Answer:
Let's call the representation of the polynomial function ƒ. We know that ƒ has zeroes at 5/2 (multiplicity 2), 3 (multiplicity 1), and 0 (multiplicity 4). We can use this to express factors of ƒ:x = 5/2, (x - 5/2) = 0 (subtract 5/2 from both sides)2x - 5 = 0 (multiply by 2 on both sides)x = 3(x - 3) = 0 (subtract 3 from both sides)and(x) = 0 (no need to rearrange)Now express ƒ as the product of the factors using exponents for multiplicity:ƒ(x) = (2x-5)2(x-3)1(x)4,This is a polynomial function with the specified zeros. Now we only need to express this in standard form, which is of the form: ax7 + bx6 + cx5 + dx4 + ex3 + fx2 + gx1 + hx0. (We now it will be degree 7 because of the number of zeros including multiplicity (2 + 1 + 4 = 7))So now we expand the polynomial to standard form by multiplying the factors together:(2x - 5)2 = (2x - 5)(2x - 5)= 4x2 - 20x + 25(4x2 - 20x + 25)(x-3) = 4x3 - 20x2 + 25x - 12x2 + 60x - 75= 4x3 - 32x2 + 85x -75(4x3 - 32x2 + 85x -75)x4 = 4x7 - 32x6 + 85x5 -75x4So our answer is:ƒ(x) = 4x7 - 32x6 + 85x5 -75x4
Explanation: