5.1 Q-10

Exercise 5.1 — Question 10

Topic: Remainder Theorem and Factor Theorem

Relevant SLOs: M-11-A-42, M-11-A-43, M-11-A-44

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Key Concepts

Polynomial Division

When a polynomial $P(x)$ of degree $\le 4$ is divided by a linear polynomial $(x-k)$, the result is:

$P(x)=(x-k)\cdot Q(x)+R$

where:

• $Q(x)$ = quotient (degree one less than $P(x)$)
• $R$ = remainder (a constant)

Division can be performed by long division or synthetic division.

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Remainder Theorem

No long division needed — simply substitute $x=a$ into $P(x)$.

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

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

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

$\therefore$ Remainder $=5$

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Factor Theorem

This is a special case of the Remainder Theorem: when the remainder is zero, $(x-a)$ divides $P(x)$ exactly.

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 $Q(x)$.
3. Factorize the quadratic $Q(x)$ by inspection, completing the square, or the quadratic formula.
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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