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Which expression is equivalent to this polynomial?

[tex]16 x^2+4[/tex]

A. [tex]\((4 x+2 i)(4 x-2 i)\)[/tex]
B. [tex]\((4 x+2)(4 x-2)\)[/tex]
C. [tex]\((4 x+2)^2\)[/tex]
D. [tex]\((4 x-2 i)^2\)[/tex]


Sagot :

To find which expression is equivalent to the polynomial [tex]\(16x^2 + 4\)[/tex], let's factor the given polynomial step-by-step.

1. Factor out the greatest common factor (GCF):
The GCF of the terms [tex]\(16x^2\)[/tex] and [tex]\(4\)[/tex] is [tex]\(4\)[/tex]. So, we can factor [tex]\(4\)[/tex] out of the polynomial:
[tex]\[ 16x^2 + 4 = 4(4x^2 + 1) \][/tex]

2. Examine the remaining quadratic expression:
The quadratic expression inside the parentheses is [tex]\(4x^2 + 1\)[/tex].

3. Interpret the complex roots:
The expression [tex]\(4x^2 + 1\)[/tex] can be written as [tex]\( (2x)^2 + 1^2 \)[/tex], which fits the form of a sum of squares. The factorization of a sum of squares involves complex numbers:
[tex]\[ 4x^2 + 1 = (2x + i)(2x - i) \][/tex]

4. Combine the results:
Substituting this factorization back into the expression we factored out earlier, we get:
[tex]\[ 4(4x^2 + 1) = 4((2x + i)(2x - i)) = 4(2x + i)(2x - i) \][/tex]

From the steps and intermediate results, we find that the correct factorization of [tex]\(16x^2 + 4\)[/tex] is:
[tex]\[ 4(2x + i)(2x - i) \][/tex]

Looking at the given choices:
- A. [tex]\((4x + 2i)(4x - 2i)\)[/tex]
- B. [tex]\((4x + 2)(4x - 2)\)[/tex]
- C. [tex]\((4x + 2)^2\)[/tex]
- D. [tex]\((4x - 2i)^2\)[/tex]

The correct equivalent expression to [tex]\(4(4x^2 + 1)\)[/tex] is [tex]\((4x + 2i)(4x - 2i)\)[/tex], corresponding to:
[tex]\[ 4(2x + i)(2x - i) \][/tex]
Therefore, the correct answer is:

A. [tex]\((4x + 2i)(4x - 2i)\)[/tex]