Christopher A. answered • 12/09/19

Math and Computer Programming Guide

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: ax^{7 }+ bx^{6} + cx^{5 }+ dx^{4} + ex^{3} + fx^{2} + gx^{1} + hx^{0}. (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)

= 4x^{2}- 20x + 25

(4x^{2}- 20x + 25)(x-3) = 4x^{3}- 20x^{2}+ 25x - 12x^{2}+ 60x - 75

= 4x^{3}- 32x^{2}+ 85x -75

(4x^{3}- 32x^{2}+ 85x -75)x^{4}= 4x^{7}- 32x^{6}+ 85x^{5}-75x^{4}

So our answer is:

ƒ(x) = 4x^{7}- 32x^{6}+ 85x^{5}-75x^{4}

Donovan B.

True, both of our solutions are correct depending on how you initialize the problem. In the end, both equations would represent the function above. Isn't math so FUN!!!12/09/19

Arturo O.

Quite true. Often, these problems include a statement like f(2) = 16, so we can get a unique solution, or maybe the leading coefficient is stated in the problem. Without information like that, there are infinitely many polynomials with this set of zeros and multiplicities.12/09/19

Christopher A.

12/09/19

Christopher A.

12/09/19