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Iranian Mathematical Olympiad

Algebra Difficulty 6.2 National Olympiad Prove it Iran

Consider the second degree polynomial $x^2 + a x + b$ with real coefficients. We know that the necessary and sufficient condition for this polynomial to have roots in real numbers is that its discriminant, $a^2 - 4b$ , be greater than or equal to zero. Note that the discriminant is also a polynomial with variables $a$ and $b$ . Prove that the same story is not true for polynomials of degree 4: Prove that there does not exist a 4 variable polynomial $P(a, b, c, d)$ such that the fourth degree polynomial $x^4 + a x^3 + b x^2 + c x + d$ can be written as the product of four 1st degree polynomials if and only if $P(a, b, c, d) \ge 0$ . (All the coefficients are real numbers.)

Solution

If we put $a = c = 0$ , polynomial $x^4 + b x^2 + d$ can be written as product of four linear terms if and only if quadratic polynomial $y^2 + b y + d$ has two nonnegative roots. Therefore $P(0, b, 0, d) \ge 0$ if and only if $b \le 0$ , $d \ge 0$ and $b^2 - 4d \ge 0$ . For a fixed $b \le 0$ let $Q_b(d) = P(0, b, 0, d)$ . Now, $Q_b(d) \ge 0$ if and only if $0 \le d \le \frac{b^2}{4}$ and hence by continuity of $Q_b$ , $Q_b(b^2/4)$ must be zero. This implies that for all $b \le 0$ , one variable polynomial $P(0, b, 0, \frac{b^2}{4}) = 0$ , and hence this polynomial is always zero. This means that polynomial $x^4 + b x^2 + \frac{b^2}{4}$ has four real roots for all values of $b$ . Contradiction! $\square$

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