find all real zeros and their multiplicity of the polynomial function
f(x)=5(x-4)squared(x+2
To find the real zeros and their multiplicity of the polynomial function f(x) = 5(x-4)^2(x+2), we can consider each factor separately and set them equal to zero.
1. For the factor (x-4)^2 to equal zero, set (x-4)^2 = 0 and solve for x:
(x-4)^2 = 0
Taking the square root of both sides:
x - 4 = 0
x = 4
So, the factor (x-4)^2 has one real zero at x = 4, with a multiplicity of 2 (since it's squared).
2. For the factor (x+2) to equal zero, set (x+2) = 0 and solve for x:
x + 2 = 0
x = -2
So, the factor (x+2) has one real zero at x = -2, with a multiplicity of 1.
Hence, the polynomial function f(x) = 5(x-4)^2(x+2) has two real zeros: x = 4 (with a multiplicity of 2) and x = -2 (with a multiplicity of 1).