Question

3(2x-y) ^2+14(2x-y) +8​

Answer 1

$\textbf{\huge{YOUR ANSWER:}}$

Let the term :-

(2x - y) be = z

Now, the previous equation, which was given to us becomes :-

$3z^{2} + 14z + 8$

By using the splitting middle term method, we get the result as :-

$3z^{2} + 12z + 2z + 8$

Some more steps to go for the result

=》 3z ( z + 4 ) + 2 ( z + 4 )

Last one step to go for the answer

=》 ( 3z + 2 ) ( z + 4 )

There's your final equation !

Now, put the value back of ( z ), which we kept previously :-

=》 ( 3 ( 2x - y ) + 2 ) ( 2x - y + 4 )

Do the required operations, and then, get the required answer :-

=》 ( 6x - 3y + 2 ) ( 2x - y + 4 )

There's your final answer !

Answer 2

Given :

$\small \tt{3(2x - y) {}^{2} + 14(2x - y) + 8 = 0}$

Here, (2x - y) is common in the equation.

Thus, Let (2x - y) be 'm'... (1)

So, The equation formed is as follows :

$\tt{3m {}^{2} + 14m + 8 = 0}$

Now, Simplifying the equation by splitting middle term.

$\tt{3m {}^{2} + 12m + 2m + 8 = 0}$

$\tt{3m(m + 4) + 2(m + 4) = 0}$

$\tt{(3m + 2)(m + 4) = 0}$

Now,

From (1),
Putting the value of m in the equation, we get :

(3( 2x - y ) + 2 )( 2x - y + 4 )

Just one more step to go !
Opening the brackets, we get :

(6x - 3y + 2)(2x - y +4 ) is the solution of the question.

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