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))
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))
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_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
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,
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))