5.3 Q-5

Exercise 5.3 — Question 5

Problem Statement

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

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

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

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Key Concepts

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

Polynomial Division: The volume equals the product of all three dimensions: $V(y)=({\text{side}}_1)\times ({\text{side}}_2)\times ({\text{side}}_3)$

Since $(y-2)$ is a known factor, dividing $V(y)$ by $(y-2)$ gives the product of the remaining two dimensions.

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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: $(y-2)$ is a factor. ✓

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Step 2 — Divide $V(y)$ by $(y-2)$

Using synthetic division with $k=2$:

$V(y)÷(y-2)=y^2+0y-1=y^2-1$

So: $V(y)=(y-2)(y^2-1)$

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Step 3 — Factor the Quadratic

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

(Difference of two squares: $a^2-b^2=(a-b)(a+b)$)

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Final Answer

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

The three dimensions of the rectangular solid are:

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

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Verification

Multiply all three factors: $(y+1)(y-1)=y^2-1$ $(y^2-1)(y-2)=y^3-2y^2-y+2✓$