Question

2(2x² + xy-z²)-xy. when x=2, y = (-1) and z = (-2)​

Answer 1

Explanation:

p = 2(2x² + xy - z²) - xy

p = 2[ 2×(2)² + 2×(-1) - (-2)²] - 2×(-1)

p = 2[ 2×4 - 2 - 4 ] + 2

p = 2[ 8 - 6 ] + 2

p = 2[ 2 ] + 2

p = 4 + 2

p = 6

NOTE :

1. (+) × (+) = (+)
2. (+) × (-) = (-)
3. (-) × (-) = (+)
4. (-) × (+) = (-)

NOTE :-

mathematical expressions with multiple operators need to be solved from left to right in the order of BODMAS.

Answer 2

6

Explanation:

substitute the given values into the expression

2(2(2)² + 2(- 1) - (- 2)² ) - 2(- 1) ← evaluate parenthesis

= 2(2(4) - 2 - 4) + 2

= 2(8 - 2 - 4) + 2

= 2(2) + 2

= 4 + 2

= 6