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 treat $x^3$ as a single variable, say $y$, to simplify the equation to a quadratic form: $y^2 + 28y + 27 = 0$. Factoring this equation, we get $(y + 1)(y + 27) = 0$, which gives us $y = -1$ and $y = -27$. Substituting back $x^3$ for $y$, we have $x^3 = -1$ and $x^3 = -27$. Solving these equations gives us the zeros $x = -1$ (since $(-1)^3 = -1$) and $x = -3$ (since $(-3)^3 = -27$).

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 $f ( x ) = x ^ { 6 } + 28 x ^ { 3 } + 27$