Help with factoring an equation?

Question

4p^2 + 4pq - 3q^2

Answer 1

It's the old FOIL gambit .
This equation contains variables to the power 2. It therefore has 2 factors, no more, no less
The 2 factors are going to look like this:(-----)(-----)
We can begin to fill in the brackets, by this
(?p-----3q)(?p-----q). We do this because the "L" in FOIL stands for "multiply last terms". Your last term is
-3q^2, and the only factors of 3 are 3 and 1,so we have to use 3q and q

Now the "F" in foil stands for "multiply first terms". Your first term is 4p^2, and the factors of 4 are 4 and 1, or 2 and 2. This gives us three options:
(4p----3q)(p----q)
(p----3q)(4p----q)
(2p----3q)(2p----q)

The "O" and "I" in FOIL stand for "multiply outside terms and inside terms to get the final middle term".
Our job is to pick one of the three options which, when accompanied by the appropriate signs, gives us +4pq when "OI" is done, AND +4p^2 for "F", AND
-3q^2 for "L".

OptionA) (4p----3q)(p----q) NO. I can't get +4pq out of this combination.
OptionB) (p----3q)(4p----q) No. Same reason.
OptionC)(2p----3q)(2p----q) Oh yesss!

(2p+3q)(2p-q) WORKS "F" is 4p^2, "O" is -2pq,
"I" is +6pq, "O" and "I" adding to +4pq, and "L" gives
-3q^2

Your factors are (2p+3q)(2p-q).
Hope this helps

Answer 2

4p^2 + 4pq - 3q^2 can be written as:

2p + 3q and 2p - q since the pq term can be written as 6 - 2.

Answer 3

4p^2 + 4pq - 3q^2

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


FOIL back out to make sure it's correct...

4p^2 + 6pq - 2pq - 3q^2
4p^2 + 4pq - 3q^2 check.

Answer 4

Hey there!

Use grouping to factor the problem.

4p^2+4pq-3q^2 -->
4p^2+6pq-2pq-3q^2 -->
2p(2p+3q)-1q(2p+3q) -->
(2p-1q)(2p+3q) -->
(2p-q)(2p+3q)

So the answer is (2p-q)(2p+3q).

Hope it helps!