Find root using Rational root theorem x⁴-5x³+6x²+4x-8

The Rational Root Theorem helps to find the possible rational roots of a polynomial equation. For the polynomial \( P(x) = x^4 - 5x3 + 6x^2 + 4x - 8 \), the Rational Root Theorem states that any rational root, when expressed as a fraction \(\frac{p}{q}\), \( p \) is a factor of the constant term \( -8 \), and \( q \) is a factor of the leading coefficient \( 1 \).

### Step 1: List Factors

**Factors of the constant term (-8):**
\[
\pm 1, \pm 2, \pm 4, \pm 8
\]

**Factors of the leading coefficient (1):**
\[
\pm 1
\]

### Step 2: Possible Rational Roots

Using the Rational Root Theorem, the possible rational roots are:
\[
\pm 1, \pm 2, \pm 4, \pm 8
\]

### Step 3: Test the Possible Rational Roots

We will use synthetic division or direct substitution to test these possible roots.

**Testing \( x = 1 \):**
\[
P(1) = 1^4 - 5 \cdot 1^3 + 6 \cdot 1^2 + 4 \cdot 1 - 8 = 1 - 5 + 6 + 4 - 8 = -2 \neq 0
\]
So, \( x = 1 \) is not a root.

**Testing \( x = -1 \):**
\[
P(-1) = (-1)^4 - 5 \cdot (-1)^3 + 6 \cdot (-1)^2 + 4 \cdot (-1) - 8 = 1 + 5 + 6 - 4 - 8 = 0
\]
So, \( x = -1 \) is a root.

### Step 4: Factor the Polynomial

Since \( x = -1 \) is a root, we can factor out \( (x + 1) \).

Perform synthetic division of \( P(x) \) by \( x + 1 \):

\[
\begin{array}{r|rrrrr}
 -1 & 1 & -5 & 6 & 4 & -8 \\
    &   & -1 & 6 & -12 & 8 \\
\hline
    & 1 & -6 & 12 & -8 & 0 \\
\end{array}
\]

The quotient is:
\[
x^3 - 6x^2 + 12x - 8
\]

### Step 5: Repeat the Process

Now we factor \( x^3 - 6x^2 + 12x - 8 \). Again, list the factors of the constant term (-8) and the leading coefficient (1).

Possible roots are again:
\[
\pm 1, \pm 2, \pm 4, \pm 8
\]

**Testing \( x = 1 \):**
\[
P(1) = 1^3 - 6 \cdot 1^2 + 12 \cdot 1 - 8 = 1 - 6 + 12 - 8 = -1 \neq 0
\]
So, \( x = 1 \) is not a root.

**Testing \( x = -1 \):**
\[
P(-1) = (-1)^3 - 6 \cdot (-1)^2 + 12 \cdot (-1) - 8 = -1 - 6 - 12 - 8 = -27 \neq 0
\]
So, \( x = -1 \) is not a root.

**Testing \( x = 2 \):**
\[
P(2) = 2^3 - 6 \cdot 2^2 + 12 \cdot 2 - 8 = 8 - 24 + 24 - 8 = 0
\]
So, \( x = 2 \) is a root.

Perform synthetic division of \( x^3 - 6x^2 + 12x - 8 \) by \( x - 2 \):

\[
\begin{array}{r|rrrr}
  2 & 1 & -6 & 12 & -8 \\
    &   & 2 & -8 & 8 \\
\hline
    & 1 & -4 & 4 & 0 \\
\end{array}
\]

The quotient is:
\[
x^2 - 4x + 4
\]

### Step 6: Solve the Remaining Quadratic Equation

Solve \( x^2 - 4x + 4 = 0 \):

\[
(x - 2)^2 = 0 \implies x = 2
\]

### Step 7: Collect All Roots

The polynomial \( P(x) = x^4 - 5x^3 + 6x^2 + 4x - 8 \) has the following roots:
\[
x = -1, x = 2 \text{ (with multiplicity 3)}
\]

### Summary
The roots of the polynomial are:
\[
x = -1, \quad x = 2, \quad x = 2, \quad x = 2
\]

Would you like more details or have any questions?

Here are some related questions:

1. How is the Rational Root Theorem derived?
2. What are the applications of the Rational Root Theorem?
3. Can the Rational Root Theorem be used for polynomials with irrational or complex coefficients?
4. What is synthetic division and how is it used in polynomial division?
5. How do you determine the multiplicity of a root?
6. What are other methods to find the roots of polynomials besides the Rational Root Theorem?
7. How does the Rational Root Theorem relate to the Fundamental Theorem of Algebra?
8. How can graphing a polynomial help in finding its roots?

**Tip:** When using synthetic division, always write down each step clearly to avoid mistakes in calculations.

Learn how to find roots using the Rational Root Theorem for the polynomial x^4 - 5x^3 + 6x^2 + 4x - 8. Detailed step-by-step solution with synthetic division and testing of possible rational roots. Suitable for advanced high school students.

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