Question

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

Answer 1

what does simple factors mean?

Answer 2

meaning (x+1) (x+2)(x+4) something like this

Answer 3

it can't be (2x+1), right?

Answer 4

simple had me too, i wanna say integers but simple can mean nothing in math :)

Answer 5

(x^3-7x)+(px^2-6)

x(x^2 - 7) + p(x^2 - 6/p)

p = 6/7 if i see it right

Answer 6

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).

Answer 7

but this is not concrete solution ..this is hit and trial...As choices were 0,1,2,3

Answer 8

try vieta formula

Answer 9

(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 :)

Answer 10

possible rational roots are pm 1, pm 2, pm 3, pm 6

Answer 11

0 is it according to the trial and error method :)

Answer 12

\[x^3+px^2-7x-6=x^x(x+p)-(7x+6)=(x^2-1)(x+p) =(x^2-1)(7x+6)\]
hence x+p=7x+6
p=6x+6

Answer 13

when p= 0 -> f(x) = x^3 - 7x - 6
= ( x + 1) ( x^2 - x - 6)
=> x = -1, x = -2, x = 3