Question

factorise (4a-2b-3c)²​

Answer 1

Answer:

The given expression (4a−2b−3c)

2

can be solved as shown below:

(4a−2b−3c)

2

=(4a)

2

+(−2b)

2

+(−3c)

2

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

(∵(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca)

=16a

2

+4b

2

+9c

2

−16ab+12bc−24ca

Hence, (4a−2b−3c)

2

=16a

2

+4b

2

+9c

2

−16ab+12bc−24ca.

Step-by-step explanation:

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