Question

If f(x) = x^3- 2x^2 - 13x - 10 and f(-2)=0 ,then find all of the zeros of f(x) algebraically
x = -5, 1, 2
x = -5, 1, 2

x = -2, -1, 5
x = -2, -1, 5

x = -2, -1, 0
x = -2, -1, 0

x = -2, 2, 3

Answer 1

We are given the polynomial function \( f(x) = x^3 - 2x^2 - 13x - 10 \) and we know that \( f(-2) = 0 \), which indicates that \( x = -2 \) is a zero of the function.

To find all zeros of \( f(x) \), we can perform polynomial long division or synthetic division to divide \( f(x) \) by \( x + 2 \) (since \( x = -2 \) is a root).

**Perform synthetic division:**

We set up synthetic division using \( -2 \):

\[
\begin{array}{r|rrrr}
-2 & 1 & -2 & -13 & -10 \\
& & -2 & 8 & 10 \\
\hline
& 1 & -4 & -5 & 0 \\
\end{array}
\]

The coefficients of the quotient are \( 1 \), \( -4 \), and \( -5 \). So we can write:

\[
f(x) = (x + 2)(x^2 - 4x - 5)
\]

Now we need to factor the quadratic \( x^2 - 4x - 5 \) further. We can do this by finding two numbers that multiply to \(-5\) (the constant term) and add to \(-4\) (the coefficient of \( x \)):

These numbers are \(-5\) and \(1\).

Thus, we can factor the quadratic as:

\[
x^2 - 4x - 5 = (x - 5)(x + 1)
\]

Now substituting this back, we have:

\[
f(x) = (x + 2)(x - 5)(x + 1)
\]

Now we can find the zeros by setting each factor equal to zero:

1. \( x + 2 = 0 \) → \( x = -2 \)
2. \( x - 5 = 0 \) → \( x = 5 \)
3. \( x + 1 = 0 \) → \( x = -1 \)

Thus, the zeros of \( f(x) \) are:

\[
\boxed{-2, 5, -1}
\]

These zeros correspond to the choice:

**x = -2, -1, 5**