Question

factorise 9(3a+2b-c)^2-4(2a-3b+4c)^2​

Answer 1

Answer:

Rewrite  

9 ( 3 a + 2 b − c ) $x^{2}$ as  ( 3 ( 3 a + 2 b − c ) )$x^{2}$ .

( 3 ( 3 a + 2 b − c ) ) $x^{2}$ − 4 ( 2 a − 3 b + 4 c ) $x^{2}$

Rewrite  

4 ( 2 a − 3 b + 4 c ) $x^{2}$ as  ( 2 ( 2 a − 3 b + 4 c ) ) $x^{2}$.

( 3 ( 3 a + 2 b − c ) ) $x^{2}$ − ( 2 ( 2 a − 3 b + 4 c ) ) $x^{2}$

Since both terms are perfect squares, factor using the difference of squares formula,  

a $x^{2}$− b $x^{2}$= ( a + b ) ( a − b )  where  a = 3 ( 3 a + 2 b − c )  and  b = 2 ( 2 a − 3 b + 4 c ) .

( 3 ( 3 a + 2 b − c ) + 2 ( 2 a − 3 b + 4 c ) )  ( 3 ( 3 a + 2 b − c ) − ( 2 ( 2 a − 3 b + 4 c ) ) )

Simplify.

( 13 a + 5 c ) ( 5 a + 12 b − 11 c )

Step-by-step explanation:

Oh and my reason is in the answer hoped this helped you out buddy