Expansion of (2a+3b+4c)2 is ___________
4a2+9b2+16c2+12ab+24bc+16ca
a2+b2+2ca
a2-b2-2ca
a2-b2+2ca
Explanation for the correct otpion:
Expanding the given expression.
We know,
(x+y+z)2=x2+y2+z2+2xy+2yz+2zx
Here,
x=2a
y=3b
z=4c
Substituting the values in the above formula, we get,
⇒(2a+3b+4c)2=(2a)2+(3b)2+(4c)2+2(2a)(3b)+2(3b)(4c)+2(4c)(2a)
On further calculation, we get,
⇒4a2+9b2+16c2+12ab+24bc+16ca
Therefore, the correct option is A
Factorize the following expression:
(2a+3b)2+2(2a+3b)(2a-3b)+(2a-3b)2
If x=2a+3b+2a-3b2a+3b-2a-3b, then show that 3bx2-4ax+3b=0.