Question 6, Exercise 5.3

Solutions of Question 6 of Exercise 5.3 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

The volume of box is $y^3-2y^2-y+2$. If the length of one side $y-2$, find the length of the other two sides.

Solution.

Let volume of box $=p(y)=y^3-2y^2-y+2$

Length of one side = $y-2$.

This gives $y-2$ is factor of $p(y)$. Thus $2$ is zeros of $p(y)$

Using synthetic division: \[ \begin{array}{r|rrrr} 2 & 1 & -2 & -1 & 2 \\ & \downarrow & 2 & 0 & -2 \\ \hline & 1 & 0 & -1 & 0 \\ \end{array} \] Thus \begin{align*} p(y) & = (y-2)(y^2-1) \\ & = (y-2)(y+1)(y-1) \end{align*} Hence the length of remaining two sides are $y+1$ and $y-1$.

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