Question -
Answer -
f(x) = x3┬а+ 6x2┬а+ 11x + 6
Construct = 6 = ┬▒1, ┬▒2, +3, ┬▒6
If x = 1, then
f(1) = (1)3┬а+ 6(1)2┬а+ 11 x 1 + 6
= 1+ 6+11+ 6 = 24
тИ╡┬аf(x) тЙа 0, +0
тИ┤ x = 1is not its zero
Similarly, f(-1) = (-1)3┬а+ 6(-1)2┬а+ 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
тИ┤┬аx = -1 is its zero
f(-2) = (-2)3┬а+ 6(-2)2┬а+ 11 (-2) + 6
= -8 + 24 тАУ 22 + 6
= -30 + 30 = 0
тИ┤ x = -2is its zero
f(-3) = (-3)3┬а+ 6(-3)2┬а+ 11 (-3) + 6
= -27 + 54 тАУ 33 + 6 = 60 тАУ 60 = 0
тИ┤┬аx = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3