Question

The function f(x)f(x) is a quartic function and the zeros of f(x)f(x) are -6−6, -5−5, -2−2 and 11. Assume the leading coefficient of f(x)f(x) is 11. Write the equation of the quartic polynomial in standard form.

Answer 1

Answer:

$\boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$

Explanation:

A quartic function is a function given by the the following equation in standard form:

$f(x)=a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0} \\ \\ \\ Where: \\ \\ a_{4},a_{3},a_{2},a_{1},a_{0} \ \text{are constant} \\ \\ \text{and} \ a_{4} \ \text{is the leading coefficient}$

From the statement we must assume the leading coefficient of f(x) is 11, so:

$a_{4}=11$

The zeros are:

$x=-6 \\ \\ x=-5 \\ \\ x=-2 \\ \\ x=11$

So, we can write:

$f(x)=11(x-(-6))(x-(-5))(x-(-2))(x-11) \\ \\ \\ So: \\ \\ \mathrm{Rule}:-\left(-a\right)=a \\ \\ \\ Then: \\ \\ f(x)=11\left(x+6\right)\left(x+5\right)\left(x+2\right)\left(x-11\right)$

$Expand:\left(x+6\right)\left(x+5\right):\ x^2+11x+30: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^2+11x+30\right)\left(x+2\right)\left(x-11\right) \\ \\ \\ Expand:\left(x^2+11x+30\right)\left(x+2\right):\ x^3+13x^2+52x+60: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^3+13x^2+52x+60\right)\left(x-11\right)$

$Expand: \left(x^3+13x^2+52x+60\right)\left(x-11\right):\ x^4+2x^3-91x^2-512x-660 \\ (Distributive \ property)$

Finally:

$f(x) = 11(x^4 + 2x^3 - 91x^2 - 512x - 660) \\ \\ \boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}$