Answer
$3x^2+11xy-4y^2 = (3x-y)(x+4y)$
Work Step by Step
Let's consider a trinomial in this form: $~ax^2+bxy+cy^2$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$
Then the next step is to write the trinomial as follows:
$~x^2+bxy+cy^2 = ax^2+rxy+sxy+cy^2$
Let's consider this trinomial: $~3x^2+11xy-4y^2$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 11$ and $r\times s = -12$. We can see that $(12)+(-1) = 11~$ and $(12)\times (-1) = -12$
We can factor the trinomial as follows:
$3x^2+11xy-4y^2 = 3x^2+12xy-xy-4y^2$
$3x^2+11xy-4y^2 = (3x^2+12xy)+(-xy-4y^2)$
$3x^2+11xy-4y^2 = (3x)(x+4y)+(-y)(x+4y)$
$3x^2+11xy-4y^2 = (3x-y)(x+4y)$
