+p)-(7x+6)=(x^2-1)(x+p) =(x^2-1)(7x+6)\] hence x+p=7x+6 p=6x+6
when p= 0 -> f(x) = x^3 - 7x - 6 = ( x + 1) ( x^2 - x - 6) => x = -1, x = -2, x = 3

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