Question

can anyone here help me out with this??????
Find the zeroes of the polynomial $3 x^{2} +x-2$

Answer 1

3x² + x - 2 = 0
⇒ 3x² + 3x - 2x -2 = 0
⇒3x(x+1) - 2(x+1) = 0
⇒(x+1)(3x-2) = 0

Thus 
x+1 = 0                                3x-2 = 0
x = -1                                   3x = 2
                                            x = 2/3

So zeroes of the polynomials are -1 and 2/3

Answer 2

3x²+x-2=0
3x²+3x-2x-2=0
3x(x+1)-2(x+1)=0
(x+1)(3x-2)=0

x+1=0
x=-1

3x-2=0
3x=2
x=2/3