$$ -24 / 25 $$ AdvAig A USL. 12 Portlowo xus.com/assessments/elm_aspxpidWebuser=57831018idAssessment=3152274\&mode=review\&idSection=1957486\&resize=true\&popup=true\&close=true 5 Adv.Alg.A.U5L12Portfolio a) Find the roots of the function $$ x^{3}-2 x^{2}-7 x+2 $$ given that $$ (x+2) $$ is a factor. Rational Root: 1 $$ x=-2 $$ Irrational roots: 2 $$ x=2 \pm \sqrt{3} $$ $$ x=2 $$ $$ x=-2 \pm \sqrt{3} $$ $$ x=2 \pm 2 \sqrt{3} $$ $$ x=-2 \pm 2 \sqrt{3} $$ Score $$ \square $$ 4 Upload your handwritten work for part a here. In your work, you should be using either long or synthetic division to simplify and then the quadratic formula to solve. $$ \square $$

Question

UpStudy AI · Accepted Answer

Certainly! Let's work through the problem step by step to find all the roots of the function:

$$
f(x) = x^{3} - 2x^{2} - 7x + 2
$$

**Given:** $$(x + 2)$$ is a factor of $$f(x)$$.

### 1. **Identify the First Root**

Since $$(x + 2)$$ is a factor, setting it equal to zero gives the first root:

$$
x + 2 = 0 \implies x = -2
$$

**First Root:** $$x = -2$$

### 2. **Factor the Polynomial**

To find the remaining roots, we need to factor out $$(x + 2)$$ from $$f(x)$$. We'll use **synthetic division** for this purpose.

#### **Synthetic Division Steps:**

1. **Setup:**
   - **Divisor (root):** $$x = -2$$
   - **Coefficients of $$f(x)$$:** 1 (for $$x^3$$), -2 (for $$x^2$$), -7 (for $$x$$), 2 (constant term)

$$
\begin{array}{r|rrrr}
-2 & 1 & -2 & -7 & 2 \
\hline
   &   & -2 & 8 & -2 \
\hline
   & 1 & -4 & 1 & 0 \
\end{array}
$$

2. **Calculation:**
   - Bring down the first coefficient: **1**
   - Multiply by $$-2$$: $$1 \times -2 = -2$$
   - Add to the next coefficient: $$-2 + (-2) = -4$$
   - Multiply by $$-2$$: $$-4 \times -2 = 8$$
   - Add to the next coefficient: $$-7 + 8 = 1$$
   - Multiply by $$-2$$: $$1 \times -2 = -2$$
   - Add to the last coefficient: $$2 + (-2) = 0$$ (Remainder)

3. **Result:**
   The quotient polynomial is:

$$
x^{2} - 4x + 1
$$

So,

$$
f(x) = (x + 2)(x^{2} - 4x + 1)
$$

### 3. **Find the Remaining Roots**

Now, solve the quadratic equation:

$$
x^{2} - 4x + 1 = 0
$$

**Using the Quadratic Formula:**

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

Where $$a = 1$$, $$b = -4$$, and $$c = 1$$.

$$
x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4 \times 1 \times 1}}{2 \times 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2}
$$

Simplify $$\sqrt{12}$$:

$$
\sqrt{12} = 2\sqrt{3}
$$

So,

$$
x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}
$$

**Remaining Roots:**

$$
x = 2 + \sqrt{3} \quad \text{and} \quad x = 2 - \sqrt{3}
$$

### 4. **Summary of All Roots**

$$
\boxed{
\begin{align*}
x &= -2 \
x &= 2 + \sqrt{3} \
x &= 2 - \sqrt{3}
\end{align*}
}
$$

These are the three roots of the function $$f(x) = x^{3} - 2x^{2} - 7x + 2$$.

### 5. **Matching with Given Options**

From the options you provided:

- $$x = 2$$
- $$x = -2 \pm \sqrt{3}$$
- $$x = 2 \pm 2\sqrt{3}$$
- $$x = -2 \pm 2\sqrt{3}$$

**Correct Roots:**

$$
x = -2 \quad \text{and} \quad x = 2 \pm \sqrt{3}
$$

So, the correct option corresponds to:

$$
x = -2 \quad \text{and} \quad x = 2 \pm \sqrt{3}
$$
The roots of the function $$ x^{3} - 2x^{2} - 7x + 2 $$ are: - $$ x = -2 $$ - $$ x = 2 + \sqrt{3} $$ - $$ x = 2 - \sqrt{3} $$

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Algebra Questions