Question

the volume of a box V varies with some variable x as V(x)= x^3-12x^2 + 44x - 48 cubic meters. if (x-a) meter is the measurement of one side of the box then find the value of a.​

Answer 1

Answer:

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Answer 2

Given: Volume of bos V, $\[V\left( x \right) = {x^3} - 12{x^2} + 44x - 48\]$ cubic meters.

To find: the value of a.

Solution:

Factorise the $V(x)$.

$\[\begin{array}{l}V\left( x \right) = {x^3} - 12{x^2} + 44x - 48\\ \Rightarrow {x^3} - 12{x^2} + 44x - 48 = {x^2}\left( {x - 2} \right) - 10x\left( {x - 2} \right) + 24\left( {x - 2} \right)\\ \Rightarrow {x^3} - 12{x^2} + 44x - 48 = \left( {x - 2} \right)\left( {{x^2} - 10x + 24} \right)\\ \Rightarrow {x^3} - 12{x^2} + 44x - 48 = \left( {x - 2} \right)\left( {x - 4} \right)\left( {x - 6} \right)\end{array}\]$

Observe that,

$\[V\left( x \right) = \left( {x - 2} \right)\left( {x - 4} \right)\left( {x - 6} \right)\]$

Compare the above expression with $(x-a)$ to find the value of a.

$\[\begin{array}{l}{a_1} = 2,{a_2} = 4,{a_3} = 6\\ \Rightarrow a \in \left\{ {2,4,6} \right\}\end{array}\]$

Hence, the value of a could be $\[\left\{ {2,4,6} \right\}\]$.