Exercise 5.1 — Question 9

Topic Overview

This question covers polynomial division, the Remainder Theorem, and the Factor Theorem — core tools for working with polynomials up to degree 4.

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Key Concepts

1. Polynomial Division

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

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

where:

• $Q(x)$ is the quotient (degree $=n-d$)
• $R(x)$ is the remainder (degree $<d$)

Division by a linear factor $(x-a)$: remainder is a constant $R$.

Division by a quadratic $(x^2+bx+c)$: remainder is of the form $Ax+B$.

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2. Remainder Theorem

$P(x)=(x-a)\cdot Q(x)+P(a)$

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

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

Remainder $=5$.

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3. Factor Theorem

This is a special case of the Remainder Theorem: when the remainder is zero, the divisor is a factor.

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

1. Find a value $a$ such that $P(a)=0$ (try $\pm 1,\pm 2,\pm 3,\dots$)
2. Divide $P(x)$ by $(x-a)$ using synthetic or long division to get a quadratic $Q(x)$.
3. Factorize $Q(x)$ using the quadratic formula or inspection.
4. Write $P(x)=(x-a)\cdot Q(x)$ fully factored.

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

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

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