Factor the polynomial completely...?

Question

10(x + y)^2 - 11(x + y) - 6

Answer 1

substituting x+y by 'a' the expression becomes
10a^2-11a-6
=10a^2-15a+4a-6
=5a(2a-3)+2(2a-3)
=(2a-3)(5a+2)
={2(x+y)-3}{5(x+y)+2} [Putting back the value of a}
=(2x+2y-3)(5x+5y+2)

Answer 2

10(x + y)^2 - 11(x + y) - 6

[5(x + y) +2] [2(x + y) -3]

Answer 3

Because (x+y) is like the variable in a quadratic equation, the easiest way to do this is temporarily replace it with a, so a=(x+y)
10a^2-11a-6

Now factor it
(2a-3)(5a+2)

Now, replace the a with (x+y)
[2(x+y)-3][5(x+y)+2]

Answer 4

Let X = x + y
10 X² - 11X - 6
(5X + 2) (2X - 3)
[ 5 (x + y) + 2 ] [ 2 (x + y) - 3 ]

Answer 5

This took me like thirty minutes to figure out:

THE ANSWER IS PIE

{I love Pie}