Factor completely [tex]$81x^4-16$[/tex].

A. [tex](3x-2)(3x-2)(9x^2+4)[/tex]

B. [tex](3x-2)(3x+2)(9x^2-4)[/tex]

C. [tex](3x-2)(3x+2)(9x^2+4)[/tex]

D. [tex](3x+2)(3x+2)(9x^2+4)[/tex]



Answer :

Certainly! Let's factor the expression [tex]\(81 x^4 - 16\)[/tex] completely, step by step.

### Step 1: Recognize the Structure
Notice that [tex]\(81 x^4 - 16\)[/tex] resembles a difference of squares form: [tex]\(a^2 - b^2\)[/tex], where:

[tex]\[ a^2 = (9x^2)^2 \quad \text{and} \quad b^2 = 4^2 \][/tex]

Thus, we can rewrite the expression as:

[tex]\[ (9x^2)^2 - 4^2 \][/tex]

### Step 2: Apply the Difference of Squares Formula
Recall the difference of squares formula:

[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]

Applying this to our expression:

[tex]\[ (9x^2)^2 - 4^2 = (9x^2 - 4)(9x^2 + 4) \][/tex]

### Step 3: Factor Further if Possible
Next, we look at each factor to determine if they can be factored further.

#### Factor [tex]\(9x^2 - 4\)[/tex]
This again is a difference of squares and can be factored further. Write:

[tex]\[ 9x^2 - 4 = (3x)^2 - 2^2 \][/tex]

Again, apply the difference of squares formula:

[tex]\[ (3x)^2 - 2^2 = (3x - 2)(3x + 2) \][/tex]

#### Factor [tex]\(9x^2 + 4\)[/tex]
The term [tex]\(9x^2 + 4\)[/tex] is a sum of squares and cannot be factored further over the real numbers. Hence, it remains as it is.

### Step 4: Combine All Factors
Combine all the factors we obtained:

[tex]\[ 81 x^4 - 16 = (3 x - 2)(3 x + 2) (9 x^2 + 4) \][/tex]

### Conclusion
Thus, the completely factored form of [tex]\(81 x^4 - 16\)[/tex] is:

[tex]\[ (3 x - 2)(3 x + 2)(9 x^2 + 4) \][/tex]