Question

como resolver (2u+3) (u-4)+ 4u(u-2)

Answer 1

(2u x u)+(2u x -4)+(3 x u)+(3 x -4) = 2u al cuadrado - 5u - 12
(4u x u)+(4u x -2) = 4u al cuadrado - 8u
rpta = 6u al cuadrado - 13u - 12

Answer 2

(2u+3) (u-4)+ 4u(u-2)

$2u^{2} -8u+3u -12 + 4 u^{2} -8$

$6u^{2} -5u-20$