Question

Let p(x) be a cubic polynomial with integer coefficients with leading coefficient 1 and with one of its roots equal to the product of the other two. Show that 2p(−1) is a multiple of p(1) + p(−1) − 2(1 + p(0))

Collected in the board: Quartic Equations

Steven Zheng posted 1 hour ago

Answer

Let p(x) = x^3 -ax^2 + bx - c and r_1 = r_2r_3

where r_1,r_2,r_3 are the three roots of the equation.

Apply Vieta's formula

r_1+r_2+r_3 = a
(1)
r_1r_2+r_2r_3+r_1r_3=b
(2)

r_1 r_2r_3=c

Since r_1 = r_2r_3, we get r_1 = \sqrt{c} . Since r_1 is an integer, c must be a perfect square.

Substitute the value of r_1 to (2), solve for r_2+r_3

r_2+r_3 = \dfrac{b}{\sqrt{c} }-1
(3)

Substitute to (1), we get

a-\sqrt{c}= \dfrac{b}{\sqrt{c} }-1

Multiply each term ny \sqrt{c} to eliminate denominator

a\sqrt{c}-c = b-\sqrt{c}

Then,

\dfrac{b+c}{a+1} = \sqrt{c}
(4)

On the other hand,

p(−1) = -1-a-b-c

p(1) = 1-a+b-c

p(0) = -c

p(1) + p(−1) − 2(1 + p(0))

=-2a-2

2p(−1) = -2-2a-2b-2c

\dfrac{2p(−1) }{p(1) + p(−1) − 2(1 + p(0))}

=\dfrac{-2-2a-2b-2c}{-2a-2}

=1+\dfrac{b+c}{a+1}

Substituting the result of (4) showa

\dfrac{2p(−1) }{p(1) + p(−1) − 2(1 + p(0))} = 1+\sqrt{c}

Therefore, 2p(−1) is a multiple of p(1) + p(−1) − 2(1 + p(0))

Steven Zheng posted 1 hour ago

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