Find the set of value of the expression (x-2)(x-4)(x-6)>0
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Answered by
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Solution :-
(x-2)(x-4)(x-6)>0
= (x^2-4x-2x+8)((x-6)
= x^3-6x^2+8x-6x^2+36x-48
= (x^3-12x^2+44x-48)ans
(x-2)(x-4)(x-6)>0
= (x^2-4x-2x+8)((x-6)
= x^3-6x^2+8x-6x^2+36x-48
= (x^3-12x^2+44x-48)ans
mysticd:
It is wrong
how
Where are the set of values ?
Answered by
0
Hi ,
( x - 2 ) ( x - 4 ) ( x - 6 ) > 0
case 1:
x>6 ,
case 2:
x < 2 , x < 4 , x > 6
case 3 :
x < 2 , x > 4 , x < 6
case 4:
x > 2 , x <4 , x < 6
I hope this helps you.
: )
( x - 2 ) ( x - 4 ) ( x - 6 ) > 0
case 1:
x>6 ,
case 2:
x < 2 , x < 4 , x > 6
case 3 :
x < 2 , x > 4 , x < 6
case 4:
x > 2 , x <4 , x < 6
I hope this helps you.
: )
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