Question

16(4a-3b) whole square + 4(4a-3b) is factorised then the factors are​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Solution:-

=)16(4a-3b)² + 4(4a-3b)

=) 4² ( 4a-3b)² + 4(4a-3b)

=) [ 4(4a-3b)]² + 4(4a-3b)

Taking 4(4a-3b) as common. we get,

=) 4( 4a-3b) [ 4(4a-3b) + 1]

=) 4(4a-3b)[ 16a - 12b + 1]

Hence Solved!