If ` x^(3) + 6 x ^(2) + 12 x = 19` , then what is the value of ` x^(3)` ?

Options

A. 8
B. 27
C. `-1`
D. 1

Answer

To solve the equation \( x^3 + 6x^2 + 12x = 19 \) and find the value of \( x^3 \), follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x^3 + 6x^2 + 12x - 19 = 0 \]

### Step 2: Check for rational roots
We can use the Rational Root Theorem to test possible rational roots. We will check \( x = 1 \):
\[ 1^3 + 6(1^2) + 12(1) - 19 = 1 + 6 + 12 - 19 = 0 \]
Since this equals zero, \( x = 1 \) is a root.

### Step 3: Factor the polynomial
Since \( x = 1 \) is a root, we can factor the polynomial using synthetic division or polynomial long division. We divide \( x^3 + 6x^2 + 12x - 19 \) by \( x - 1 \).

Performing synthetic division:
- Coefficients: \( 1, 6, 12, -19 \)
- Root: \( 1 \)

The synthetic division yields:
\[ 1 | 1 \quad 6 \quad 12 \quad -19 | \quad 1 \quad 7 \quad 19 ------------------------- 1 \quad 7 \quad 19 \quad 0 \]
This gives us:
\[ x^3 + 6x^2 + 12x - 19 = (x - 1)(x^2 + 7x + 19) \]

### Step 4: Solve the quadratic
Now we have:
\[ (x - 1)(x^2 + 7x + 19) = 0 \]
Setting each factor to zero gives us:
1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( x^2 + 7x + 19 = 0 \)

To find the roots of the quadratic \( x^2 + 7x + 19 \), we calculate the discriminant:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 1 \cdot 19 = 49 - 76 = -27 \]
Since the discriminant is negative, there are no real roots from this quadratic.

### Step 5: Conclusion
The only real solution is \( x = 1 \). Now we can find \( x^3 \):
\[ x^3 = 1^3 = 1 \]

Thus, the value of \( x^3 \) is:
\[ \boxed{1} \]