Question 61890
{{{f(x)=3x^2(x-5)(2x+9)^3(x-4)}}}
The 3 will not give a zero
The {{{x^2}}} gives a zero of 0 with a multiplicity of 2 because of the square.
The (x-5) gives a zero of 5 with a multiplicity of 1.
The {{{(2x + 9)^3}}} gives a zero of -4.5 with a multiplicity of 3 because of the cube
The (x-4) gives a zero of 4 with a multiplicity of 1