Factor Theorem

Factor theorem is used for finding the roots of the given polynomial. This theorem is very helpful in finding the factors of the polynomial equation without actually solving them.

According to the factor theorem, for any polynomial f(x) of degree n ≥ 1 a linear polynomial (x – a) is the factor of the polynomial if f(a) is zero.

Let’s learn about the factor theorem, its proof, and others in detail in this article.

What is the Factor Theorem?

A special theorem that links polynomials with their zeros and helps to find the factors of the polynomial is called the Factor Theorem. Factor theorem along with the remainder theorem is very helpful in solving complex polynomial equations. For any polynomial of a higher degree factor, the theorem removes all the known zeros and reduces the polynomial to a lesser degree, and then its factors are easily calculated. 

Factor Theorem Statement

According to the factor theorem, any polynomial f(x) of degree greater than or equal to 1 its factor can be easily found if we know the root of the polynomial equation. The root of the polynomial equation is the number that satisfies the polynomial equation. Suppose the root of f(x) is x = a then according to the factor theorem (x- a) is the factor of the polynomial f(x).

The image given below shows the factor theorem statement.

Factor Theorem

Factor Theorem Formula

According to the factor theorem x – a can be considered the factor of the polynomial f(x) if f(a) is zero. Now the Factor Theorem Formula is,

If (x – a) is a factor of f(x) then,

f(a) = 0, 

Now f(x) can be expanded as,

f(x) = (x-a)q(x)

where,
a is the root of the equation
q(x) is the quotient when f(x) is divided by q(a)

Also Read: Real Life Applications of Factor Theorem

Zero of a Polynomial

The concept of zero of the polynomials is very useful for understanding the factor theorem. Zero of any Polynomial is the real number that satisfies the polynomial equation. i.e. For polynomial f(x) ‘a’ is the zero of f(x) if f(a) is zero.

For example, the zero of the polynomial x² + 4x – 12 is calculated as,

Solution:

x² + 4x – 12

=  x² + 6x -2x – 12

= x(x + 6) – 2(x + 6)

= (x -2)(x+6)

zeros of the polynomial x² + 4x – 12 are,

x – 2 = 0

⇒ x = 2, and

x + 6 = 0

⇒ x = -6

Verification: To verify the zero of the polynomial x² + 4x – 12

f(2) = 2² + 4(2) -12

⇒ f(2) = 4 + 8 -12 = 0

f(-6) = (-6)² + 4(-6) – 12

⇒ f(-6) = 36 – 24 -12 = 0

Thus, zero of the polynomial x² + 4x – 12 is verified.

Factor Theorem Proof

Consider a polynomial p(x) that is being divided by (x – b) only if p(b) = 0. 

Given Polynomial can be written as,

Dividend = (Divisor × Quotient) + Remainder

By using the division algorithm,

p(x) = (x – b) q(x) + remainder

Where,

• p(x) is the dividend,
• (x – b) is the divisor, and
• q(x) is the quotient.

From the remainder theorem, 

p(x) = (x – b) q(x) + p(b).

Suppose that p(b) =0.

p(x) = (p – b) q(x) + 0

⇒ p(x) = (x – b) q(x)

Thus, we can say that (x – b) is a factor of the polynomial p(x). 

Here we can see that the factor theorem is actually a result of the remainder theorem, which states that a polynomial (x) has a factor (x – a), if and only if, a is a root i.e., p(b) = 0.

How to Use Factor Theorem?

Factor Theorem is used widely in solving the polynomial equation.

We can learn about the use of factor theorem by going through the example discussed below,

Example: Find the factor of x² – 4x + 3.

Solution:

Given polynomial,

f(x) =  x² – 4x + 3

by hit and try method,

taking x = 1

f(1) = 1² -4(1) + 3       = 0

Thus, x -1 is the factor of x² – 4x + 3

again taking x = 3

f(3) = 3² – 4(3) + 3 = 0

Thus, x -3 is the factor of x² – 4x + 3

Using the Factor Theorem to Factor a Cubic Polynomial

Factor Theorem To Factor a Third-Degree Polynomial is widely used for solving the three-degree polynomial also called a cubic polynomial. Generally, for quadratic polynomial factorization method is used and the factor theorem is used when we have to find the factors of any cubic polynomial. 

Use the steps given below to find the Factor of a Cubic Polynomial

Factorize the cubic polynomial f(x) = ax³ + bx² + cx + d.

• Step 1: By hit and try method find one of the zeros of the polynomial f(x). Say the zero of the polynomials is “a” such that f(a) is zero.
• Step 2: Divide the polynomial f(x) by x-a using synthetic division of the polynomial method.
• Step 3: Using the division algorithm, write the given cubic polynomial as the f(x) = (x-a)g(x), where g(x) is a quadratic polynomial.
• Step 4: Factorize the cubic polynomial g(x). As, g(x) = (x-b)(x-c) for any real numbers b and c.
• Step 5: Express the given cubic polynomial as the product of these factors.

f(x) = (x -a)(x-b)(x-c)

Example Using factor theorem factorize f(x) = x³ + 10x² + 23x + 14

Solution:

f(x) = x³ + 10x² + 23x + 14

Now applying Factor Theorem.

(x – a) is a factor of f(x) if f(a) = 0.

And g(y) = (y-a)q(a)

Now, by hit and try method we get,

f(-1) = (-1)³ + 10(-1)² + 23(-1) + 14

⇒ f(-1) = 24 – 24 = 0

Thus, x + 1 is a factor of  x³ + 10x² + 23x + 14

Now, by dividing x + 1 by  x³ + 10x² + 23x + 14 we get,

f(x) = (x + 1)(x²  + 9x + 14)

⇒ f(x) = (x + 1)(x²  + 7x + 2x + 14)

⇒ f(x) = (x + 1){x(x +7) + 2(x + 2)}

⇒ f(x) = (x + 1)(x + 2)(x + 7)

Factor Theorem and Remainder Theorem

Remainder Theorem and Factor Theorem are the basic theorems that help in mathematics and the basic difference between them is stated in the table given below,

Difference Between Factor Theorem and Remainder Theorem

Key difference between factor and remainder theorem are listed as follows:

For example, for the given polynomial f(x) = x² + 3x – 4 the remainder when f(x) is divided by x – 2 = 0 is,

Solution:

Given,

x – 2 = 0 ⇒ x = 2

Thus, f(2) = 2² + 3(2) – 4 = 6

Hence, the remainder when we divide x² + 3x – 4  by x – 2 is 6.

Similarly, for the given polynomial f(x) = x² + 3x – 4 the remainder when f(x) is divided by x – 2 = 0 is “0” then we can say that x -2 is a factor of x² + 3x – 4

Given,

x – 1 = 0 ⇒ x = 1

f(2) = 1² + 3(1) – 4 = 0

Hence, the remainder when we divide x² + 3x – 4  by x – 2 is 0.

Thus, according to the factor theorem x -2 is a factor of x² + 3x – 4.

Related Article:

• Division
• Divisibility Rules
• Chinese Remainder Theorem
• Quotient Remainder Theorem
• Factoring Polynomials
• Practice Questions on Remainder theorem

Factor Theorem Examples

Example 1: Find the root of the polynomial f(x) = x² – 5x + 6 using the factor theorem.

Solution:

f(x) = x² – 5x + 6.

Now applying factor theorem.

If (x – a) is a factor of f(x) then f(a) = 0

By hit and try method,

f(2) = 4 – 10 + 6 = 0

Thus, (x – 2) is a factor of f(x).

Example 2: Suppose a polynomial f(x) = x³ + 3x² – 6x – 18. If x = -3 is a root show that x + 3 is a factor.

Solution:

f(x) = x³ + 3x² – 6x – 18.

Now applying factor theorem.

If x = a is the root of f(x) then x – a is the factor of the f(x)

Now for, f(x) at x = -3

f(-3) = -27 + 27 + 18 – 18 = 0

Thus, (x + 3) is a factor of f(x).

Example 3: Using the factor theorem check if y + 2 is a factor of the polynomial 2y⁴ + y³ – 2y² + 4y -8, or not.

Solution:

Given y + 2

If, y + 2 = 0, then y = -2

Substituting y = -2 in the given polynomial 2y⁴ + y³ – 2y² + 4y -8

= 2(-2)⁴ + (-2)³ – 2(-2)² + 4(-2) – 8

= 32 – 8 – 8 – 8 – 8

= 0

Thus, we can say that y + 2 is a factor of the polynomial 2y⁴ + y³ – 2y² + 4y -8

Example 4: Check if x – 1 is a factor of the polynomial f(x) = 2x⁴ + 3x² – 5x + 7

Solution:

Given x – 1

If, x – 1 = 0, then x = 1

Substituting x = 1 in the given polynomial 2x⁴ + 3x² – 5x + 7

= 2(1)⁴ + 3(1)² – 5(1) + 7

= 2 + 3 – 5 + 7

= 7

As, f(-1) ≠ 0

We can say that x – 1 is not a factor of the polynomial  2x⁴ + 3x² – 5x + 7

Example 5: Given a polynomial f(x) = x³ – x ² – 2x + 1. Factorize and find its roots.

Solution:

f(x) = x³ – x ² – x + 1

Now applying Factor Theorem.

(x – a) is a factor of f(x) if f(a) = 0.

And g(y) = (y-a)q(a)

Now, by hit and try method we get,

f(-1) = (-1)³ – (-1)² -(-1)

⇒ f(-1) = -8 + 4 + 4 = 0

Thus, x + 1 is a factor of x³ – x ² – x + 1

Now, by dividing x + 1 byx³ – x ² – x + 1 we get,

f(x) = (x + 1)(x² – 2x + 2)

⇒ f(x) = (x + 1)(x²  – x – x -2)

⇒ f(x) = (x + 1){x(x – 1) – 1(x – 1)}

⇒ f(x) = (x + 1)(x – 1)(x – 1)