# Appendix A - A.4 - Quadratic Equations - Exercises - Page 556: 21

$2ax^2+3ax-35a = (a)~(2x-7)~(x+5)$

#### 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: $2ax^2+3ax-35a = (a)~(2x^2+3x-35)$ $2ax^2+3ax-35a = (a)~(2x^2+10x-7x-35)$ $2ax^2+3ax-35a = (a)~[~(2x^2+10x)+(-7x-35)~]$ $2ax^2+3ax-35a = (a)~[~(2x)(x+5)+(-7)(x+5)~]$ $2ax^2+3ax-35a = (a)~(2x-7)~(x+5)$

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