Find the zeros of the function below algebraically, if any exist. $$f ( x ) = x ^ { 6 } + 28 x ^ { 3 } + 27$$
A) $- 3$ and $- 1$
B) $- 2$ and $- 1$
C) $- 3$ and $- 2$
D) $- 2 , - 1,1$, and 2
E) $- 3 , - 2,2$, and 3

Question

Quizplus · Accepted Answer

To find the zeros of the function $f(x) = x^6 + 28x^3 + 27$, we can substitute $y = x^3$, turning the equation into a quadratic form: $y^2 + 28y + 27 = 0$. Factoring this equation gives $(y + 1)(y + 27) = 0$, leading to $y = -1$ and $y = -27$. Substituting back $x^3$ for $y$, we get $x^3 = -1$ and $x^3 = -27$, which have the real roots $x = -1$ and $x = -3$, respectively.

Find the Zeros of the Function Below Algebraically, If Any f(x)=x6+28x3+27f ( x ) = x ^ { 6 } + 28 x ^ { 3 } + 27f(x)=x6+28x3+27

Find the Zeros of the Function Below Algebraically, If Any $f ( x ) = x ^ { 6 } + 28 x ^ { 3 } + 27$