The product of uncommon real roots of the polynomials ` p(x)=x^4+2x^3-8x^2-6x+15` and ` q(x) ` `= x^3+4x^2-x-10 ` is :

Options

A. 4
B. 6
C. 8
D. 12

Answer

To solve the problem of finding the product of the uncommon real roots of the polynomials \( p(x) = x^4 + 2x^3 - 8x^2 - 6x + 15 \) and \( q(x) = x^3 + 4x^2 - x - 10 \), we will follow these steps:

### Step 1: Identify the common root
We will first find the common root of the two polynomials. Let \( \alpha \) be the common root. This means:
\[ p(\alpha) = 0 \]
\[ q(\alpha) = 0 \]

### Step 2: Use the trial and error method to find the common root
We will substitute various values into both polynomials to find a common root.

Let's try \( \alpha = -2 \):
1. For \( p(-2) \):
\[ p(-2) = (-2)^4 + 2(-2)^3 - 8(-2)^2 - 6(-2) + 15 \]
\[ = 16 - 16 - 32 + 12 + 15 = 0 \]
Thus, \( -2 \) is a root of \( p(x) \).

2. For \( q(-2) \):
\[ q(-2) = (-2)^3 + 4(-2)^2 - (-2) - 10 \]
\[ = -8 + 16 + 2 - 10 = 0 \]
Thus, \( -2 \) is also a root of \( q(x) \).

### Step 3: Factor out the common root
Since \( -2 \) is a common root, we can factor both polynomials by \( x + 2 \).

#### Factor \( q(x) \):
Using synthetic division or polynomial long division to divide \( q(x) \) by \( x + 2 \):
\[ q(x) = (x + 2)(x^2 + 2x - 5) \]

#### Factor \( p(x) \):
Using synthetic division or polynomial long division to divide \( p(x) \) by \( x + 2 \):
\[ p(x) = (x + 2)(x^3 - 3x^2 - 12) \]

### Step 4: Find the roots of the remaining factors
Now, we need to find the roots of the remaining factors \( x^2 + 2x - 5 \) and \( x^3 - 3x^2 - 12 \).

1. **For \( x^2 + 2x - 5 = 0 \)**:
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = -1 \pm \sqrt{6} \]
Thus, the roots are \( -1 + \sqrt{6} \) and \( -1 - \sqrt{6} \).

2. **For \( x^3 - 3x^2 - 12 = 0 \)**:
We can use numerical methods or further factorization to find the roots. Let's use the Rational Root Theorem or numerical approximation to find one root, which we can find to be \( 3 \) (by trial).

Performing synthetic division of \( x^3 - 3x^2 - 12 \) by \( x - 3 \):
\[ x^3 - 3x^2 - 12 = (x - 3)(x^2 + 0x + 4) \]
The quadratic \( x^2 + 4 = 0 \) gives complex roots \( 2i \) and \( -2i \).

### Step 5: Identify the uncommon roots
The uncommon real roots are:
- From \( p(x) \): \( -1 + \sqrt{6} \) and \( -1 - \sqrt{6} \).
- From \( q(x) \): \( 3 \).

### Step 6: Calculate the product of the uncommon real roots
The uncommon real roots are \( -1 + \sqrt{6} \), \( -1 - \sqrt{6} \), and \( 3 \).

The product is:
\[ (-1 + \sqrt{6})(-1 - \sqrt{6}) \cdot 3 = (1 - 6) \cdot 3 = -5 \cdot 3 = -15 \]

### Final Answer
The product of the uncommon real roots is \( -15 \).