Question

5. From the sum of 4b2 + 5bc, -2b2 - 2bc - 2z2 and 2bc + 4c2, subtract the sum of 562 - c2 and 362 + 2bc + c2.​

Answer 1

Given :- From the sum of 4b2 + 5bc, -2b2 - 2bc - 2z2 and 2bc + 4c2, subtract the sum of 562 - c2 and 362 + 2bc + c2. ?

Answer :-

→ sum = (4b² + 5bc) + (-2b² - 2bc - 2z²) + (2bc + 4c²)

→ sum = 4b² - 2b² + 5bc - 2bc + 2bc - 2z² + 4c²

→ sum = 2b² + 5bc - 2z² + 4c²

similarly,

→ sum = (5b² - c²) + (3b² + 2bc + c²)

→ sum = 5b² + 3b² - c² + c² + 2bc

→ sum = 8b² + 2bc .

now,

→ from sum of first subtracting second sum we get = (2b² + 5bc - 2z² + 4c²) - (8b² + 2bc) = 2b² - 8b² + 5bc - 2bc - 2z² + 4c² = (-6b²) + 3bc - 2z² + 4c² = (4c² - 6b² - 2z² + 3bc)

Learn more :-

JEE mains Question :-

https://brainly.in/question/22246812

. Find all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

https://brainly.in/question/39026698