Question 6 (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.

a # of complex roots: 6
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:
b # of complex roots: 6
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:
c # of complex roots: 3
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots
d # of complex roots: 3
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:

Ask by Zimmerman Crawford. in the United States

Mar 17,2025

Question

Question 6 (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.

a # of complex roots: 6
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:
b # of complex roots: 6
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:
c # of complex roots: 3
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots
d # of complex roots: 3
Possible # of real roots: 3 or 1
Possible # of imaginary roots: 2 or 0
Possible rational roots:
Roots:

Ask by Zimmerman Crawford. in the United States

Mar 17,2025

UpStudy AI · Accepted Answer

To analyze the polynomial equation $$x^{3}-2x^{2}-3x+6=0$$, we will follow the steps outlined in your request.

### Step 1: Total Number of Complex Roots
According to the Fundamental Theorem of Algebra, a polynomial of degree $$n$$ has exactly $$n$$ roots in the complex number system (counting multiplicities). Since our polynomial is of degree 3, it has a total of 3 complex roots.

### Step 2: Possible Number of Real and Imaginary Roots
- The possible number of real roots for a cubic polynomial can be 3, 1, or none (which would imply 2 complex roots).
- The possible number of imaginary roots can be 0 (if all roots are real), or 2 (if one root is real and the other two are complex).

### Step 3: Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots of the polynomial $$x^{3}-2x^{2}-3x+6$$ are the factors of the constant term (6) divided by the factors of the leading coefficient (1). Thus, the possible rational roots are:
$$
\pm 1, \pm 2, \pm 3, \pm 6
$$

### Step 4: Finding All Roots
We will now find the roots of the polynomial $$x^{3}-2x^{2}-3x+6=0$$ using synthetic division and/or the Rational Root Theorem.

1. **Testing Possible Rational Roots:**
   - Let's test $$x = 1$$:
   $$
   1^3 - 2(1^2) - 3(1) + 6 = 1 - 2 - 3 + 6 = 2 \quad (\text{not a root})
   $$
   - Testing $$x = -1$$:
   $$
   (-1)^3 - 2(-1)^2 - 3(-1) + 6 = -1 - 2 + 3 + 6 = 6 \quad (\text{not a root})
   $$
   - Testing $$x = 2$$:
   $$
   2^3 - 2(2^2) - 3(2) + 6 = 8 - 8 - 6 + 6 = 0 \quad (\text{is a root})
   $$

2. **Synthetic Division by $$x - 2$$:**
   We will divide $$x^{3}-2x^{2}-3x+6$$ by $$x - 2$$:
   $$
   \begin{array}{r|rrrr}
   2 & 1 & -2 & -3 & 6 \
     &   & 2 & 0 & -6 \
   \hline
     & 1 & 0 & -3 & 0 \
   \end{array}
   $$
   The quotient is $$x^2 - 3$$.

3. **Finding Roots of the Quotient:**
   Now we solve $$x^2 - 3 = 0$$:
   $$
   x^2 = 3 \implies x = \pm \sqrt{3}
   $$

### Summary of Roots
Thus, the roots of the polynomial $$x^{3}-2x^{2}-3x+6=0$$ are:
- $$x = 2$$
- $$x = \sqrt{3}$$
- $$x = -\sqrt{3}$$

### Final Results
- **Total number of complex roots:** 3
- **Possible number of real roots:** 3
- **Possible number of imaginary roots:** 0
- **Possible rational roots:** $$ \pm 1, \pm 2, \pm 3, \pm 6 $$
- **Roots:** $$ 2, \sqrt{3}, -\sqrt{3} $$
The polynomial $$x^{3}-2x^{2}-3x+6=0$$ has 3 complex roots. It can have 3 real roots or 1 real root and 2 imaginary roots. The possible rational roots are $$ \pm 1, \pm 2, \pm 3, \pm 6 $$. The roots are $$2$$, $$\sqrt{3}$$, and $$-\sqrt{3}$$.

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