Exercise 5.1 — Question 4

This exercise covers polynomial division, the Remainder Theorem, and the Factor Theorem.

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Key Concepts

Polynomial Division

When a polynomial $P(x)$ of degree $n$ is divided by a divisor $D(x)$ of degree $m<n$, we obtain:

$P(x)=D(x)\cdot Q(x)+R(x)$

where $Q(x)$ is the quotient and $R(x)$ is the remainder, with $\mathrm{deg}(R)<\mathrm{deg}(D)$.

• Dividing by a linear factor $(x-a)$: remainder is a constant.
• Dividing by a quadratic factor: remainder is at most linear (i.e., $ax+b$).

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Remainder Theorem

Example: Find the remainder when $P(x)=2x^3+3x^2-4x+1$ is divided by $(x+2)$.

Here $(x+2)=(x-(-2))$, so $a=-2$:

$P(-2)=2(-2{)}^3+3(-2{)}^2-4(-2)+1=-16+12+8+1=5$

Remainder $=5$.

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Factor Theorem

This is a special case of the Remainder Theorem: if the remainder $P(a)=0$, then $(x-a)$ divides $P(x)$ exactly.

Steps to factorize a cubic polynomial $P(x)$:

1. Find a value $a$ such that $P(a)=0$ (try factors of the constant term).
2. Divide $P(x)$ by $(x-a)$ using synthetic or long division to get a quadratic $Q(x)$.
3. Factorize $Q(x)$ using standard methods.
4. Write $P(x)=(x-a)\cdot Q(x)$.

Example: Factorize $P(x)=x^3-6x^2+11x-6$.

• Test $x=1$: $P(1)=1-6+11-6=0$ ✓
• Divide: $P(x)=(x-1)(x^2-5x+6)$
• Factorize quadratic: $x^2-5x+6=(x-2)(x-3)$
• Result: $P(x)=(x-1)(x-2)(x-3)$