Question -
Answer -
f(x) = x3 + 6x2 + 11x + 6
Construct = 6 = ±1, ±2, +3, ±6
If x = 1, then
f(1) = (1)3 + 6(1)2 + 11 x 1 + 6
= 1+ 6+11+ 6 = 24
∵ f(x) ≠ 0, +0
∴ x = 1is not its zero
Similarly, f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
∴ x = -1 is its zero
f(-2) = (-2)3 + 6(-2)2 + 11 (-2) + 6
= -8 + 24 – 22 + 6
= -30 + 30 = 0
∴ x = -2is its zero
f(-3) = (-3)3 + 6(-3)2 + 11 (-3) + 6
= -27 + 54 – 33 + 6 = 60 – 60 = 0
∴ x = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3