5.3 Q-2

Exercise 5.3 — Question 2

Background: Factor Theorem

The Factor Theorem states that a polynomial $P(x)$ has $(x-a)$ as a factor if and only if $P(a)=0$.

This is a direct consequence of the Remainder Theorem: when $P(x)$ is divided by $(x-a)$, the remainder is $P(a)$. If $P(a)=0$, the remainder is zero, so $(x-a)$ divides $P(x)$ exactly.

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Question 2

The volume of a rectangular solid is given by:

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

One dimension (side length) of the solid is $(y-2)$. Find the remaining two dimensions.

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Step 1 — Verify $(y-2)$ is a Factor

By the Factor Theorem, check $V(2)$:

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

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

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Step 2 — Divide $V(y)$ by $(y-2)$

Use synthetic division with $k=2$:

The quotient is $y^2+0y-1=y^2-1$, with remainder $0$.

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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