Show that 2x + 7 is a factor of `2x^3 + 7x^2 - 4x - 14` Hence, solve the equation: `2x^3 + 7x^2 - 4x - 14 = 0.`

Answer

To show that \(2x + 7\) is a factor of the polynomial \(2x^3 + 7x^2 - 4x - 14\), we can use the Factor Theorem. According to the Factor Theorem, if \(f(a) = 0\), then \(x - a\) is a factor of the polynomial \(f(x)\).

### Step 1: Define the polynomial
Let \(f(x) = 2x^3 + 7x^2 - 4x - 14\).

### Step 2: Find the root corresponding to the factor
The factor we are testing is \(2x + 7\). To find the root, we set \(2x + 7 = 0\):
\[ 2x = -7 \implies x = -\frac{7}{2} \]

### Step 3: Substitute the root into the polynomial
Now we will substitute \(x = -\frac{7}{2}\) into \(f(x)\) to check if \(f\left(-\frac{7}{2}\right) = 0\):
\[ f\left(-\frac{7}{2}\right) = 2\left(-\frac{7}{2}\right)^3 + 7\left(-\frac{7}{2}\right)^2 - 4\left(-\frac{7}{2}\right) - 14 \]

### Step 4: Calculate each term
Calculating each term:
1. \(2\left(-\frac{7}{2}\right)^3 = 2 \cdot -\frac{343}{8} = -\frac{343}{4}\)
2. \(7\left(-\frac{7}{2}\right)^2 = 7 \cdot \frac{49}{4} = \frac{343}{4}\)
3. \(-4\left(-\frac{7}{2}\right) = 14\)
4. The constant term is \(-14\).

### Step 5: Combine the results
Now, substituting these back into \(f\):
\[ f\left(-\frac{7}{2}\right) = -\frac{343}{4} + \frac{343}{4} + 14 - 14 \]
This simplifies to:
\[ f\left(-\frac{7}{2}\right) = 0 \]

### Step 6: Conclusion about the factor
Since \(f\left(-\frac{7}{2}\right) = 0\), we conclude that \(2x + 7\) is indeed a factor of the polynomial \(2x^3 + 7x^2 - 4x - 14\).

### Step 7: Polynomial Division
Now we will divide \(f(x)\) by \(2x + 7\) to find the other factor.

1. Divide \(2x^3\) by \(2x\) to get \(x^2\).
2. Multiply \(x^2\) by \(2x + 7\) to get \(2x^3 + 7x^2\).
3. Subtract this from \(f(x)\):
\[ (2x^3 + 7x^2 - 4x - 14) - (2x^3 + 7x^2) = -4x - 14 \]
4. Now divide \(-4x\) by \(2x\) to get \(-2\).
5. Multiply \(-2\) by \(2x + 7\) to get \(-4x - 14\).
6. Subtract this from \(-4x - 14\):
\[ (-4x - 14) - (-4x - 14) = 0 \]

The quotient is \(x^2 - 2\).

### Step 8: Factor the quotient
Now we can factor \(x^2 - 2\):
\[ x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}) \]

### Final Factorization
Thus, the complete factorization of the polynomial is:
\[ 2x^3 + 7x^2 - 4x - 14 = (2x + 7)(x - \sqrt{2})(x + \sqrt{2}) \]

### Step 9: Solve the equation
To solve \(2x^3 + 7x^2 - 4x - 14 = 0\), we set each factor to zero:
1. \(2x + 7 = 0 \implies x = -\frac{7}{2}\)
2. \(x - \sqrt{2} = 0 \implies x = \sqrt{2}\)
3. \(x + \sqrt{2} = 0 \implies x = -\sqrt{2}\)

### Solutions
The solutions to the equation \(2x^3 + 7x^2 - 4x - 14 = 0\) are:
\[ x = -\frac{7}{2}, \quad x = \sqrt{2}, \quad x = -\sqrt{2} \]