find all real zeros of the polynomial f(x)-x^4-97x^2+1296 and determine the multiplicity of each

Question

I am thrown off because of the multiplicity part

Answer 1

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

Answer 2

P(x) = 6x3 − 11x2 − 14x − 2