To find the zeroes of the function \( f(x) = x^3 + x^2 + 4x + 4 \) using synthetic division, we will first make a guess about the possible rational roots. We can use the Rational Root Theorem, which suggests that the possible rational roots could be the factors of the constant term divided by the factors of the leading coefficient.
In our case:
- The constant term is \( 4 \). Its factors are \( \pm 1, \pm 2, \pm 4 \).
- The leading coefficient is \( 1 \). Its factors are just \( \pm 1 \).
Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 4 \).
Next, we'll check these possible roots one by one through synthetic division to find an acceptable root.
Step 1: Testing possible roots
Testing \( x = -1 \):
Set up synthetic division with \( -1 \):
\[ \begin{array}{r|rrrr} -1 & 1 & 1 & 4 & 4 \ & & -1 & 0 & -4 \ \hline & 1 & 0 & 4 & 0 \ \end{array} \]
The last number is \( 0 \), which indicates that \( x + 1 \) is a factor of \( f(x) \).
Step 2: Factor the polynomial
The result of the synthetic division shows that: \[ f(x) = (x + 1)(x^2 + 4) \]
Step 3: Finding the zeroes
Now, we need to find the zeroes of \( x^2 + 4 \).
Set the equation equal to zero: \[ x^2 + 4 = 0 \] \[ x^2 = -4 \] \[ x = \pm 2i \]
Final Results
Thus, the zeroes of the function \( f(x) = x^3 + x^2 + 4x + 4 \) are:
- \( x = -1 \)
- \( x = 2i \)
- \( x = -2i \)
In summary, we have found the zeroes \( \boxed{-1, 2i, -2i} \).
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