Answer
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(3a-2b)~(2a+5b)$
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: $~6a^2c^2+11abc^2-10b^2c^2$
First we can factor by using the GCF:
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+11ab-10b^2)$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 11$ and $r\times s = -60$. We can see that $(15)+(-4) = 11~$ and $(15)\times (-4) = -60$
We can factor the trinomial as follows:
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+11ab-10b^2)$
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(6a^2+15ab-4ab-10b^2)$
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~[~(6a^2+15ab)+(-4ab-10b^2)~]$
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~[~(3a)(2a+5b)+(-2b)(2a+5b)~]$
$6a^2c^2+11abc^2-10b^2c^2 = (c^2)~(3a-2b)~(2a+5b)$
