determine all the real roots of the following polynomial x^5-6x^4+3x^3+26x^2-24x

To determine all the real roots of the polynomial \( f(x) = x^5 - 6x^4 + 3x^3 + 26x^2 - 24x \), we can follow these steps:

### Step 1: Factor out the greatest common factor (GCF)
The polynomial has a common factor of \( x \) in all terms. Let's factor it out:

\[
f(x) = x(x^4 - 6x^3 + 3x^2 + 26x - 24)
\]

### Step 2: Solve for \( x = 0 \)
The first root is immediately \( x = 0 \).

Now, we need to find the roots of the quartic polynomial:

\[
x^4 - 6x^3 + 3x^2 + 26x - 24 = 0
\]

### Step 3: Use the Rational Root Theorem
The Rational Root Theorem suggests that possible rational roots are the factors of the constant term (\(-24\)) divided by the factors of the leading coefficient (\(1\)).

Possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \).

### Step 4: Test possible rational roots
Let's test these roots by substituting them into the polynomial.

**Test \( x = 2 \):**

\[
f(2) = 2^4 - 6(2)^3 + 3(2)^2 + 26(2) - 24 = 16 - 48 + 12 + 52 - 24 = 8 \quad \text{(not a root)}
\]

**Test \( x = -2 \):**

\[
f(-2) = (-2)^4 - 6(-2)^3 + 3(-2)^2 + 26(-2) - 24 = 16 + 48 + 12 - 52 - 24 = 0 \quad \text{(root)}
\]

So, \( x = -2 \) is a root.

### Step 5: Factor using synthetic division
Now, let's factor \( x^4 - 6x^3 + 3x^2 + 26x - 24 \) by dividing it by \( x + 2 \) using synthetic division:

\[
\begin{array}{r|rrrrr}
-2 & 1 & -6 & 3 & 26 & -24 \\
   &   & -2 & 16 & -38 & 24 \\
\hline
   & 1 & -8 & 19 & -12 & 0 \\
\end{array}
\]

The quotient is:

\[
x^3 - 8x^2 + 19x - 12
\]

So, the factorization so far is:

\[
f(x) = x(x + 2)(x^3 - 8x^2 + 19x - 12)
\]

### Step 6: Find the roots of the cubic polynomial
We now solve \( x^3 - 8x^2 + 19x - 12 = 0 \). We can apply the Rational Root Theorem again to test possible roots:

**Test \( x = 1 \):**

\[
f(1) = 1^3 - 8(1)^2 + 19(1) - 12 = 1 - 8 + 19 - 12 = 0 \quad \text{(root)}
\]

So, \( x = 1 \) is a root.

### Step 7: Factor using synthetic division again
Factor \( x^3 - 8x^2 + 19x - 12 \) by dividing it by \( x - 1 \):

\[
\begin{array}{r|rrrr}
1 & 1 & -8 & 19 & -12 \\
  &   & 1 & -7 & 12 \\
\hline
  & 1 & -7 & 12 & 0 \\
\end{array}
\]

The quotient is:

\[
x^2 - 7x + 12
\]

So, the factorization is:

\[
f(x) = x(x + 2)(x - 1)(x^2 - 7x + 12)
\]

### Step 8: Solve the quadratic equation
We solve \( x^2 - 7x + 12 = 0 \) using the quadratic formula:

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

Where \( a = 1 \), \( b = -7 \), and \( c = 12 \):

\[
x = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm \sqrt{1}}{2}
\]

\[
x = \frac{7 \pm 1}{2}
\]

This gives us:

\[
x = 4 \quad \text{or} \quad x = 3
\]

### Final Answer
The real roots of the polynomial \( f(x) = x^5 - 6x^4 + 3x^3 + 26x^2 - 24x \) are:

\[
\boxed{0, -2, 1, 3, 4}
\]

Do you have any questions about this process or need further details? Here are some related questions:

1. How do you use the Rational Root Theorem to find possible roots?
2. What are the advantages of factoring polynomials?
3. How can synthetic division help in factoring polynomials?
4. What if a polynomial has complex roots—how would you find them?
5. How does the quadratic formula work for solving quadratic equations?
6. Can you explain the difference between real and complex roots?
7. How can you graphically verify the roots of a polynomial?
8. What strategies can be used if the polynomial is difficult to factor directly?

**Tip:** When working with higher-degree polynomials, starting with the Rational Root Theorem and synthetic division can often simplify the process of finding roots significantly.

Learn how to find all real roots of the polynomial x^5 - 6x^4 + 3x^3 + 26x^2 - 24x using step-by-step methods including factoring, Rational Root Theorem, and synthetic division. This problem covers key algebraic concepts and theorems suitable for advanced high school students.

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