Question

If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?

Answer 1

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$.

Step-by-step explanation:

The given is,

                        $2 x^{3} + 4 x^{2} -30x$................................(1)

Step:1

    Check for option A,

             x = 1, dimensions 8 by 9 by 1  

            From the equation (1),

                      Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

                                    $=2+4-30$ = -24...................(2)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                     = 72............................................(3)

            From equation (2) and (3)

                                -24 ≠ 72

            So, X=1; dimensions 8 by 9 by 1 is not possible.

   Check for option B,

             x = 1, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

                                    $=2+4-30$ = -24...................(4)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30.........................................(5)

            From equation (4) and (5)

                                -24 ≠ 30

            So, X=1; dimensions 2 by 5 by 3 is not possible.

   Check for option C,

            x = 4, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

                                    $=2(64)+4(16)-30(4)$

                                    $= 128+64-120$

                                    = 72.............................................(6)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30............................................(7)

            From equation (6) and (7)

                               72 ≠ 30

            So, X=4; dimensions 2 by 5 by 3 is not possible.

    Check for option C,

            x = 4, dimensions 8 by 9 by 1

            From the equation (1),

                      Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

                                    $=2(64)+4(16)-30(4)$

                                    $= 128+64-120$

                                    = 72............................................(8)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                    = 72............................................(9)

            From equation (8) and (9)

                               72 = 72

            So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

           The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$.