Question 4 (1 point) State the total number of complex roots, the possible number of real and imaginary roots and the possible rational roots for each equation. Then find all roots. $$ \begin{array}{l}x^{3}+3 x^{2}-x-3=0 \ ext { a of complex roots: } 3 \ ext { Possible \# of real roots: } 3 ext { or } 1 \ ext { Possible \# of imaginary roots: } 2 ext { or } 0 \ ext { Possible rational roots: } \pm 1, \pm 3 \ ext { Roots: }\{-3,1,-1\} \ ext { \# of complex roots: } 3 \ ext { Possible \# of real roots: } 3 ext { or } 1 \ ext { Possible \# of imaginary roots: } 2 ext { or } 0 \ ext { Possible rational roots: } \pm 1, \pm 3 \ ext { Roots: }\{3,1,-1\} \ ext { \# of complex roots: } 3 \ ext { Possible \# of real roots: } 3 ext { or } 1 \ ext { Possible \# of imaginary roots: } 2 ext { or } 0 \ ext { Possible rational roots: } \pm 1, \pm 3 \ ext { Roots: }\{-3,-1,-1\} \ ext { \# of complex roots: } 3 \ ext { Possible \# of real roots: } 3 ext { or } 1 \ ext { Possible \# of imaginary roots: } 2 ext { or } 0 \ ext { Possible rational roots: } \pm 1, \pm 3 \ ext { Roots: }\{3, ext { double root at } 1\}\end{array} $$ $$ \begin{array}{l} ext { d }\end{array} $$

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Question 4 (1 point) State the total number of complex roots, the possible number of real and imaginary roots and the possible rational roots for each equation. Then find all roots. $$ \begin{array}{l}x^{3}+3 x^{2}-x-3=0 \ 	ext { a of complex roots: } 3 \ 	ext { Possible \# of real roots: } 3 	ext { or } 1 \ 	ext { Possible \# of imaginary roots: } 2 	ext { or } 0 \ 	ext { Possible rational roots: } \pm 1, \pm 3 \ 	ext { Roots: }\{-3,1,-1\} \ 	ext { \# of complex roots: } 3 \ 	ext { Possible \# of real roots: } 3 	ext { or } 1 \ 	ext { Possible \# of imaginary roots: } 2 	ext { or } 0 \ 	ext { Possible rational roots: } \pm 1, \pm 3 \ 	ext { Roots: }\{3,1,-1\} \ 	ext { \# of complex roots: } 3 \ 	ext { Possible \# of real roots: } 3 	ext { or } 1 \ 	ext { Possible \# of imaginary roots: } 2 	ext { or } 0 \ 	ext { Possible rational roots: } \pm 1, \pm 3 \ 	ext { Roots: }\{-3,-1,-1\} \ 	ext { \# of complex roots: } 3 \ 	ext { Possible \# of real roots: } 3 	ext { or } 1 \ 	ext { Possible \# of imaginary roots: } 2 	ext { or } 0 \ 	ext { Possible rational roots: } \pm 1, \pm 3 \ 	ext { Roots: }\{3, 	ext { double root at } 1\}\end{array} $$ $$ \begin{array}{l}	ext { d }\end{array} $$

UpStudy AI · Accepted Answer

To analyze the polynomial equation $$ x^3 + 3x^2 - x - 3 = 0 $$, we will follow the steps outlined in your request.

### Step 1: Total Number of Complex Roots
The degree of the polynomial is 3, which means it has a total of 3 complex roots (counting multiplicities).

### Step 2: Possible Number of Real and Imaginary Roots
According to the Fundamental Theorem of Algebra:
- The possible number of real roots can be 3 or 1 (since complex roots come in conjugate pairs).
- The possible number of imaginary roots can be 2 or 0.

### Step 3: Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. For the polynomial $$ x^3 + 3x^2 - x - 3 $$:
- Possible rational roots: $$ \pm 1, \pm 3 $$.

### Step 4: Finding All Roots
We will test the possible rational roots to find the actual roots of the polynomial.

1. **Testing $$ x = 1 $$**:
   $$
   1^3 + 3(1^2) - 1 - 3 = 1 + 3 - 1 - 3 = 0
   $$
   So, $$ x = 1 $$ is a root.

2. **Performing Synthetic Division** with $$ x - 1 $$:
   $$
   \begin{array}{r|rrrr}
   1 & 1 & 3 & -1 & -3 \
     &   & 1 & 4 & 3 \
   \hline
     & 1 & 4 & 3 & 0 \
   \end{array}
   $$
   The quotient is $$ x^2 + 4x + 3 $$.

3. **Factoring the Quadratic**:
   $$
   x^2 + 4x + 3 = (x + 1)(x + 3)
   $$

4. **Finding the Roots**:
   The roots from the factorization are:
   - $$ x + 1 = 0 $$ gives $$ x = -1 $$
   - $$ x + 3 = 0 $$ gives $$ x = -3 $$

### Summary of Roots
The roots of the polynomial $$ x^3 + 3x^2 - x - 3 = 0 $$ are:
- $$ x = 1 $$
- $$ x = -1 $$
- $$ x = -3 $$

### Final Results
- **Total number of complex roots**: 3
- **Possible number of real roots**: 3
- **Possible number of imaginary roots**: 0
- **Possible rational roots**: $$ \pm 1, \pm 3 $$
- **Actual roots**: $$ \{1, -1, -3\} $$

Thus, the polynomial has 3 real roots and no imaginary roots.
The polynomial $$ x^3 + 3x^2 - x - 3 = 0 $$ has 3 real roots: $$ x = 1 $$, $$ x = -1 $$, and $$ x = -3 $$. There are no imaginary roots. Possible rational roots are $$ \pm 1 $$ and $$ \pm 3 $$.

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