Factor Theorem: Statement & Applications

Factor Theorem: Statement and Application in Factorization

The Factor Theorem is a powerful tool in algebra used to find the factors of a polynomial. It establishes a direct relationship between the roots of a polynomial and its linear factors. Essentially, if substituting a value into a polynomial results in zero, then $(x – \text{value})$ is a factor of that polynomial. This theorem greatly simplifies the process of factoring polynomials, especially those of higher degrees.

Formulae

Let $P(x)$ be a polynomial.

• Factor Theorem: If $P(a) = 0$, then $(x – a)$ is a factor of $P(x)$.
• Converse: If $(x – a)$ is a factor of $P(x)$, then $P(a) = 0$.

Examples

Example-1: Factor the polynomial $P(x) = x^2 – 5x + 6$.

• We can try a few values to find a root. Let’s try $x = 2$: $P(2) = (2)^2 – 5(2) + 6 = 4 – 10 + 6 = 0$.
• Since $P(2) = 0$, then $(x – 2)$ is a factor.
• Now, perform polynomial division or synthetic division to divide $P(x)$ by $(x – 2)$. This results in $(x – 3)$.
• Therefore, $P(x) = (x – 2)(x – 3)$.

Example-2: Factor the polynomial $P(x) = x^3 – 7x + 6$.

• Let’s try $x = 1$: $P(1) = (1)^3 – 7(1) + 6 = 1 – 7 + 6 = 0$.
• Since $P(1) = 0$, then $(x – 1)$ is a factor.
• Dividing $P(x)$ by $(x – 1)$ gives us $x^2 + x – 6$.
• Now, factor the quadratic: $x^2 + x – 6 = (x + 3)(x – 2)$.
• Therefore, $P(x) = (x – 1)(x + 3)(x – 2)$.

Theorem with Proof

Theorem: If $P(a) = 0$, then $(x – a)$ is a factor of $P(x)$.

Proof:

1. By the Remainder Theorem, if a polynomial $P(x)$ is divided by $(x – a)$, the remainder is $P(a)$.
2. If $P(a) = 0$, then the remainder is 0.
3. If the remainder is 0, then $(x – a)$ divides $P(x)$ evenly, meaning $(x – a)$ is a factor of $P(x)$.
4. Therefore, if $P(a) = 0$, then $(x – a)$ is a factor of $P(x)$.

Common mistakes by students

Common mistakes include:

• Incorrectly applying the theorem: Students often make mistakes by not identifying the ‘a’ correctly (e.g., if $x + 2$ is a factor, they might incorrectly assume $a = 2$ instead of $a = -2$).
• Not checking the solution: After finding a potential root, students sometimes fail to substitute the value back into the original polynomial to verify if it indeed results in zero.
• Incomplete factorization: They may stop after finding one factor and not fully factorize the polynomial into its linear factors.
• Confusing Factor Theorem with Remainder Theorem: Sometimes the two are mixed up. The Remainder Theorem deals with the remainder after division; Factor Theorem specifically deals with factors and when the remainder is zero.

Real Life Application

The Factor Theorem has applications in various fields:

• Engineering: Used in analyzing and designing systems where polynomials model physical phenomena (e.g., signal processing, control systems).
• Computer Graphics: Used in modeling curves and surfaces, where polynomial equations are fundamental.
• Economics: Applied in analyzing economic models and optimizing various parameters.

Fun Fact

The Factor Theorem is a special case of the Remainder Theorem, which is a more general theorem about the remainders when dividing polynomials.

Recommended YouTube Videos for Deeper Understanding

Q.1 If $(x-2)$ is a factor of the polynomial $x^3 – 5x^2 + 8x – 4$, then the other factor is:

Check Solution

Ans: A

By the Factor Theorem, if $(x-2)$ is a factor, then $x=2$ should make the polynomial equal to zero. Dividing the polynomial by $(x-2)$ gives the other factor. $(x^3 – 5x^2 + 8x – 4) / (x-2) = x^2 – 3x + 2$

Q.3 Which of the following is a factor of $x^3 – 6x^2 + 11x – 6$?

Check Solution

Ans: A

Applying the factor theorem. If $x=1$, $1^3 – 6(1)^2 + 11(1) – 6 = 1-6+11-6=0$, therefore $(x-1)$ is a factor.

Q.5 Factorize the polynomial $x^3 – 8$ completely.

Check Solution

Ans: B

Using the identity $a^3 – b^3 = (a-b)(a^2 + ab + b^2)$, we have $x^3 – 8 = x^3 – 2^3$. Applying the formula with $a=x$ and $b=2$, we get $(x-2)(x^2 + 2x + 4)$. But this is not the answer.

Next Topic: Factorization of Polynomials