5.3 Q-6

Exercise 5.3 — Question 6

Problem Statement

The volume of a rectangular solid is given by the polynomial:

$V(y)=y^3-2y^2-y+2$

If one dimension of the solid is $(y-2)$, find the remaining two dimensions.

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Key Concepts

Factor Theorem: If $(y-a)$ is a factor of $P(y)$, then $P(a)=0$.

Polynomial Division: Since Volume = length × width × height, dividing $V(y)$ by the known factor $(y-2)$ gives the product of the remaining two dimensions.

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Solution

Step 1: Verify $(y-2)$ is a factor

By the Factor Theorem, substitute $y=2$:

$V(2)=(2{)}^3-2(2{)}^2-(2)+2=8-8-2+2=0$

Since $V(2)=0$, confirmed that $(y-2)$ is a factor. ✓

Step 2: Divide $V(y)$ by $(y-2)$ using synthetic division

Coefficients of $V(y)$: $1,-2,-1,2$

Using $y=2$:

$\begin{array}{rrrrr}2& 1& -2& -1& 2\\ & & 2& 0& -2\\ & 1& 0& -1& 0\end{array}$

Quotient: $y^2+0y-1=y^2-1$

Remainder: $0$ ✓

Step 3: Factor the quadratic quotient

$y^2-1=(y-1)(y+1)$

Step 4: Write the complete factorization

$V(y)=(y-2)(y-1)(y+1)$

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Answer

The three dimensions of the rectangular solid are:

$\boxed{(y-2),(y-1),(y+1)}$

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