Evaluate the zeroes of the polynomial `2x^(2)-16` ?

Options

A. `2sqrt2,-2sqrt2`
B. `sqrt2,-sqrt2`
C. `4,-4`
D. `2,-2`

Answer

To evaluate the zeroes of the polynomial \(2x^2 - 16\), we will follow these steps:

### Step 1: Set the polynomial equal to zero
We start by setting the polynomial equal to zero:
\[ 2x^2 - 16 = 0 \]

### Step 2: Add 16 to both sides
Next, we add 16 to both sides of the equation:
\[ 2x^2 = 16 \]

### Step 3: Divide both sides by 2
Now, we divide both sides by 2 to isolate \(x^2\):
\[ x^2 = \frac{16}{2} \]
\[ x^2 = 8 \]

### Step 4: Take the square root of both sides
Next, we take the square root of both sides to solve for \(x\):
\[ x = \pm \sqrt{8} \]

### Step 5: Simplify \(\sqrt{8}\)
We can simplify \(\sqrt{8}\) further. Since \(8 = 4 \times 2\), we can write:
\[ \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2} \]
Thus, we have:
\[ x = \pm 2\sqrt{2} \]

### Conclusion
The zeroes of the polynomial \(2x^2 - 16\) are:
\[ x = 2\sqrt{2} \quad \text{and} \quad x = -2\sqrt{2} \]

### Final Answer
The zeroes of the polynomial are \(2\sqrt{2}\) and \(-2\sqrt{2}\).

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