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Factor completely. Select "Prime" if the polynomial cannot be factored.

[tex]\[x^2 - 4x - 12\][/tex]

A. (x - 6)(x + 2)
B. (x - 2)(x + 6)
C. Prime


Sagot :

To factor the quadratic polynomial [tex]\(x^2 - 4x - 12\)[/tex] completely, we can follow these steps:

1. Identify the quadratic polynomial:
The given polynomial is [tex]\(x^2 - 4x - 12\)[/tex].

2. Look for factors of the constant term (-12) that will add up to the coefficient of the middle term (-4):
We need two numbers that multiply to [tex]\(-12\)[/tex] and add up to [tex]\(-4\)[/tex]. Those numbers are [tex]\(6\)[/tex] and [tex]\(-2\)[/tex]. Let's check:

- Factors of [tex]\(-12\)[/tex]: [tex]\((1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4)\)[/tex]
- The pair [tex]\((-6)\)[/tex] and [tex]\(+2\)[/tex] multiply to [tex]\(-12\)[/tex] and add up to [tex]\(-4\)[/tex].

3. Write the expression using these factors:
Rewrite the middle term [tex]\(-4x\)[/tex] using [tex]\(-6x\)[/tex] and [tex]\(2x\)[/tex]:
[tex]\[ x^2 - 4x - 12 = x^2 - 6x + 2x - 12 \][/tex]

4. Factor by grouping:
Group the terms to make factoring easier:
[tex]\[ x^2 - 6x + 2x - 12 = x(x - 6) + 2(x - 6) \][/tex]

5. Factor out the common term:
Factor out the common binomial factor [tex]\((x - 6)\)[/tex] from both groups:
[tex]\[ x(x - 6) + 2(x - 6) = (x + 2)(x - 6) \][/tex]

Thus, the quadratic polynomial [tex]\(x^2 - 4x - 12\)[/tex] factors completely into [tex]\((x - 6)(x + 2)\)[/tex].

So, the final answer is:
[tex]\[ (x - 6)(x + 2) \][/tex]