Let f(x) = a12x^12+ a10x^10 + a8x^8 +.... +a2x²+a0, where a0, a2....a12 are all real. If there are 3 sign changes in f(x) and f(x) has exactly 4 non-real roots, then which of the following is definitely true? a0 = 0 a10 > 0 a10 < 0 a10 = 0

f(x)=a12x12+a10x10+a8x8+…+a2x2+a0 � ( � ) = � 12 � 12 + � 10 � 10 + � 8 � 8 + … + � 2 � 2 + � 0 Replace x2 � 2 by y � F(y))=a12y6+a10y5+a8y4+…+a2y+a0 � ( � ) ) = � 12 � 6 + � 10 � 5 + � 8 � 4 + … + � 2 � + � 0 F(y)=0 � ( � ) = 0 has 6 6 roots. since thee are three sign changes, at the most there could be 3 3 positive roots for y. For every negative or complex root of y there are 2 2 complex roots of x � . Sincef(x)=0 � ( � ) = 0 has 4 4 non real zeroes , there could at the most be 2 2 negative or complex roots of y � . Hence it follows there is at least one root y equal to 0. 0. Since a0 is product of all roots ,a0 must be 0 .