I am thrown off because of the multiplicity part
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f(x) = x^4 - 97x^2 + 1296

The multiplicity part just means 'how many of each zero is there?'

If the problem was g(x) = (x-2)^2, factoring to g(x) = (x-2)(x-2) there would be just one zero (x=2), with a multiplicity of 2.

 

x^4 - 97x^2 + 1296 = 0

x^2 = (97 +- sqrt(97^2 - 4*1*1296) ) /2

x^2 = (97 +- sqrt(9409 - 5184) ) /2

x^2 = (97 +- sqrt(4225) )/2

x^2 = (97 +- 65) / 2

x^2 = 162/2 or 32/2

x^2 = 81 or 16

x = -9, 9, -4, or 4

f(x) has zeros of -9, 9, -4, and 4, each with a multiplicity of 1
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P(x) = 6x3  11x2  14x  2

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