Answer
-3,-1/3, 2, 5
Work Step by Step
(x^2+x-6)(3x^2-14x-5)=0
let each part equal to zero separately:
(x^2+x-6)=0
(3x^2-14x-5)=0
Lets do the first part first
(x^2+x-6)=0
Using ac method ac=-6 b= 1
two numbers are 3 and -2 then
x^2+3x-2x-6=0
x(x+3)-2(x+3)=0
(x-2)(x+3)=0
x-2=0 x=2
x+3=0 x=-3
Then let's do the second part
(3x^2-14x-5)=0
Using ac method ac=-15 b= -14
two numbers are -15 and 1 then
(3x^2-15x+x-5)=0
3x(x-5)+(x-5)=0
(3x+1)(x-5)=0
3x+1=0 3x=-1 x=-1/3
x-5=0 x=5
