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

.