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This is a Maths Algebraic Expressions 8th Standard ICSE BOARD question.
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Answer 1

Answer:

2a+3b-5c

Step-by-step explanation:

=(4a²+12ab+9b²-25c²)/2a+3b+5c

=[{(2a)²+(2*2*3ab)+(3b)²}-(5c)²]/2a+3b+5c

=[(2a+3b)²-(5c)²}/2a+3b+5c (by formula: (a+b)²= a²+b²+2ab)

={(2a+3b-5c)(2a+3b+5c)}/2a+3b+5c (by formula:a²-b²=(a+b)(a-b)

=2a+3b-5c

Answer 2

Solution:

                     

2a + 3b + 5c ) 4a² + 12ab + 9b² - 25c² ( 2a + 3b - 5c

                   {-}     4a² + 6ab                      + 10ac

                                      6ab + 9b² - 25c² - 10ac

                            {-}       6ab + 9b²                       + 15bc

                                                       - 25c² - 10ac - 15bc

                                           {-}         - 25c² - 10ac - 15bc

                                                                     0

Therefore,

• Quotient ➝ 2a + 3b - 5c
• Remainder ➝ 0

EXTRAS:

We considered -10ac and -15bc because while subtracting the below part from the above part, we change the signs. The same is applied here.

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