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Polynomial

Show that p(x...

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Show that p(x)=x^3 - 3x^2 + 2x - 6 has only one real zero.

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x^{3} + x

has only one real zero. Reason: A polynomial of nth degree must have n real zeroes.

If α, β, γ be the zeros of the polynomial p(x) such that α + β + γ = 3, αβ + βγ + γα = -10 and αβγ =24, then p(x) is

Q. If the equation

x^{3} + p x + q = 0

has three real roots, then show that

4 p^{3} + 27 q^{2} < 0

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$

Find all zeros of the polynomial $P (x) = x^{3} - 3 x^{2} - 6 x + 8$ knowing that x = 1 is a zero of the polynomial.

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