Factor: 2xy + 6x - 4y - 12
a)(2x + 4) (y + 3)
b)(2x - 4) (y + 3)^2
c)(2x - 4) (y - 3)
d)(2x - 4) (y + 3)
What are the factors of the following trinomial?
16y^2 - 12y + 2
a) (4y - 1) (4y - 2)
b) (2y - 1) (8y + 2)
c) (16y - 1) (y - 2)
d) (8y - 1) (2y - 2)
i think a but i can be wrong
See http://www.jiskha.com/display.cgi?id=1275520227 for the first post.
For the second post, the choice is correct.
(4y-1)(4y-2)
=16y² - 4y - 2(4y) + (-1)(-2)
=16y² -12y +2
To factor the expression 2xy + 6x - 4y - 12, we can look for common terms that can be factored out from each term in the expression.
First, let's find the common factor for the first two terms, 2xy and 6x. The highest common factor between 2xy and 6x is 2x. We can factor out 2x from these terms:
2xy + 6x = 2x(y + 3)
Now let's find the common factor for the last two terms, -4y and -12. The highest common factor between -4y and -12 is -4. We can factor out -4 from these terms:
-4y - 12 = -4(y + 3)
Combining the factored terms, we have:
2xy + 6x - 4y - 12 = 2x(y + 3) - 4(y + 3)
Notice that we now have a common factor of (y + 3) in both terms. We can factor out (y + 3) from these terms:
2xy + 6x - 4y - 12 = (2x - 4)(y + 3)
Therefore, the correct factorization of the expression 2xy + 6x - 4y - 12 is (2x - 4)(y + 3). So the answer is option (d).