Question

The zeros of a function are `-1, 2, -sqrt3`, and `2+sqrt5`. If `f(0)=-12`, what is the leading coefficient of this function?

Answer 1

`4sqrt(3)-2sqrt(15)`

Explanation:

The fact that the question speaks of the function having a "leading coefficient" suggests to me that we are talking about a polynomial function.

Since the question says "the" zeros, I will assume that these are the only zeros (including complex zeros) of the function and that they are of multiplicity `1`. As a result the function will have irrational coefficients.

Each zero `x=a` corresponds to a factor `(x-a)`.

So we can write:

`f(x) = k(x+1)(x-2)(x+sqrt(3))(x-2-sqrt(5))`

for some constant `k` (which is also the leading coefficient) to be determined.

Then:

`-12 = f(0) = k((color(blue)(0))+1)((color(blue)(0))-2)((color(blue)(0))+sqrt(3))((color(blue)(0))-2-sqrt(5))`

`color(white)(-12 = f(0)) = 2sqrt(3)(2+sqrt(5))k`

So:

`k = -12/(2sqrt(3)(2+sqrt(5)))`

`color(white)(k) = -(2sqrt(3))/(2+sqrt(5))`

`color(white)(k) = -(2sqrt(3)(2-sqrt(5)))/((2+sqrt(5))(2-sqrt(5)))`

`color(white)(k) = -(2sqrt(3)(2-sqrt(5)))/(4-5)`

`color(white)(k) = 4sqrt(3)-2sqrt(15)`