Table of Contents

Last modified on June 10th, 2024

The factor theorem states that if f(x) is a polynomial of degree n (≥ 1) and ‘a’ is a real number, then (x – a) is a factor of f(x), if and only if f(a) = 0

Thus, if we substitute a number for x in a polynomial and get zero, then x minus that number is a factor of the polynomial.

Mathematically,

If f(a) = 0, then f(x) = (x – a) × q(x)

Thus, (x – a) is a factor of f(x)

Also, if (x – a) is a factor of f(x)

Such that f(x) = (x – a) × q(x), then f(a) = 0

This is, thus, an extension of the remainder theorem and is commonly used to factor a polynomial and find its roots.

Let us consider a polynomial f(x), which is being divided by another polynomial (x – a)

If the quotient is q(x), and the remainder is f(a).

From the remainder theorem, we know

f(x) = (x – a) × q(x) + f(a)

By substituting f(a) = 0, we get

f(x) = (x – a) × q(x) + 0

⇒ f(x) = (x – a) × q(x)

Since (x – a) completely divides f(x), (x – a) is a factor of f(x)

Thus, if f(a) = 0, then (x – a) is a factor of f(x)

Now, let (x – a) be a factor of f(x)

Thus, f(x) = (x – a) × q(x)

⇒ f(x) = (x – a) × q(x) + 0

On comparing with the remainder theorem, we get

The remainder = f(a) = 0

Thus, if (x – a) is a factor of f(x), then f(a) = 0

**How to Use Factor Theorem**

Let us find the factors of the polynomial f(x) = x^{2} + 4x – 5

Assuming f(x) = 0 and solving further, we get

x^{2} + 4x – 5 = 0

⇒ x^{2} + 5x – x – 5 = 0

⇒ x(x + 5) – 1(x + 5) = 0

⇒ (x + 5)(x – 1) = 0

Thus, we get either (x + 5) = 0 or (x – 1) = 0

⇒ x = -5 or x = 1

If f(x) = 0, the values of x are called the roots of the function.

Here, the roots or the zeros of f(x) = x^{2} + 4x – 5 are 1, -5

The factors of f(x) = x^{2} + 4x – 5 are (x – 1), (x + 5)

Now, using the factor theorem, we get

If x = -5 is a root of f(x), then

f(-5) = (-5)^{2} + 4(-5) – 5 = 25 – 20 – 5 = 0

If x = 1 is a root of f(x), then

f(1) = (1)^{2} + 4(1) – 5 = 1 + 4 – 5 = 0

Since the remainders f(-5) and f(1) are zero, (x + 5) and (x – 1) are the factors of f(x), which proves that -5 and 1 are the zeros of the polynomial f(x).

Also, if x = a is a root of the polynomial, then (x – a) is a factor of it, and vice versa.

**Verify whether p + 1 is a factor of p ^{4} + 2p^{2} + 1**

Solution:

Here, f(p) = p^{4} + 2p^{2} + 1

By putting (p + 1) = 0

⇒ p = -1

At p = -1, we get f(p) = f(-1)

⇒ f(-1) = (-1)^{4} + 2(-1)^{2} + 1

⇒ f(-1) = 1 + 2 + 1

⇒ f(-1) = 4

Thus, p + 1 is not a factor of p^{4} + 2p^{2} + 1

**Verify whether x = ${\dfrac{2}{5}}$ is a root of f(x) = 125x ^{3} – 20x + 4**

Solution:

For x = ${\dfrac{2}{5}}$,

We get,

${f\left( x\right) =f\left( \dfrac{2}{5}\right)}$

= ${125\left( \dfrac{2}{5}\right) ^{3}-20\left( \dfrac{2}{5}\right) +4}$

= (2)^{3} – 4(2) + 4

= 4

Thus, x = ${\dfrac{2}{5}}$ is not a root of f(x) = 125x^{3} – 20x + 4

**Use the factor theorem to verify if 2y + 3 is a factor of the polynomial 4y ^{2} + 12y + 9**

Solution:

Here, f(y) = 4y^{2} + 12y + 9

By evaluating for y, we get

(2y + 3) = 0

⇒ 2y = -3

⇒ y = ${\dfrac{-3}{2}}$

Using the factor theorem, at y = ${\dfrac{-3}{2}}$, we get

${f\left( y\right) =f\left( \dfrac{-3}{2}\right)}$

= ${4\left( \dfrac{-3}{2}\right) ^{2}+12\left( \dfrac{-3}{2}\right) +9}$

= (-3)^{2} + 12(-3) + 9

= 9 – 18 + 9

= 0

Thus, 2y + 3 is a factor of the polynomial 4y^{2} + 12y + 9

**Use the factor theorem to verify if 3n + 4 is a factor of the polynomial 6n ^{2} + 11n + 4**

Solution:

Here, f(n) = 6n^{2} + 11n + 4

By evaluating for n, we get

(3n + 4) = 0

⇒ 3n = -4

⇒ n = ${\dfrac{-4}{3}}$

Using the factor theorem, at n = ${\dfrac{-4}{3}}$, we get

${f\left( n\right) =f\left( \dfrac{-4}{3}\right)}$

= ${6\left( \dfrac{-4}{3}\right) ^{2}+11\left( \dfrac{-4}{3}\right) +4}$

= ${6\times \dfrac{16}{9}-\dfrac{44}{3}+4}$

= ${\dfrac{32}{3}-\dfrac{44}{3}+4}$

= ${\dfrac{32-44+12}{3}}$

= 0

Thus, 3n + 4 is a factor of the polynomial 6n^{2} + 11n + 4

**More Resources**

Last modified on June 10th, 2024