x^2+x-12 find its zeros
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x^2+x-12
x^2+4x-3x-12
x(x+4)-3(x+4)
(x-3)(x+4)
for zeroes
x-3=0, x+4=0
x= 3, x= -4
zeroes are 3,-4
x^2+4x-3x-12
x(x+4)-3(x+4)
(x-3)(x+4)
for zeroes
x-3=0, x+4=0
x= 3, x= -4
zeroes are 3,-4
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