Rational Root Theorem online exercise for - Free Printable

Problem Analysis:

The task involves solving two polynomial equations and determining their roots, followed by providing the complete factorization of each polynomial. Let's solve each part step by step.

---

Part 1: Polynomial Equation

#### Given Polynomial:
\[ x^6 + 7x^5 + 20x^4 + 30x^3 + 25x^2 + 12x + 2 = 0 \]

#### Step 1: Use the Rational Root Theorem
The Rational Root Theorem suggests that any rational root of the polynomial is a factor of the constant term (2) divided by a factor of the leading coefficient (1). Thus, the possible rational roots are:
\[ \pm 1, \pm 2 \]

#### Step 2: Test Possible Roots
We test these possible roots using synthetic division or direct substitution.

- Test \( x = -1 \):
\[
(-1)^6 + 7(-1)^5 + 20(-1)^4 + 30(-1)^3 + 25(-1)^2 + 12(-1) + 2 = 1 - 7 + 20 - 30 + 25 - 12 + 2 = -1
\]
\( x = -1 \) is not a root.

- Test \( x = -2 \):
\[
(-2)^6 + 7(-2)^5 + 20(-2)^4 + 30(-2)^3 + 25(-2)^2 + 12(-2) + 2 = 64 - 224 + 320 - 240 + 100 - 24 + 2 = 0
\]
\( x = -2 \) is a root.

#### Step 3: Perform Synthetic Division
Divide the polynomial by \( x + 2 \):
\[
\begin{array}{r|rrrrrrr}
-2 & 1 & 7 & 20 & 30 & 25 & 12 & 2 \\
& & -2 & -10 & -20 & -20 & -10 & -4 \\
\hline
& 1 & 5 & 10 & 10 & 5 & 2 & 0 \\
\end{array}
\]
The quotient is:
\[ x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 2 \]

#### Step 4: Repeat the Process
We now solve:
\[ x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 2 = 0 \]

- Test \( x = -1 \):
\[
(-1)^5 + 5(-1)^4 + 10(-1)^3 + 10(-1)^2 + 5(-1) + 2 = -1 + 5 - 10 + 10 - 5 + 2 = 1
\]
\( x = -1 \) is not a root.

- Test \( x = -2 \):
\[
(-2)^5 + 5(-2)^4 + 10(-2)^3 + 10(-2)^2 + 5(-2) + 2 = -32 + 80 - 80 + 40 - 10 + 2 = 0
\]
\( x = -2 \) is a root.

Perform synthetic division again:
\[
\begin{array}{r|rrrrrr}
-2 & 1 & 5 & 10 & 10 & 5 & 2 \\
& & -2 & -6 & -8 & -4 & -2 \\
\hline
& 1 & 3 & 4 & 2 & 1 & 0 \\
\end{array}
\]
The quotient is:
\[ x^4 + 3x^3 + 4x^2 + 2x + 1 \]

#### Step 5: Solve the Quartic Polynomial
We now solve:
\[ x^4 + 3x^3 + 4x^2 + 2x + 1 = 0 \]

Notice that this polynomial can be factored as:
\[ (x^2 + x + 1)^2 = 0 \]

Thus, the roots of \( x^2 + x + 1 = 0 \) are:
\[ x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

#### Final Roots:
The roots of the original polynomial are:
\[ x = -2, -2, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \]

#### Complete Factorization:
\[ (x + 2)^2 \left( x^2 + x + 1 \right)^2 \]

---

Part 2: Polynomial Equation

#### Given Polynomial:
\[ x^4 + 2x^3 - 5x^2 - 6x = 0 \]

#### Step 1: Factor Out the Greatest Common Factor (GCF)
Factor out \( x \):
\[ x(x^3 + 2x^2 - 5x - 6) = 0 \]

This gives one root:
\[ x = 0 \]

#### Step 2: Solve the Cubic Polynomial
Solve:
\[ x^3 + 2x^2 - 5x - 6 = 0 \]

Use the Rational Root Theorem. Possible rational roots are:
\[ \pm 1, \pm 2, \pm 3, \pm 6 \]

- Test \( x = -1 \):
\[
(-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0
\]
\( x = -1 \) is a root.

Perform synthetic division:
\[
\begin{array}{r|rrrr}
-1 & 1 & 2 & -5 & -6 \\
& & -1 & -1 & 6 \\
\hline
& 1 & 1 & -6 & 0 \\
\end{array}
\]
The quotient is:
\[ x^2 + x - 6 \]

#### Step 3: Factor the Quadratic Polynomial
Factor:
\[ x^2 + x - 6 = (x + 3)(x - 2) \]

#### Final Roots:
The roots of the original polynomial are:
\[ x = 0, -1, -3, 2 \]

#### Complete Factorization:
\[ x(x + 1)(x + 3)(x - 2) \]

---

Final Answers:

1. For \( x^6 + 7x^5 + 20x^4 + 30x^3 + 25x^2 + 12x + 2 = 0 \):
- Roots: \( x = -2, -2, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \)
- Complete Factorization: \( (x + 2)^2 \left( x^2 + x + 1 \right)^2 \)

2. For \( x^4 + 2x^3 - 5x^2 - 6x = 0 \):
- Roots: \( x = 0, -1, -3, 2 \)
- Complete Factorization: \( x(x + 1)(x + 3)(x - 2) \)

Boxed Final Answer:

\[
\boxed{
\begin{aligned}
&\text{Part 1:} \\
&\text{Roots: } x = -2, -2, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \\
&\text{Complete Factorization: } (x + 2)^2 \left( x^2 + x + 1 \right)^2 \\
&\text{Part 2:} \\
&\text{Roots: } x = 0, -1, -3, 2 \\
&\text{Complete Factorization: } x(x + 1)(x + 3)(x - 2)
\end{aligned}
}
\]