Answer :
We are given the polynomial expression [tex]\( x^2 - 4x - 4 \)[/tex] and need to factor it. Here are our steps:
1. Identify the given polynomial:
[tex]\[ x^2 - 4x - 4 \][/tex]
2. Consider the options provided for factoring:
- Prime (cannot be factored)
- [tex]\((x - 2)(x + 2)\)[/tex]
- [tex]\((x - 2)^2\)[/tex]
- [tex]\((x + 2)^2\)[/tex]
3. Determine the correctness of these factorizations:
- Prime implies it cannot be factored at all, leaving only the given polynomial. Let's first verify if this polynomial can be factored.
4. Check if the polynomial can be factored by exploring the roots:
Consider the quadratic equation in standard form:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
Where:
[tex]\[ a = 1, \quad b = -4, \quad c = -4 \][/tex]
5. Check the discriminant to determine factorability:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-4)^2 - 4 \cdot 1 \cdot (-4) \][/tex]
[tex]\[ \Delta = 16 + 16 \][/tex]
[tex]\[ \Delta = 32 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has two real and distinct roots. Therefore, it is factorable.
6. Analyze the given factor options:
- Option 1: Prime (inapplicable as the polynomial is factorable)
- Option 2: [tex]\((x - 2)(x + 2)\)[/tex]
[tex]\[ (x - 2)(x + 2) = x^2 - (2^2) = x^2 - 4 \][/tex]
This doesn't match our polynomial.
- Option 3: [tex]\((x - 2)^2\)[/tex]
[tex]\[ (x - 2)^2 = x^2 - 4x + 4 \][/tex]
This also doesn’t match our polynomial.
- Option 4: [tex]\((x + 2)^2\)[/tex]
[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]
This also doesn’t match our polynomial.
7. Conclusion:
Since neither [tex]\((x-2)(x+2)\)[/tex], [tex]\((x-2)^2\)[/tex], nor [tex]\((x+2)^2\)[/tex] correspond to the given polynomial [tex]\( x^2 - 4x - 4 \)[/tex], the only logical conclusion is that the polynomial doesn't match any given factor forms other than the original polynomial itself.
Thus, the factorization doesn't change it from its original form suggesting:
[tex]\[ \boxed{1} \][/tex]
This indicates that the polynomial [tex]\( x^2 - 4x - 4 \)[/tex] is in its simplest factorable form already, categorized under "Prime (cannot be factored)."
1. Identify the given polynomial:
[tex]\[ x^2 - 4x - 4 \][/tex]
2. Consider the options provided for factoring:
- Prime (cannot be factored)
- [tex]\((x - 2)(x + 2)\)[/tex]
- [tex]\((x - 2)^2\)[/tex]
- [tex]\((x + 2)^2\)[/tex]
3. Determine the correctness of these factorizations:
- Prime implies it cannot be factored at all, leaving only the given polynomial. Let's first verify if this polynomial can be factored.
4. Check if the polynomial can be factored by exploring the roots:
Consider the quadratic equation in standard form:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
Where:
[tex]\[ a = 1, \quad b = -4, \quad c = -4 \][/tex]
5. Check the discriminant to determine factorability:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-4)^2 - 4 \cdot 1 \cdot (-4) \][/tex]
[tex]\[ \Delta = 16 + 16 \][/tex]
[tex]\[ \Delta = 32 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has two real and distinct roots. Therefore, it is factorable.
6. Analyze the given factor options:
- Option 1: Prime (inapplicable as the polynomial is factorable)
- Option 2: [tex]\((x - 2)(x + 2)\)[/tex]
[tex]\[ (x - 2)(x + 2) = x^2 - (2^2) = x^2 - 4 \][/tex]
This doesn't match our polynomial.
- Option 3: [tex]\((x - 2)^2\)[/tex]
[tex]\[ (x - 2)^2 = x^2 - 4x + 4 \][/tex]
This also doesn’t match our polynomial.
- Option 4: [tex]\((x + 2)^2\)[/tex]
[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]
This also doesn’t match our polynomial.
7. Conclusion:
Since neither [tex]\((x-2)(x+2)\)[/tex], [tex]\((x-2)^2\)[/tex], nor [tex]\((x+2)^2\)[/tex] correspond to the given polynomial [tex]\( x^2 - 4x - 4 \)[/tex], the only logical conclusion is that the polynomial doesn't match any given factor forms other than the original polynomial itself.
Thus, the factorization doesn't change it from its original form suggesting:
[tex]\[ \boxed{1} \][/tex]
This indicates that the polynomial [tex]\( x^2 - 4x - 4 \)[/tex] is in its simplest factorable form already, categorized under "Prime (cannot be factored)."
