\begin{tabular}{|l} I can find all the factors and roots of a polynomial function given one root. \ Completely factor $$ 3 x^{3}+8 x^{2}+3 x-2 $$ given -1 is a root. Find all the roots of the polynomial. \ Completely factor $$ x^{3}+x^{2}-5 x-2 $$ given 2 is a root. Find all the roots of the polynomial. \ \hline\end{tabular}

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**Problem 1:** Factor and find the roots of  
$$
3x^3+8x^2+3x-2
$$  
given that $$-1$$ is a root.

1. **Divide by $$(x+1)$$:**  
   Since $$-1$$ is a root, $$(x+1)$$ is a factor. Divide the polynomial by $$(x+1)$$ using synthetic division with $$-1$$:

- Coefficients: $$3,\ 8,\ 3,\ -2$$.
   - Synthetic division steps:
     - Bring down $$3$$.
     - Multiply: $$3 \times (-1) = -3$$; add to $$8$$ to get $$5$$.
     - Multiply: $$5 \times (-1) = -5$$; add to $$3$$ to get $$-2$$.
     - Multiply: $$-2 \times (-1) = 2$$; add to $$-2$$ to get $$0$$.

The quotient is:
   $$
   3x^2+5x-2.
   $$

2. **Factor the quadratic $$3x^2+5x-2$$:**  
   Look for two numbers that multiply to $$3 \times (-2) = -6$$ and add to $$5$$. The numbers $$6$$ and $$-1$$ work since $$6 \times (-1) = -6$$ and $$6+(-1)=5$$.  
   Rewrite the middle term:
   $$
   3x^2+6x-x-2.
   $$
   Factor by grouping:
   $$
   (3x^2+6x) + (-x-2) = 3x(x+2) -1(x+2) = (x+2)(3x-1).
   $$

3. **Final factorization and roots:**  
   Including the factor $$(x+1)$$, the complete factorization is:
   $$
   (x+1)(3x-1)(x+2).
   $$  
   The roots are obtained by setting each factor to zero:
   - $$x+1=0$$ gives $$x=-1$$,
   - $$3x-1=0$$ gives $$x=\frac{1}{3}$$,
   - $$x+2=0$$ gives $$x=-2$$.

---

**Problem 2:** Factor and find the roots of  
$$
x^3+x^2-5x-2
$$  
given that $$2$$ is a root.

1. **Divide by $$(x-2)$$:**  
   Since $$2$$ is a root, $$(x-2)$$ is a factor. Divide the polynomial by $$(x-2)$$ using synthetic division with $$2$$:

- Coefficients: $$1,\ 1,\ -5,\ -2$$.
   - Synthetic division steps:
     - Bring down $$1$$.
     - Multiply: $$1 \times 2 = 2$$; add to $$1$$ to get $$3$$.
     - Multiply: $$3 \times 2 = 6$$; add to $$-5$$ to get $$1$$.
     - Multiply: $$1 \times 2 = 2$$; add to $$-2$$ to get $$0$$.

The quotient is:
   $$
   x^2+3x+1.
   $$

2. **Solve the quadratic $$x^2+3x+1$$:**  
   Use the quadratic formula:
   $$
   x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
   $$
   where $$a=1$$, $$b=3$$, and $$c=1$$.  
   Calculate the discriminant:
   $$
   \Delta = 3^2 - 4(1)(1) = 9-4=5.
   $$
   The roots are:
   $$
   x=\frac{-3\pm\sqrt{5}}{2}.
   $$

3. **Final factorization and roots:**  
   The complete factorization is:
   $$
   (x-2)(x^2+3x+1).
   $$  
   The roots of the polynomial are:
   - $$x=2$$,
   - $$x=\frac{-3+\sqrt{5}}{2}$$,
   - $$x=\frac{-3-\sqrt{5}}{2}$$.
**Problem 1:** Factor and find the roots of $$3x^3 + 8x^2 + 3x - 2$$ given that $$-1$$ is a root. - **Factors:** $$(x+1)(3x-1)(x+2)$$ - **Roots:** $$x = -1$$, $$x = \frac{1}{3}$$, $$x = -2$$ **Problem 2:** Factor and find the roots of $$x^3 + x^2 - 5x - 2$$ given that $$2$$ is a root. - **Factors:** $$(x-2)(x^2 + 3x + 1)$$ - **Roots:** $$x = 2$$, $$x = \frac{-3 + \sqrt{5}}{2}$$, $$x = \frac{-3 - \sqrt{5}}{2}$$

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