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### Advanced Algebra Concepts and Connections - Unit 4

#### Problem 1

Since the 1930s, physicists have known that the 'vacuum' of space is not empty. It contains particles and energy that come and go and cannot be directly detected. Moments after the Big Bang, this vacuum energy was large enough to cause the universe to expand by trillions of times in size! Astronomers call this Cosmological Inflation.

Several theoretical studies of the vacuum state have focused attention on a polynomial function:

[tex]\[ V(x) = \frac{L}{6} x^4 - m^2 x^2 \][/tex]

This function, called the Coleman-Weinberg Potential, allows physicists to calculate the energy of the vacuum state, [tex]\( V(x) \)[/tex], in terms of the mass, [tex]\( x \)[/tex], of a new kind of yet-to-be-discovered particle called the [tex]\( X \)[/tex] Boson.

Task:
Factor [tex]\( V(x) \)[/tex] and determine the location of all [tex]\( x \)[/tex]-intercepts for the general case where [tex]\( m \)[/tex] and [tex]\( L \)[/tex] are not specified.

Sagot :

To solve the problem of factoring the polynomial [tex]\( V(x) = \frac{L}{6} x^4 - m^2 x^2 \)[/tex] and determining the locations of the [tex]\( x \)[/tex]-intercepts, we will follow these steps:

### Step 1: Factoring [tex]\( V(x) \)[/tex]

Given the polynomial:
[tex]\[ V(x) = \frac{L}{6} x^4 - m^2 x^2 \][/tex]

We first note that we can factor out a common term, [tex]\( x^2 \)[/tex], from both terms in the polynomial:
[tex]\[ V(x) = x^2 \left( \frac{L}{6} x^2 - m^2 \right) \][/tex]

Now we have:
[tex]\[ V(x) = x^2 \left( \frac{L}{6} x^2 - m^2 \right) \][/tex]

To simplify further, we leave it in this factored form since it is compact and already shows the expression in terms of products.

### Step 2: Finding the [tex]\( x \)[/tex]-intercepts

The [tex]\( x \)[/tex]-intercepts occur where [tex]\( V(x) \)[/tex] equals zero:
[tex]\[ V(x) = 0 \][/tex]

Thus,
[tex]\[ x^2 \left( \frac{L}{6} x^2 - m^2 \right) = 0 \][/tex]

We solve this equation by setting each factor to zero separately:
1. [tex]\( x^2 = 0 \)[/tex]
2. [tex]\( \frac{L}{6} x^2 - m^2 = 0 \)[/tex]

#### Solving [tex]\( x^2 = 0 \)[/tex]

This implies:
[tex]\[ x = 0 \][/tex]

#### Solving [tex]\( \frac{L}{6} x^2 - m^2 = 0 \)[/tex]

To solve this, we isolate [tex]\( x^2 \)[/tex]:
[tex]\[ \frac{L}{6} x^2 = m^2 \][/tex]

Multiplying both sides by [tex]\( \frac{6}{L} \)[/tex] to solve for [tex]\( x^2 \)[/tex]:
[tex]\[ x^2 = \frac{6m^2}{L} \][/tex]

Taking the square root of both sides, we find:
[tex]\[ x = \pm \sqrt{\frac{6m^2}{L}} \][/tex]

This simplifies to:
[tex]\[ x = \pm \sqrt{6} m \sqrt{\frac{1}{L}} \][/tex]

### Summary of [tex]\( x \)[/tex]-intercepts

Combining these results, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x = 0, x = -\sqrt{6} m \sqrt{\frac{1}{L}}, x = \sqrt{6} m \sqrt{\frac{1}{L}} \][/tex]

### Final Answer

Thus, the factored form of [tex]\( V(x) \)[/tex] and the locations of the [tex]\( x \)[/tex]-intercepts are:

Factored Form:
[tex]\[ V(x) = x^2 \left( \frac{L}{6} x^2 - m^2 \right) \][/tex]

[tex]\( x \)[/tex]-intercepts:
[tex]\[ x = 0, \quad x = -\sqrt{6} m \sqrt{\frac{1}{L}}, \quad x = \sqrt{6} m \sqrt{\frac{1}{L}} \][/tex]