Question

factorise (5a+3b-7c)*2 using identity 5 of Algerbic Identities. with steps. ​

Answer 1

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${(5a + 3b - 7c)}^{2} \\ = {(5a + 3b + (- 7c))}^{2} \\ = {(5a}^{2}) + ( {3b}^{2}) + {( - 7c)}^{2} + 2((5a)(3b) + (3b)( - 7c) + ( - 7c)(5a)) \\ = {25a}^{2} + {9b}^{2} + {21c}^{2} + 2(15ab - 21bc - 35ac) \\ = {25a}^{2} + {9b}^{2} + {21c}^{2} + 30ab - 42bc - 70ac$

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Answer 2

AnsweR :-

→ (5a + 3b - 7c)²

→ (5a)² + (3b)² + (-7c)² + 2(5a×3b) + 2(3b×(-7c)) + 2((-7c) × 5a)

→ (25a)² + (9b)² - (49c)² + 30ab - 42bc - 75ca

Procedure :-

$\implies$ As per the required question we have to simplify by using 5th Indentity of Algerbric identites. You can see that the third number is in negetive this will effect in the steps.

$\implies$ First we will square all the three numbers and then we will keep the a,b and c in brackets and then multiply it with 2 which is outside the bracket.

$\implies$ Then after equating the whole expression we will write - (49c)² because 7 is in negetive, and in the brackets one we will convert positive into negetive.

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