(3p+2q)(3p-2q)?

Question

I need help solving this math problem.

Answer 1

9p² - 6pq + 6pq - 2q²
9p² - 2q²

Answer 2

(a +b)(a -b) = a^2 - b^2. That rule is called "The Difference of Squares," and it can be proven easily enough by doing the multiplication.

In this case, a = 3p and b = 2q
(3p)^2 = 3^2*p^2 = 9(p^2)
(2q)^2 = 2^2*q^2 = 4(q^2)
So the answer is 9(p^2) - 4(q^2)

Answer 3

This is just FOIL.
You must distribute each term in the first set of parenthesis in the second set of parenthesis.

Essentially, it is as follows:
(3p x 3p) + (3p x -2q) + (2q x 3p) + (2q x -2q), which equals:
9p^2 + -6pq + 6pq + -4q^2, which can be simplified into:
9p^2 - 4q^2 (That Is Your Final Answer)...

Answer 4

Expand and combine like terms:

3p*3p + 3p*-2q + 2q*3p + 2q*-2q
9p^2 - 6pq + 6pq - 4q^2
9p^2 - 4q^2

Notice that the middle terms cancel out. This demonstrates that (a + b)*(a - b) = a^2 - b^2

Answer 5

the answer is 9p^2-4q^2.

Multiply 3p and 3p to give you 9p^2
Multiply 2q and -2q to get -4q^2
3p x -2q to get -6qp
3p x 2q to get 6qp

-6qp and 6qp cancel eachother out since they are opposites. Since 9p^2 and -4q^2 are not like terms, you can only simplify them to 9p^2-4q^2

Answer 6

3p square minus 2q square

Answer 7

The answer is 9p-4q
This is according to a identity,a2-b2=(a+b)(a-b)
So take a=3p & b=2q.Now the ques.isthe RHS side. Therefore put value of a & b in LHS side &square them.Here u r with the answer.

Answer 8

= (3p + 2q) (3p - 2q)
Simplify:
= (3p + 2q) * 3p
= 9p^2 + 6pq

= (3p + 2q) * - 2q
= - 6pq - 4q^2

= 9p^2 + 6pq - 6pq - 4q^2
= 9p^2 - 4q^2

Answer: 9p^2 - 4q^2

Answer 9

=(3p)(3p) + (3p)(-2q) + (2q)(3p) + (2q)(-2q)
=9p^2 -6pq +6pq -4q^2
=9p^2-4q^2

Answer 10

I'm not great with maths, but I think the answer is 9p-4q