Question

Find the value of – 4x2y2 + 6xy + 8x + 12y-1,
when x = -2, y=-1.​

Answer 1

$4 {x}^{2} {y}^{2} + 6xy + 8x + 12y - 1 \\ x = - 2 \: \: \: and \: \: \: y = - 1$

$= 4 {( - 2)}^{2} {( - 1)}^{2} + 6( - 2)( - 1) + \\ 8( - 2) + 12( - 1) - 1 \\ \\ = 4(4)(1) + 6(2) - 16 - 12 - 1 \\ \\ = 16 + 12 - 28 - 1 \\ = 28 - 29 \\ = - 1$

Answer 2

-4(-2)^2(-1)^2+6(-2)-(-1)+(-2)+12(-1)-1

-4(-4)(1)+6(-2)-(-1)+8(-2)+12(-1)-1

16(1)-12(1)-16-12-1

-24-1

-25