Exercise 5.3 — Question 3

Topic: Polynomial Division, Remainder Theorem & Factor Theorem

---

Key Concepts

Remainder Theorem

If a polynomial $P(x)$ is divided by a linear factor $(x-k)$, the remainder is $P(k)$.

$P(x)=(x-k)\cdot Q(x)+R\text{where }R=P(k)$

Factor Theorem

$(x-k)$ is a factor of $P(x)$ if and only if $P(k)=0$.

This is a special case of the Remainder Theorem where the remainder equals zero.

---

Question 3 — Finding Dimensions from a Volume Polynomial

Problem: The volume of a rectangular solid is given by $V(y)=y^3-2y^2-y+2$ If one dimension is $(y-2)$, find the remaining two dimensions.

---

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

Apply the Factor Theorem — substitute $y=2$: $V(2)=(2{)}^3-2(2{)}^2-(2)+2=8-8-2+2=0$

Since $V(2)=0$, by the Factor Theorem, $(y-2)$ is a factor of $V(y)$. ✓

---

Step 2: Divide $V(y)$ by $(y-2)$

Use synthetic division with $k=2$:

$V(y)÷(y-2)=y^2+0y-1=y^2-1$

Remainder $=0$ ✓ (confirms $(y-2)$ is a factor)

---

Step 3: Factor the Quotient

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

This is a difference of squares: $a^2-b^2=(a-b)(a+b)$.

---

Step 4: Write the Complete Factorization

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

The three dimensions of the rectangular solid are: $\boxed{(y-1),(y+1),(y-2)}$

---

Verification

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

---

Summary of Method