The zeros of a function are -1, 2, -sqrt3−1,2,−√3, and 2+sqrt52+√5. If f(0)=-12f(0)=−12, what is the leading coefficient of this function?

1 Answer
George C.
Feb 27, 2018

4sqrt(3)-2sqrt(15)4√3−2√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 11. As a result the function will have irrational coefficients.

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

So we can write:

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

for some constant kk (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))−12=f(0)=k((0)+1)((0)−2)((0)+√3)((0)−2−√5)

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

So:

k = -12/(2sqrt(3)(2+sqrt(5)))k=−122√3(2+√5)

color(white)(k) = -(2sqrt(3))/(2+sqrt(5))k=−2√32+√5

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

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

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

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