Question 10, Exercise 5.1

Solutions of Question 10 of Exercise 5.1 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.

A rectangular room has a volume of $\left(x^{3}+11 x^{2}+34 x+24\right)$ cubic feet. The height of the room is $(x+1)$ feet. Find the area of its floor.

Solution.

Suppose $p(x)=x^{3}+11 x^{2}+34 x+24$.

By using synthetic division: \begin{align} \begin{array}{r|rrrr} -1 & 1 & 11 & 34 & 24 \\ & \downarrow & -1 & -10 & -24 \\ \hline & 1 & 10 & 24 & 0 \\ \end{array}\end{align}

This gives $$ p(x) = (x+1)(x^2+10x+24)$$

We have volume of room = area of floor $\times$ height.

$\implies x^{3}+11 x^{2}+34 x+24=(x^2+10x+24)(x+1)$

Hence area of floor = $x^2+10x+24$ square feet.

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