b(c-d)²-a(d-c) +5c-5d
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given:-
b(c-d)2-a(d-c)+5c-5d
b (c-d)2-a(d-c)+5(c+d)
now here we can write -a(d-c) as a(c-d)
b(c-d)2+a(c-d)+5(c-d)
take (c-d) common
(c-d)[b(c-d)+a+5]
(c-d)[bc-bd +a+5]
hope helps
given:-
b(c-d)2-a(d-c)+5c-5d
b (c-d)2-a(d-c)+5(c+d)
now here we can write -a(d-c) as a(c-d)
b(c-d)2+a(c-d)+5(c-d)
take (c-d) common
(c-d)[b(c-d)+a+5]
(c-d)[bc-bd +a+5]
hope helps
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The person who answered before me's answer is correct. I am Confirming it.
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