Question

expand each of the following using suitable identities
$(4a + 2b - 3c) {}^{2}$
$(5 + a) {}^{2}$
$(2a - 3b + 5) {}^{2}$
please answer this question

Answer 1

Hello dear,

(4a + 2b - 3c)² we can expand by using the identity =>
(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

So,
(4a + 2b - 3c)² = (4a)² + (2b)² + (-3c)² + 2 [ 4a*2b + 2b*(-3c) + (-3c)*4a ]
= 16a² + 4b² + 9c² + 2 [ 8ab - 6bc - 12ca]
= 16a² + 4b² + 9c² + 16ab - 12bc - 24ca

Now,
(5+a)² we can expand by using the identity
=> (a+b)² = a² + b² + 2ab
So,
(5+a)² =(5)² + (a)² + 2*5*a
= 25 + a² + 10a

Now,
(2a - 3b + 5)² we can expand by using the identity
=> (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

So,
(2a - 3b + 5)² = (2a)² + (-3b)² + (5)² + 2 [ 2a*(-3b) + (-3b)*5 + 5*2a ]
= 4a² + 9b² + 25 + 2 [ (-6ab) + (-15b) + 10a ]
= 4a² + 9b² + 25 - 12ab - 30b + 20a

Hope it helps.

Answer 2

hey!!!

your answer is here...

given :-

1. ( 4a + 2b - 3c )²

we will use this identity - ( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac

a = 4a, b = 2b and c = -3c

>> (4a)² + (2b)² + (-3c)² + 2 (4a) (2b) + 2 (2b) (-3c) + 2 (4a) ( -3c)
= 16a² + 4b² + 9c² + 16ab - 12bc - 23ac

2. ( 5 + a )²

here we will use first identity that is - ( a + b )² = a² + 2ab + b²

a = 5 and b = a

>> (5)² + 2 (5) (a) + (a)²
=> 25 + 10a + a²

3. ( 2a - 3b + 5 )²

we will use the same identity which I used in ( 4a + 2b - 3c )² that is - ( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac

a = 2a, b = -3b and c = 5

>> (2a)² + (-3b)² + (5)² + 2 (2a) (-3b) + 2 (-3b) (5) + 2 (2a) (5)
= 4a² + 9b² + 25 - 12ab - 30b + 20a

cheers!!!