Question
Enter your answer and show all the steps that you use to solve this problem in the space provided.
Use synthetic division to find the zeroes of the function f(x)=x3+x2+4x+4
.
Use synthetic division to find the zeroes of the function f(x)=x3+x2+4x+4
.
Answers
GPT-4o mini
To find the zeroes of the function \( f(x) = x^3 + x^2 + 4x + 4 \) using synthetic division, we will first need to guess a possible rational root using the Rational Root Theorem and then perform synthetic division.
### Step 1: Identify potential rational roots
The Rational Root Theorem states that potential roots of a polynomial \( f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \) can be given by the factors of the constant term divided by the factors of the leading coefficient.
In this case:
- The constant term \( a_0 = 4 \): its factors are \( \pm 1, \pm 2, \pm 4 \).
- The leading coefficient \( a_3 = 1 \): its factors are \( \pm 1 \).
Thus, possible rational roots are \( \pm 1, \pm 2, \pm 4 \).
### Step 2: Test possible roots using synthetic division
Let’s test \( x = -2 \):
Set up synthetic division:
\[
\begin{array}{r|rrrr}
-2 & 1 & 1 & 4 & 4 \\
& & -2 & 2 & -12 \\
\hline
& 1 & -1 & 6 & -8 \\
\end{array}
\]
The bottom row shows the coefficients of the quotient polynomial and the remainder:
- Quotient: \( x^2 - x + 6 \)
- Remainder: \( -8 \)
Since the remainder is not zero, \( x = -2 \) is not a root.
Next, let’s test \( x = -1 \):
Set up synthetic division:
\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & 4 & 4 \\
& & -1 & 0 & -4 \\
\hline
& 1 & 0 & 4 & 0 \\
\end{array}
\]
The bottom row shows the coefficients of the quotient polynomial and the remainder:
- Quotient: \( x^2 + 0x + 4 = x^2 + 4 \)
- Remainder: \( 0 \)
Since the remainder is zero, we have found that \( x = -1 \) is a root.
### Step 3: Factor the polynomial
From our synthetic division, we can factor \( f(x) \) as:
\[
f(x) = (x + 1)(x^2 + 4)
\]
### Step 4: Find the remaining roots
Next, we need to find the roots of \( x^2 + 4 = 0 \):
\[
x^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm 2i
\]
### Step 5: State the zeroes
The zeroes of the function \( f(x) = x^3 + x^2 + 4x + 4 \) are:
\[
x = -1, \quad x = 2i, \quad x = -2i
\]
### Summary of Roots
- The real root is \( x = -1 \).
- The complex roots are \( x = 2i \) and \( x = -2i \).
### Step 1: Identify potential rational roots
The Rational Root Theorem states that potential roots of a polynomial \( f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \) can be given by the factors of the constant term divided by the factors of the leading coefficient.
In this case:
- The constant term \( a_0 = 4 \): its factors are \( \pm 1, \pm 2, \pm 4 \).
- The leading coefficient \( a_3 = 1 \): its factors are \( \pm 1 \).
Thus, possible rational roots are \( \pm 1, \pm 2, \pm 4 \).
### Step 2: Test possible roots using synthetic division
Let’s test \( x = -2 \):
Set up synthetic division:
\[
\begin{array}{r|rrrr}
-2 & 1 & 1 & 4 & 4 \\
& & -2 & 2 & -12 \\
\hline
& 1 & -1 & 6 & -8 \\
\end{array}
\]
The bottom row shows the coefficients of the quotient polynomial and the remainder:
- Quotient: \( x^2 - x + 6 \)
- Remainder: \( -8 \)
Since the remainder is not zero, \( x = -2 \) is not a root.
Next, let’s test \( x = -1 \):
Set up synthetic division:
\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & 4 & 4 \\
& & -1 & 0 & -4 \\
\hline
& 1 & 0 & 4 & 0 \\
\end{array}
\]
The bottom row shows the coefficients of the quotient polynomial and the remainder:
- Quotient: \( x^2 + 0x + 4 = x^2 + 4 \)
- Remainder: \( 0 \)
Since the remainder is zero, we have found that \( x = -1 \) is a root.
### Step 3: Factor the polynomial
From our synthetic division, we can factor \( f(x) \) as:
\[
f(x) = (x + 1)(x^2 + 4)
\]
### Step 4: Find the remaining roots
Next, we need to find the roots of \( x^2 + 4 = 0 \):
\[
x^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm 2i
\]
### Step 5: State the zeroes
The zeroes of the function \( f(x) = x^3 + x^2 + 4x + 4 \) are:
\[
x = -1, \quad x = 2i, \quad x = -2i
\]
### Summary of Roots
- The real root is \( x = -1 \).
- The complex roots are \( x = 2i \) and \( x = -2i \).