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--> We try at f ( 2 ) to get f(x) = 0 at x = 2 ;
=> f(x) = ( x - 2 )( 12x³ - 32x² + 25x - 6 )
Now, We try at --> x = 2 ; x = 3 and x = 1 , x = 0 , x = -1
We see that the roots are non-negative....
By real root theorem, we have p = 12 , q = -6
One possible root = 6/12 = 1/2
So, we try at 6/12 ---> f(x) = 0
Hence, ( 2x - 1 ) is also a factor
=> f ( x ) = ( x - 2 )( 2x - 1 )( 6x² - 13x + 6 )
=> f ( x ) = ( x - 2 )( 2x - 1 )( 3x - 2 )( 2x - 3 ) √√
Now, for the roots, we have -->
-----> f ( x ) = 0
=> x = 2 || x = 1/2 ||| x = 2/3 |||| x = 3/2 = 1.5 √√
--> We try at f ( 2 ) to get f(x) = 0 at x = 2 ;
=> f(x) = ( x - 2 )( 12x³ - 32x² + 25x - 6 )
Now, We try at --> x = 2 ; x = 3 and x = 1 , x = 0 , x = -1
We see that the roots are non-negative....
By real root theorem, we have p = 12 , q = -6
One possible root = 6/12 = 1/2
So, we try at 6/12 ---> f(x) = 0
Hence, ( 2x - 1 ) is also a factor
=> f ( x ) = ( x - 2 )( 2x - 1 )( 6x² - 13x + 6 )
=> f ( x ) = ( x - 2 )( 2x - 1 )( 3x - 2 )( 2x - 3 ) √√
Now, for the roots, we have -->
-----> f ( x ) = 0
=> x = 2 || x = 1/2 ||| x = 2/3 |||| x = 3/2 = 1.5 √√
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