Find the zeroes of $p (y) = y^{3} - 6 y^{2} + 11 y - 6$

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Solution

$y^{3} - 6 y^{2} + 11 y - 6 = 0$

$y^{3} - 3 y^{2} - 3 y^{2} + 9 y + 2 y - 6 = 0$

$(y^{2} - 3 y + 2) (y - 3) = 0$

$(y - 1) (y - 2) (y - 3) = 0$

$∴ y = 1, 2, 3$

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Similar questions

If $p (y) = 2 y^{3} - 6 y^{2} - 5 y + 7$ then find $p (2)$ .

P (y) = (y^{3} - 27) (y^{2} - 3 y + 2)

Q (y) = (y^{2} + 3 y + 9) (y^{2} - 5 y + 6)

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y^{3} - 6 y^{2} + 6 y + 1; y - 1

p (0), p (1)

p (2)

p (y) = 2 + y + 2 y^{2} - y^{3}

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