Question

find the possible dimensions for the cuboids whose volume is
1). x3 - 25x
2). 6Kt2 - 21Kt - 12K

Answer 1

we can write ...
1)
=) x^3-25x = x(x-5)(x+5)

=) volume of cuboid = lbh

so dimensions are x , x-5 , x+5 ...ans

2)

6kt^2 - 21kt -12k

taking out common 3k

3k(2t^2 - 7t - 4)

= 3k(2t^2 - 8t +t -4)

= 3k[ 2t(t - 4) +1(t-4)]

= 3k(t-4)(2t+1)

dimensions are 3k , t-4 , 2t+1

__________________ hope it help u

Answer 2

The possible dimensions are

(i) $\bf x.(x - 5).(x + 5)$ for cuboid; having volume $\bf V= {x}^{3} - 25x$

and

(ii)$\bf 3K.(2t + 1).(t - 4)$ for cuboid; having volume $\bf V = 6K {t}^{2} - 21Kt - 12K$

Given:

• Cuboids whose volume is
• 1)x³ - 25x
• 2)6Kt² - 21Kt - 12K

To find:

• Find the possible dimensions for the cuboids.

Solution:

Formula/Concept to be used:

1. Volume of cuboid= length×breadth×height
2. Factorise the polynomial.

Step 1:

Factorise the cubic polynomial for volume of first cube.

$V = {x}^{3} - 25x$

take x common

$V = x( {x}^{2} - 25)$

Apply identity

$\bf ( {a}^{2} - {b}^{2} ) = (a - b)(a + b)$

So,

$V = x( {x}^{2} - {5}^{2} )$

or

$\bf V = x(x - 5)(x + 5)$

As

$\text{\bf length . breadth. height} = x(x - 5)(x + 5)$

(can be taken in any order)

Thus,

Dimensions of cuboid are x, (x-5) and (x+5).

Step 2:

Factorise the volume expression of second cuboid.

$V = 6K {t}^{2} - 21Kt - 12K$

take 3K common

$V = 3K(2 {t}^{2} - 7t - 4)$

Factorise the quadratic polynomial.

$V = 3K(2 {t}^{2} - 8t + t - 4)$

or

$V = 3K(2t( {t} - 4) + 1( t - 4))$

or

$V = 3K(2t + 1)( {t} - 4)$

As

$\text{\bf length . breadth. height} =3K(2t + 1)(t - 4)$

(can be taken in any order)

Thus,

Dimensions of cuboid are 3K, (2t+1) and (t-4).

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