simple factorisation
if
f
(
x
)
=
6
4
x
−
6
4
−
x
and it can be factorised into the form
f
(
x
)
=
(
a
x
+
a
−
x
)
(
(
a
x
+
a
−
x
)
2
+
c
)
,
find a .
The answer is 4.
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2 solutions
A^3-B^3=(A-B)(A^2+AB+B^2) f(x)= 4^3x - 4^-3x =(4^x + 4^-x)(4^2x + 4^x * 4^-x + 4^-2x) = (4^x + 4^-x)(4^2x + 1 + 4^-2x) = (4^x + 4^-x)(4^2x + 2 + 4^-2x -1) = (4^x + 4^-x)((4^x + 4^-x)^2 -1) thus a=4 c=-1
Your first factor is ( 4 x + 4 − x ) . It should be ( 4 x − 4 − x ) since you specified that A 3 − B 3 = ( A − B ) ( A 2 + A B + B 2 )
Also, in your question, it seems f ( x ) should be ( a x − a − x ) ( ( a x + a − x ) 2 + c )
The factor form of f(x) in the question is wrong.