I find the Zeroes of the polynomial
p(x) = 6x^2+x-1
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6x² + x - 1 = 0
6x² + 3x - 2x - 1 = 0
3x (2x + 1) - 1 (2x + 1) = 0
(3x - 1) = 0 or (2x + 1) = 0
3x = 1 or 2x = - 1
x = or x =
_________________________
Answered by
1
Answer:
P(0) = 6x² + x - 1
6x² + x - 1 = 0
6x² + 3x - 2x - 1
3x(2x + 1) - 1(2x + 1)
(3x - 1)(2x +1)
Therefore x = 1/3, - 1/2
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