Exercise 5.1 — Question 5

Concepts Covered

This exercise applies three key theorems from Chapter 5 (Polynomials):

• Polynomial Division (M-11-A-42)
• Remainder Theorem (M-11-A-43)
• Factor Theorem (M-11-A-44)

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Key Theorems

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 $\mathrm{deg}(R)<\mathrm{deg}(D)$.

For a linear divisor $(x-a)$, the remainder is a constant $R$.

Remainder Theorem

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

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

The remainder is $5$.

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 $P(x)$ using the Factor Theorem:

1. Find a root $a$ by testing integer values (factors of the constant term) until $P(a)=0$.
2. Divide $P(x)$ by $(x-a)$ using synthetic or long division to get a quadratic quotient $Q(x)$.
3. Factorize $Q(x)$ using the quadratic formula or inspection.
4. Write the complete factorization: $P(x)=(x-a)\cdot Q(x)$.

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

• Test $x=1$: $P(1)=1-6+11-6=0$ ✓ → $(x-1)$ 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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