Show that the cubic function f(x) = x^3 - 7x - 6 has a root x = -1 and hence factorise it fully.
f(-1) = (-1)^3 - (-1)*7 - 6 = -1 + 7 - 6 = 0
Hence f(x) = (x+1)(x^2 + ax - 6)
Expand this out
f(x) = x^3 + ax^2 - 6x + x^2 + ax - 6
= x^3 + (a+1)x^2 + (a-6)x -6
By comparing co-efficients
a + 1 = 0
a - 6 = -7
a = -1
Thus
f(x) = (x + 1)(x^2 - x - 6)
= (x + 1)(x - 3)(x + 2)
JB
