If #f(x)=u(x-a)^m(x-b)^n(x-c)^p(x-d)^q#, a multiplication of number of binomials of degree one, then zeros of polynomials are #a#, #b#, #c# and #d#, as any of them when put in place of #x# will make #f(x)=0#. Here #u# is just a constant.
Hence, zeros of #f(x)=4(x+7)^2(x-7)^2# are #-7# and #7#.