Question

expand using suitable identity (2p+3q+4r)^2​

Answer 1

Answer:

(2p+3q+4r)^2=4p^2+9q^2+16r^2+12pq+24qr+16rp.

Step-by-step explanation:

We know that,(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca.

Using this identity,

(2p+3q+4r)^2=(2p)^2+(3q)^2+(4r)^2+2(2p)(3q)+2(3q)(4r)+2(4r)(2p)

                     =4p^2+9q^2+16r^2+12pq+24qr+16rp.

Hope it helps you..

Answer 2

$(2p + 3q + 4r) +( 2p + 3q+ 4r)$

$2p(2p + 3q+ 4r) + 3q(2p + 3q + 4r) + 4r(2p+ 3q+ 4r)$

$4p^{2} + 6pq + 8pr + 3q(2p+ 3q + 4r) + 4r(2p + 3q + 4r)$

$4p^{2} + 6pq+ 8pr + 6pq + 9q^{2} + 12qr+ 4r (2p + 3q + 4r)$

$4p^{2} + 6pq+ 8pr + 6pq + 9q^{2} + 12qr + 8rp + 12rp + 16r^{2}$

$4p^{2} + (6pq + 6pq) + (8prp + 8pr ) + 9q^{2} + (12qr + 12qr)16r^{2}$

$4p^{2} + 12pq + 16pr + 9q^{2} + 24qr + 16r^{2}$

solve ypur question mate plz thnx.... •••••••