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Answer 1

Hey Mate !

Here is your solution :

We have to find the maximum measure which can measure both the quantities.

I.e , we have to find its H.C.F.

Let's do it !

For container 1,

= 2x³ + 2x² + 3x + 3

By taking 2x² common from first 2 terms and 3 from last 2 terms.

= 2x² ( x + 1 ) + 3 ( x + 1 )

By taking out ( x + 1 ) as common ,

= ( x + 1 ) ( 2x² + 3 )

Now for container 2 ,

= 4x³ - 2x² + 6x - 3

By taking 2x² as common from first 2 terms and 3 from last two terms,

= 2x² ( 2x - 1 ) + 3 ( 2x - 1 )

Taking out ( 2x -1 ) as common,

= ( 2x - 1 ) ( 2x² + 3 )

Now,

=> 2x³ + 2x² + 3x + 3 = ( x + 1 ) ( 2x² + 3 )

=> 4x³ - 2x² + 6x - 3 = ( 2x - 1 ) ( 2x² + 3 )

Hence , the required answer is ( 2x² + 3 ).

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Hope it helps !! ^_^