Chapter MPS, Problem 6.12MPS

To determine

To Find: The dimension of the cube

Answer to Problem 6.12MPS

  $\begin{array}{l}\text{width}=5\\ \text{length}=w-1=4\\ \text{Height}=w+3=8\end{array}$

Explanation of Solution

Given:

  $Zeros=-1\&6-3i$

Formula Used:

  $Volume=length\times breadth\times height$

  $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}\text{Width of solid}=w\\ \text{Length of solid}=w-1\\ \text{Height of the solid}=w+3\\ \text{Volume of a cube}=160i{n}^3\\ Volume=length\times breadth\times height\\ \text{⇒}w\left(w-1\right)\left(w+3\right)=160\\ \text{⇒}\left(w^2-w\right)\left(w+3\right)=160\\ \text{⇒}{w}^3-w^2+3w^2-3w=160\\ \text{⇒}{w}^3+2w^2-3w-160=0\\ factorof160=2\&5\\ \text{then zeroes may be -2,+2,-5\&+5}\\ \text{putting 5 we get the polynomial to be zero}\\ \text{⇒}\frac{w^3+-163w+160w-160}{w-5}=w^2+7w+32\\ \text{⇒}{w}^2+7w+32=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=1\\ b=7\\ c=32\\ w=\frac{-7\pm \sqrt{49-128}}{2}\\ \text{⇒}w=\frac{-7\pm i\sqrt{79}}{2}\\ \text{⇒}w=\frac{-7-i\sqrt{79}}{2},\frac{-7+i\sqrt{79}}{2}or5\\ \text{since, dimension cannot of imaginary }\\ \text{so, }\\ \text{width}=5\\ \text{length}=w-1=4\\ \text{Height}=w+3=8\end{array}$

Conclusion:

Thus the dimension of the box is

  $\begin{array}{l}\text{width}=5\\ \text{length}=w-1=4\\ \text{Height}=w+3=8\end{array}$ .