Question

Write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros. Write the function in standard form.
Given zeros: −6,0,3−√5

Answer:
f(x)= ???

Answer 1

Answer:

$f(x)=x^3+(3+\sqrt{5})x^2+(6\sqrt{5}-18)x$

or

$f(x)=x^3+3x^2+\sqrt{5}x^2-18x+6\sqrt{5}x$

Step-by-step explanation:

A polynomial function with 3 zeroes is going to be of degree 3, so we're going to have at least least 3 factors of the form $x-a$, where a is a zero of the function. For three zeroes a, b, and c, we'd have the factors $x-a,\ x-b,$ and $x-c$, making our function look like

$f(x)=(x-a)(x-b)(x-c)$

Substituting our roots for a, b, and c:

$f(x)=(x-(-6))(x-0)(x-(3-\sqrt{5}))\\=(x+6)(x)(x-3+\sqrt{5})$

To write this function in standard form, we can expand from left to right:

$f(x)=(x^2+6x)(x-3+\sqrt{5})\\=(x^2+6x)x-(x^2+6x)3+(x^2+6x)\sqrt{5}\\=x^3+6x^2-3x^2-18x+\sqrt{5}x^2+6\sqrt{5}x\\=x^3+3x^2+\sqrt{5}x^2-18x+6\sqrt{5}x$

At this point, we could either leave it as is, or group the coefficients for the $x^2$ and $x$ terms to get

$f(x)=x^3+(3+\sqrt{5})x^2+(6\sqrt{5}-18)x$