https://file.aiquickdraw.com/mathbot/file/1727801814420cbb0944p.jpg

We are tasked with finding the real zeros (roots) of the polynomial \( x^6 + 7x^3 - 8 \) and factorizing it into factors of degree 1 and 2.

### Step 1: Substitution to simplify the equation
Let’s introduce a substitution to simplify the problem:
- Set \( y = x^3 \). Therefore, \( x^6 = (x^3)^2 = y^2 \).

This transforms the given polynomial into a quadratic equation in terms of \( y \):
\[
y^2 + 7y - 8 = 0
\]

### Step 2: Solve the quadratic equation
We can now solve this quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 7 \), and \( c = -8 \). Substituting these values into the formula:
\[
y = \frac{-7 \pm \sqrt{7^2 - 4(1)(-8)}}{2(1)}
\]
\[
y = \frac{-7 \pm \sqrt{49 + 32}}{2}
\]
\[
y = \frac{-7 \pm \sqrt{81}}{2}
\]
\[
y = \frac{-7 \pm 9}{2}
\]

Thus, we have two solutions for \( y \):
\[
y_1 = \frac{-7 + 9}{2} = 1 \quad \text{and} \quad y_2 = \frac{-7 - 9}{2} = -8
\]

### Step 3: Find the values of \( x \)
Recall that \( y = x^3 \), so now we solve for \( x \) from the equations \( x^3 = 1 \) and \( x^3 = -8 \).

1. \( x^3 = 1 \) has a real solution \( x = 1 \).
2. \( x^3 = -8 \) has a real solution \( x = -2 \) (since \( (-2)^3 = -8 \)).

Thus, the real roots of the polynomial are \( x = 1 \) and \( x = -2 \).

### Step 4: Factorization
Now that we know the real roots, we can factor the polynomial. Using the substitution \( x^3 = y \), the quadratic equation factored as:
\[
(y - 1)(y + 8) = 0
\]
Returning to \( x \), we can write:
\[
(x^3 - 1)(x^3 + 8) = 0
\]

Both of these factors can be further factored:
1. \( x^3 - 1 = (x - 1)(x^2 + x + 1) \)
2. \( x^3 + 8 = (x + 2)(x^2 - 2x + 4) \)

Thus, the factorization of the polynomial \( x^6 + 7x^3 - 8 \) is:
\[
(x - 1)(x^2 + x + 1)(x + 2)(x^2 - 2x + 4)
\]

### Final Answer:
- The real roots of the polynomial are \( x = 1 \) and \( x = -2 \).
- The factorization in terms of degree 1 and 2 factors is:
\[
(x - 1)(x^2 + x + 1)(x + 2)(x^2 - 2x + 4)
\]

Would you like further details or have any questions about this solution?

---

Here are five related questions to explore:

1. How can the Rational Root Theorem be used to find real roots of polynomials?
2. What are the complex roots of \( x^3 = 1 \) and \( x^3 = -8 \)?
3. How do you perform polynomial division to factor higher degree polynomials?
4. What is the significance of the quadratic formula in solving higher-degree polynomials?
5. How would the factorization change if the coefficients of the polynomial were complex?

**Tip:** When solving higher-degree polynomials, substitution can often reduce the problem to a simpler form, like a quadratic equation.

Determine the real zeros (roots) of the polynomial x^6 + 7x^3 - 8 and give a factorization of this polynomial in factors of degree 1 and 2.

Determine the real zeros of the polynomial x^6 + 7x^3 - 8 and find its factorization into degree 1 and 2 factors. This problem involves solving a quadratic equation using substitution and factorization. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation