Factor completely [tex]$2x^2 - 6x - 56$[/tex].

A. [tex]$(2x - 14)(x + 4)$[/tex]
B. [tex][tex]$(x - 7)(2x + 8)$[/tex][/tex]
C. [tex]$2(x - 7)(x + 4)$[/tex]
D. [tex]$2(x - 4)(x + 7)$[/tex]



Answer :

Sure, let's factor the given polynomial [tex]\(2x^2 - 6x - 56\)[/tex] completely.

### Step-by-Step Solution:

1. Write down the polynomial:
[tex]$2x^2 - 6x - 56$[/tex]

2. Factor out the greatest common factor (GCF) from all terms:
Notice that each term in the polynomial has a common factor of 2.
[tex]$2(x^2 - 3x - 28)$[/tex]

3. Factor the quadratic expression inside the parentheses:
Now, we need to factor [tex]\(x^2 - 3x - 28\)[/tex].

To do this, we look for two numbers that multiply to [tex]\(-28\)[/tex] (the constant term) and add up to [tex]\(-3\)[/tex] (the coefficient of [tex]\(x\)[/tex]).

These two numbers are [tex]\(-7\)[/tex] and [tex]\(4\)[/tex], because:
[tex]$-7 \cdot 4 = -28 \quad \text{and} \quad -7 + 4 = -3$[/tex]

4. Rewrite the quadratic expression using these factors:
[tex]$x^2 - 3x - 28 = (x - 7)(x + 4)$[/tex]

5. Combine with the factored out GCF:
Now, we include the GCF we factored out initially:
[tex]$2(x - 7)(x + 4)$[/tex]

Thus, the completely factored form of the polynomial [tex]\(2x^2 - 6x - 56\)[/tex] is:
[tex]$2(x - 7)(x + 4)$[/tex]

Among the given options, the correct one is:
[tex]$\boxed{2(x-7)(x+4)}$[/tex]