Answer
$2ax^2+3ax-35a = (a)~(2x-7)~(x+5)$
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: $~2ax^2+3ax-35a$
First we can factor by using the GCF:
$2ax^2+3ax-35a = (a)~(2x^2+3x-35)$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 3$ and $r\times s = -70$. We can see that $(10)+(-7) = 3~$ and $(10)\times (-7) = -70$
We can factor the trinomial as follows:
$2ax^2+3ax-35a = (a)~(2x^2+3x-35)$
$2ax^2+3ax-35a = (a)~(2x^2+10x-7x-35)$
$2ax^2+3ax-35a = (a)~[~(2x^2+10x)+(-7x-35)~]$
$2ax^2+3ax-35a = (a)~[~(2x)(x+5)+(-7)(x+5)~]$
$2ax^2+3ax-35a = (a)~(2x-7)~(x+5)$
