To factor the polynomial \(2x^2 + 6x - 80\) completely, let's go through the steps methodically.
1. Identify the polynomial:
\(2x^2 + 6x - 80\)
2. Look for common factors:
In this case, all the terms do not have a common numerical factor other than 1, and all terms contain \(x\).
3. Attempt to factor by grouping:
We'll use the method of splitting the middle term. We need two numbers that multiply to \(2 \cdot (-80) = -160\) and add up to \(6\).
The pair that fits is \(16\) and \(-10\) because:
[tex]\[16 \times (-10) = -160\][/tex]
[tex]\[16 + (-10) = 6\][/tex]
4. Rewrite the polynomial by splitting the middle term:
[tex]\[2x^2 + 16x - 10x - 80\][/tex]
5. Group and factor each part:
Group the terms to create common factors:
[tex]\[2x^2 + 16x - 10x - 80 = 2x(x + 8) - 10(x + 8)\][/tex]
6. Factor out the common binomial factor:
We observe a common binomial factor \((x + 8)\):
[tex]\[2x(x + 8) - 10(x + 8) = (2x - 10)(x + 8)\][/tex]
7. Simplify further:
Notice that \(2x - 10\) can be factored further as:
[tex]\[2(x - 5)\][/tex]
8. Combine the factors:
Combine these factored parts:
[tex]\[2(x - 5)(x + 8)\][/tex]
Thus, the polynomial \(2x^2 + 6x - 80\) factors completely to \(2(x - 5)(x + 8)\).
So, the correct option among the given choices is:
[tex]\[ \boxed{2(x - 5)(x + 8)} \][/tex]
