Factorization of Polynomials

Chapter in a nutshell: The Remainder Theorem gives the remainder when a polynomial $p(x)$ is divided by $(x-a)$ as simply $p(a)$. The Factor Theorem says $(x-a)$ is a factor of $p(x)$ exactly when $p(a)=0$. Together they let us factorise cubic (and higher) polynomials by finding one root, then reducing to a quadratic.

1. Key Ideas

• A polynomial $p(x)$ can be divided by a linear factor $(x-a)$ leaving a remainder.
• Finding a value of x that makes $p(x)=0$ gives a factor — the basis of factorisation.

2. Formulas / Theorems (quick reference)

• Remainder Theorem: when $p(x)$ is divided by $(x-a)$, remainder $=p(a)$.

• Factor Theorem: $(x-a)$ is a factor of $p(x)$ iff $p(a)=0$.

3. Method (factorising a cubic)

1. Guess a root $a$ (try factors of the constant term: ±1, ±2, …) so that $p(a)=0$.
2. Then $(x-a)$ is a factor; divide $p(x)$ by $(x-a)$ (long division or synthetic) to get a quadratic.
3. Factorise the quadratic (splitting the middle term / formula).
4. Write $p(x)$ as the product of all three linear factors.

4. Prerequisite Formulas (earlier classes)

• Identities: $(a\pm b)^2$, $a^2-b^2=(a+b)(a-b)$, $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$.
• Splitting the middle term for quadratics; polynomial long division.

5. Worked Example (one, for the method)

6. Common Mistakes to Avoid

• Using $p(a)$ for divisor $(x+a)$ — for $(x+a)$ the remainder is $p(-a)$.
• Confusing the two theorems: remainder gives a value; factor needs that value to be 0.
• Forgetting to also factorise the resulting quadratic.
• Trying roots that are not factors of the constant term first.

7. Likely Exam Questions (with crisp answers)

1. State the Remainder Theorem. → Dividing $p(x)$ by $(x-a)$ leaves remainder $p(a)$.
2. State the Factor Theorem. → $(x-a)$ is a factor of $p(x)$ iff $p(a)=0$.
3. Remainder when $p(x)$ is divided by $(x+2)$? → $p(-2)$.
4. Is $(x-1)$ a factor of $x^3-1$? → Yes, since $p(1)=0$.
5. Remainder of $(2x-1)$ dividing $p(x)$? → $p(1/2)$.
6. First step to factorise a cubic? → Find a value $a$ with $p(a)=0$.
7. Factorise $a^3-b^3$. → $(a-b)(a^2+ab+b^2)$.
8. If $p(a)=5$, is $(x-a)$ a factor? → No (the remainder is 5, not 0).
9. Which values of $a$ should you try first? → Factors of the constant term.
10. After removing one linear factor from a cubic, what remains? → A quadratic to factorise.

Formulas — Current & Prerequisite

Current

• Remainder Theorem: when $p(x)$ is divided by $(x-a)$, remainder $=p(a)$.
• Factor Theorem: $(x-a)$ is a factor of $p(x)$ $\iff p(a)=0$; for $(ax-b)$ use $p\!\left(\tfrac{b}{a}\right)=0$.
• Factorise $p(x)$ (degree ≤ 3) by finding one factor, then dividing to get a quadratic and factorising it.

Prerequisite

• Polynomial division; quadratic factorisation; identities $(a\pm b)^3$, $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$.