Question

I find the Zeroes of the polynomial
p(x) = 6x^2+x-1​

Answer 1

${\tt{\red{\huge{Answer:}}}}$

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 =$\dfrac{1}{3}$ or x =$-\dfrac{1}{2}$

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Answer 2

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