Question

The expression
6
x
4
+
19
x
2
+
8
6x
4
+19x
2
+8 can be rewritten as
(
3
x
2
+
a
)
(
2
x
2
+
b
)
(3x
2
+a)(2x
2
+b), where
a
a and
b
b are positive integers, or as
(
3
x
2
+
c
)
(
2
x
2
+
d
)
(3x
2
+c)(2x
2
+d), where
c
c and
d
d are positive nonintegers. What is the value of
a
+
c
a+c?

Answer 1

Let y = x^2. Then 6x^4+19x^2+8 = 6y^2+19y+8 = (3y+a)(2y+b) with ab = 8 and 3b+2a = 19. Testing factor pairs of 8 gives a = 8, b = 1.

For the general factorization (3y+c)(2y+d) we have cd = 8 and 3d+2c = 19. Substitute d = 8/c:
24/c + 2c = 19 → 2c^2 - 19c + 24 = 0 → c = (19 ± 13)/4.
The noninteger positive root is c = 3/2. Thus a + c = 8 + 3/2 = 19/2.