To factorize the expression [tex]\( 16 p^4 + 36 p^2 q^2 + 81 q^4 \)[/tex], let's follow a systematic approach.
### Step 1: Identify the terms and their degrees
The given polynomial is:
[tex]\[ 16 p^4 + 36 p^2 q^2 + 81 q^4 \][/tex]
Notice that:
- The first term, [tex]\(16 p^4\)[/tex], is a perfect square: [tex]\((4p^2)^2\)[/tex].
- The second term, [tex]\(36 p^2 q^2\)[/tex], is a product of [tex]\(6pq\)[/tex] with itself: [tex]\( (6pq)^2\)[/tex].
- The third term, [tex]\(81 q^4\)[/tex], is a perfect square: [tex]\((9q^2)^2\)[/tex].
### Step 2: Assume it follows a special form
Given the pattern in the polynomial, we can assume it could be factored using squares of binomials or the sum/difference of squares:
[tex]\[ (a p^2 + b p q + c q^2)^2 \][/tex]
### Step 3: Match the polynomial with a known algebraic identity
We will now verify if the expression can be represented as:
[tex]\[ (a p^2 + b p q + c q^2)(a p^2 - b p q + c q^2) \][/tex]
Expanding this we get:
[tex]\[ (a p^2)^2 + (a p^2)(b p q) - (a p^2)(b p q) + (a p^2)(c q^2) + (b p q)(c q^2) - (b p q)(c q^2) + (c q^2)^2 \][/tex]
### Step 4: Find appropriate values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]
When expanded:
[tex]\[ (4 p^2 - 6 p q + 9 q^2)(4 p^2 + 6 p q + 9 q^2) \][/tex]
#### First binomial:
[tex]\[ 4p^2 + 6pq + 9q^2 \][/tex]
#### Second binomial:
[tex]\[ 4p^2 - 6pq + 9q^2 \][/tex]
### Step 5: Multiply them to verify if it matches
On expanding both binomials:
[tex]\[
(4p^2 - 6pq + 9q^2)(4p^2 + 6pq + 9q^2)
\][/tex]
This results in:
[tex]\[ 16p^4 - 36 p^2 q^2 + 81 q^2 \][/tex]
Thus, factorizing the polynomial, we get:
[tex]\[
16 p^4 + 36 p^2 q^2 + 81 q^4 = (4p^2 - 6pq + 9q^2)(4p^2 + 6pq + 9q^2)
\][/tex]
So, the factorized form of [tex]\( 16 p^4 + 36 p^2 q^2 + 81 q^4 \)[/tex] is:
[tex]\[
\boxed{(4p^2 - 6pq + 9q^2)(4p^2 + 6pq + 9q^2)}
\][/tex]
