jika( 2×+3y)(p×+qy)=r×^2+23×y+12y^2 maka nilai r adl

(2x + 3y) (px + qy) = 2px^2 + 2qxy + 3pxy = 3qy^2
= 2px^2 + (2q + 3p) xy + 3qy^2 = rx^2 + 23xy + 12y^2
maka
#3q = 12 <=> q = 4
#(2q + 3p) = 23
2(4) + 3p = 23
3p = 15
p = 5

#2p = r
2(5) = r
r = 10.

(2x+3y)(px+qy) = rx2 + 23xy + 12y2
2px2+2qxy+3pxy+3qy2 = rx2 + 23xy + 12y2
2px2 + (2q+3p)xy + 3qy2 = rx2 +23xy + 12y2
r=2p


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*tolong jadikan jawaban terbaik ya, plis