Answer:
The binomial that completes the factorization is 3p + 4q.
Explanation:
We can start by recognizing that the given expression has two terms, which are both perfect cubes:
27p^4 is the cube of (3p)^2, and
64pq^3 is the cube of (4q)^2.
Therefore, we can use the formula for the sum of two cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
to factorize the given expression:
27p^4 + 64pq^3 = (3p)^2 + (4q)^2)(9p^2 - 12pq + 16q^2)
Now, we want to find the binomial that completes the factorization. We notice that the first factor (3p + 4q)(9p^2 - 12pq + 16q^2) is not a binomial, but a product of two binomials. So we focus on the second factor, which is a trinomial. We want to see if we can factor it further.
We can try to factor out a common factor of 3 from the first two terms and a common factor of 4 from the last two terms:
9p^2 - 12pq + 16q^2 = 3(3p^2 - 4pq) + 4(4q^2)
Now we can see that we have the difference of two squares:
3p^2 - 4pq = (p(3p - 4q))
4q^2 = (2q)^2
So we can write:
9p^2 - 12pq + 16q^2 = 3p(3p - 4q) + 4q^2
= (3p - 4q)(3p + 4q) + 4q^2
Now we can substitute this back into our original expression:
27p^4 + 64pq^3 = (3p + 4q)((3p)^2 - 3p(4q) + (4q)^2) + 4q^2
= (3p + 4q)(9p^2 - 12pq + 16q^2) + 4q^2
So the binomial that completes the factorization is 3p + 4q.