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P^2/4+ Q^2/9+36+pq/3+4q+6p

P^2/4+ Q^2/9+36+pq/3+4q+6p-example-1
User QuantIbex
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1 Answer

4 votes

Answer:

The given expression is:

p^2 + 4q^2 + 9 + 36 + pq + 3 + 4q + 6p

Rearranging the terms, we get:

p^2 + 6p + 4q^2 + pq + 4q + 48

Now, we can group the first three terms and the last three terms together as follows:

(p^2 + 6p + 9) + (4q^2 + pq + 4q + 39)

The first group can be factorized as a perfect square trinomial:

(p + 3)^2

The second group can be factorized by grouping the first two terms and the last two terms:

(pq + 4q) + (4q^2 + 39)

We can factor out q from the first two terms, and 4 from the last two terms:

q(p + 4) + 4( q^2 + 9)

Now, we can factor the second group as a sum of squares:

q(p + 4) + 4(q + 3)(q - 3)

Therefore, the fully factorized form of the expression is:

(p + 3)^2 + q(p + 4) + 4(q + 3)(q - 3) + 48

Note that this expression cannot be simplified further.

User Andrew Einhorn
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