(x), (x-2), (x-5), (x-5), (x-7) and (x-9) are NOT zeros, they are FACTORS. The zeros are 0, 2, 5, 5, 7, and 9. Since 5 is listed twice it is said to be a zero of multiplicity 2. To find a polynomial with those zero, we start by setting x equal to each zero: x=0, x=2, x=5, x=5, x=7, and x=9. Then we subtract the zero on the right from both sides of each of those 6 equations: x-0=0, x-2=0, x-5=0, x-7=0, x-9=0 Then we use the principle that equals multiplied by equals gives equals. So we multiply all the left sides together and all the right sides together. (Multiplying all the right sides together just gives 0.) (x-0)(x-2)(x-5)(x-5)(x-7)(x-9) = 0 And since x-0 is just x, we have (x)(x-2)(x-5)(x-5)(x-7)(x-9) = 0 Those are the factors you gave. But again, the zeros are 0,2,5,5,7,9