Question The function $$ f(x)=3 x^{3}-2 x^{2}-3 x+2 $$ has at least one integer root. Use the integer root theorem to find that root, then proceed to find all complex roots. (Note: roots may be integer, rational, irrational, and/or complex.) Answer Attempt sout of 10

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To find the integer roots of the polynomial function $$ f(x) = 3x^3 - 2x^2 - 3x + 2 $$ using the Integer Root Theorem, we will follow these steps:

1. **Identify the possible integer roots**: According to the Integer Root Theorem, the possible rational roots of a polynomial $$ \frac{p}{q} $$ are such that $$ p $$ is a factor of the constant term and $$ q $$ is a factor of the leading coefficient.

2. **List the factors**:
   - The constant term is $$ 2 $$, whose factors are $$ \pm 1, \pm 2 $$.
   - The leading coefficient is $$ 3 $$, whose factors are $$ \pm 1, \pm 3 $$.

3. **Possible integer roots**: The possible integer roots are the combinations of the factors of the constant term divided by the factors of the leading coefficient:
   - Possible roots: $$ \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3} $$

4. **Test the possible integer roots**: We will evaluate $$ f(x) $$ at each of the integer candidates $$ \pm 1 $$ and $$ \pm 2 $$.

Let's calculate $$ f(1) $$, $$ f(-1) $$, $$ f(2) $$, and $$ f(-2) $$:

- $$ f(1) = 3(1)^3 - 2(1)^2 - 3(1) + 2 $$
- $$ f(-1) = 3(-1)^3 - 2(-1)^2 - 3(-1) + 2 $$
- $$ f(2) = 3(2)^3 - 2(2)^2 - 3(2) + 2 $$
- $$ f(-2) = 3(-2)^3 - 2(-2)^2 - 3(-2) + 2 $$

Now, I will calculate these values.
Calculate the value by following steps:
- step0: Calculate:
  $$3\left(-2\right)^{3}-2\left(-2\right)^{2}-3\left(-2\right)+2$$
- step1: Multiply the terms:
  $$-24-2\left(-2\right)^{2}-3\left(-2\right)+2$$
- step2: Multiply the terms:
  $$-24-2^{3}-3\left(-2\right)+2$$
- step3: Multiply the numbers:
  $$-24-2^{3}+6+2$$
- step4: Evaluate the power:
  $$-24-8+6+2$$
- step5: Calculate:
  $$-24$$
Calculate or simplify the expression $$ 3*(2)^3 - 2*(2)^2 - 3*(2) + 2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$3\times 2^{3}-2\times 2^{2}-3\times 2+2$$
- step1: Multiply the terms:
  $$24-2\times 2^{2}-3\times 2+2$$
- step2: Calculate:
  $$24-2^{3}-3\times 2+2$$
- step3: Multiply the numbers:
  $$24-2^{3}-6+2$$
- step4: Evaluate the power:
  $$24-8-6+2$$
- step5: Calculate:
  $$12$$
Calculate or simplify the expression $$ 3*(-1)^3 - 2*(-1)^2 - 3*(-1) + 2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$3\left(-1\right)^{3}-2\left(-1\right)^{2}-3\left(-1\right)+2$$
- step1: Evaluate the power:
  $$3\left(-1\right)^{3}-2\times 1-3\left(-1\right)+2$$
- step2: Multiply the terms:
  $$-3-2\times 1-3\left(-1\right)+2$$
- step3: Multiply:
  $$-3-2-3\left(-1\right)+2$$
- step4: Simplify:
  $$-3-2+3+2$$
- step5: Simplify:
  $$-2+2$$
- step6: Add the numbers:
  $$0$$
Calculate or simplify the expression $$ 3*(1)^3 - 2*(1)^2 - 3*(1) + 2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$3\times 1^{3}-2\times 1^{2}-3\times 1+2$$
- step1: Evaluate the power:
  $$3\times 1-2\times 1^{2}-3\times 1+2$$
- step2: Evaluate the power:
  $$3\times 1-2\times 1-3\times 1+2$$
- step3: Multiply:
  $$3-2\times 1-3\times 1+2$$
- step4: Multiply:
  $$3-2-3\times 1+2$$
- step5: Multiply:
  $$3-2-3+2$$
- step6: Simplify:
  $$-2+2$$
- step7: Add the numbers:
  $$0$$
The evaluations of the function $$ f(x) $$ at the integer candidates are as follows:

- $$ f(1) = 0 $$
- $$ f(-1) = 0 $$
- $$ f(2) = 12 $$
- $$ f(-2) = -24 $$

Since both $$ f(1) = 0 $$ and $$ f(-1) = 0 $$, we have found two integer roots: $$ x = 1 $$ and $$ x = -1 $$.

Next, we will proceed to find all complex roots of the polynomial. Since we have found $$ x = 1 $$ as a root, we can perform polynomial long division to factor $$ f(x) $$ by $$ (x - 1) $$.

Let's divide $$ f(x) $$ by $$ (x - 1) $$ to find the quotient polynomial. The polynomial division will yield a quadratic polynomial, which we can then solve for its roots.

The polynomial division can be set up as follows:

1. Divide $$ 3x^3 $$ by $$ x $$ to get $$ 3x^2 $$.
2. Multiply $$ 3x^2 $$ by $$ (x - 1) $$ and subtract from $$ f(x) $$.
3. Repeat the process for the resulting polynomial.

Let's perform the polynomial long division.

The polynomial division will be:

$$
\begin{array}{r|rr}
3x^2 & x - 1 \
\hline
3x^3 & -2x^2 & -3x & +2 \
- (3x^3 - 3x^2) & \
\hline
& x^2 - 3x + 2 \
- (x^2 - x) & \
\hline
& -2x + 2 \
- (-2x + 2) & \
\hline
& 0 \
\end{array}
$$

The quotient polynomial is $$ 3x^2 + x - 2 $$.

Now, we will find the roots of the quadratic polynomial $$ 3x^2 + x - 2 $$ using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 3 $$, $$ b = 1 $$, and $$ c = -2 $$.

Let's calculate the discriminant and then the roots.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(-1\right)-\sqrt{1^{2}-4\times 3\left(-2\right)}$$
- step1: Remove the parentheses:
  $$-1-\sqrt{1^{2}-4\times 3\left(-2\right)}$$
- step2: Evaluate the power:
  $$-1-\sqrt{1-4\times 3\left(-2\right)}$$
- step3: Multiply:
  $$-1-\sqrt{1+24}$$
- step4: Add the numbers:
  $$-1-\sqrt{25}$$
- step5: Simplify the root:
  $$-1-5$$
- step6: Subtract the numbers:
  $$-6$$
Calculate or simplify the expression $$ (-1) + \sqrt(1^2 - 4*3*(-2)) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(-1\right)+\sqrt{1^{2}-4\times 3\left(-2\right)}$$
- step1: Remove the parentheses:
  $$-1+\sqrt{1^{2}-4\times 3\left(-2\right)}$$
- step2: Evaluate the power:
  $$-1+\sqrt{1-4\times 3\left(-2\right)}$$
- step3: Multiply:
  $$-1+\sqrt{1+24}$$
- step4: Add the numbers:
  $$-1+\sqrt{25}$$
- step5: Simplify the root:
  $$-1+5$$
- step6: Add the numbers:
  $$4$$
Calculate or simplify the expression $$ 1^2 - 4*3*(-2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$1^{2}-4\times 3\left(-2\right)$$
- step1: Evaluate the power:
  $$1-4\times 3\left(-2\right)$$
- step2: Multiply:
  $$1+24$$
- step3: Add the numbers:
  $$25$$
The calculations yield the following results:

1. The discriminant $$ b^2 - 4ac = 25 $$.
2. The roots of the quadratic polynomial $$ 3x^2 + x - 2 $$ are:
   - $$ x = \frac{-1 + \sqrt{25}}{2 \cdot 3} = \frac{-1 + 5}{6} = \frac{4}{6} = \frac{2}{3} $$
   - $$ x = \frac{-1 - \sqrt{25}}{2 \cdot 3} = \frac{-1 - 5}{6} = \frac{-6}{6} = -1 $$

Thus, the roots of the polynomial $$ f(x) = 3x^3 - 2x^2 - 3x + 2 $$ are:

1. $$ x = 1 $$ (integer root)
2. $$ x = -1 $$ (integer root)
3. $$ x = \frac{2}{3} $$ (rational root)

Since $$ -1 $$ is already counted as a root, we can summarize the roots as:

- Integer roots: $$ 1, -1 $$
- Rational root: $$ \frac{2}{3} $$

There are no irrational or complex roots in this case.

Final roots of the polynomial $$ f(x) $$ are:
- $$ x = 1 $$
- $$ x = -1 $$
- $$ x = \frac{2}{3} $$
The integer roots of the polynomial $$ f(x) = 3x^3 - 2x^2 - 3x + 2 $$ are $$ x = 1 $$ and $$ x = -1 $$. The rational root is $$ x = \frac{2}{3} $$. There are no irrational or complex roots.

Question
The function has at least one integer root.
Use the integer root theorem to find that root, then proceed to find all
complex roots. (Note: roots may be integer, rational, irrational, and/or
complex.)
Answer Attempt sout of 10

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Question The function has at least one integer root. Use the integer root theorem to find that root, then proceed to find all complex roots. (Note: roots may be integer, rational, irrational, and/or complex.) Answer Attempt sout of 10