If (x-a) is the factor of f(x) then,
f(a)=0 and vice-versa
or,3x³-12x²-4x-55=0 .. .eqn(1)
Compare (x-a) with the binomials (x + 1), (x + 4), (x - 5), and (x - 2).....then
CHECK f(a)=0 by putting a=-1,-4,5,2 then
we get
f(-1)=3×(-1)³-12×(-1)²-4×(-1)-55=-66
f(-4)=............................,..............=other value than 0
f(5)= 3×(5)³-12×(5)²-4×(5)-55=0 (satisfies eqn1)
f(2)=...............=other value than 0(checkyourself)
Here, only f(5)=0
the factor of f(x)=(x-5)
