# Using the Factor Theorem in Finding the Factors of Polynomials (With Examples)

*Ray is a Licensed Engineer in the Philippines. He loves to write any topic about mathematics and civil engineering.*

## What Is the Factor Theorem?

Factor Theorem** **is a particular case of the Remainder Theorem that states that if

*f(x)*= 0 in this case, then the binomial (x – c) is a factor of polynomial

*f(x)*. It is a theorem linking factors and zeros of a polynomial equation.

Factor Theorem is a method that allows the factoring of polynomials of higher degrees. Consider a function f(x). If *f(1)* = 0, then (x-1) is a factor of *f(x).* If *f(-3)* = 0 then (x + 3) is a factor of *f(x).* The factor theorem can produce the factors of an expression in a trial and error manner. The Factor Theorem is useful for finding factors of polynomials.

There are two ways to interpret the Factor Theorem's definition, but both imply the same meaning.

**Definition 1**

A polynomial *f(x)* has a factor x – c if and only if *f(c) *= 0.

**Definition 2**

If (x – c) is a factor of *P(x)*, then c is a root of the equation *P(x)* = 0, and conversely.

## Factor Theorem Proof

If (x – c) is a factor of *P(x)*, then the remainder R obtained by dividing *f(x) *by (x – r) will be 0.

*P*(*x)* = (x – c) x *Q(x)*

Divide both sides by (x – c). Since the remainder is zero, then *P(r) *= 0.

*P*(*x) */ (x – c) = *Q(x) *+ [0 / (x – c)

Conversely, if c is a root of *P(x) *= 0, *P(c)* = 0, then R = 0. Then,

*P*(*x) */ (x – c) = *Q(x) *+ [0 / (x – c)]

*P*(*x) *= (x – c) x *Q(x)* + 0

*P*(*x) *= (x – c) x *Q(x)*

Therefore, (x – c) is a factor of *P(x).*

## Example 1: Factorizing a Polynomial by Applying the Factor Theorem

Factorize 2x^{3} + 5x^{2} – x – 6.

**Solution**

Substitute any value to the given function. Say, substitute 1, -1, 2, -2, and -3/2.

f(1) = 2(1)^{3} + 5(1)^{2} - 1 – 6

f(1) = 0

f(-1) = 2(-1)^{3} + 5(-1)^{2} – (-1) - 6

f(-1) = -2

f(2) = 2(2)^{3} + 5(2)2 – (2) - 6

f(2) = 28

f(-2) = 2(-2)^{3} + 5(-2)^{2} – (-2) - 6

f(-2) = 0

f(-3/2) = 2(-3/2)^{3} + 5(-3/2)^{2} – (-3/2) - 6

f(-3/2) = 0

The function resulted to zero for values 1, -2, and -3/2. Hence using the Factor Theorem, (x – 1), (x + 2), and 2x +3 are factors of the given polynomial equation.

**Final Answer**

(x – 1), (x + 2), (2x + 3)

## Example 2: Using the Factor Theorem

Using the Factor Theorem, show that x – 2 is a factor of *f(x)* = x^{3} – 4x^{2} + 3x + 2.

**Solution**

We need to show that x – 2 is a factor of the given cubic equation. Start by identifying the value of c. From the given problem, the variable c is equal to 2. Substitute the value of c to the given polynomial equation.

*f(x)* = x^{3} – 4x^{2} + 3x + 2

*f(2)* = (2)^{3} – 4(2)^{2} + 3(2) + 2

*f(2)* = 8 – 16 + 6 + 2

*f(2)* = 0

**Final Answer**

Since *f(2)* = 0, we see from the factor theorem that x – 2 is a factor of *f(x)*. Another method would be to divide *f(x)* by x – 2 and show that the remainder is zero. The quotient in the division would be another factor of *f(x)*.

## Example 3: Finding a Polynomial with Prescribed Zeros

Find a polynomial *f(x)* of degree 3 with zeros 2, -1, and 3.

**Solution**

By the factor theorem, *f(x)* has factors x – 2, x + 1, x -3.

*f(x)* = a (x – 2)(x + 1)(x – 3)

Thus, where any nonzero value may be assigned to a. If we let a = 1 and multiply to the factors, we obtain a polynomial of degree three. Same goes as you use different values of a since it will still be reduced to lowest and most simple equation.

*f(x)* = (x – 2)(x + 1)(x – 3)

*f(x)* = (x^{2} + x – 2x – 2)(x – 3)

*f(x)* = x^{3} + x^{2} – 2x^{2} – 2x – 3x^{2} – 3x + 6x + 6

*f(x)* = x^{3} – 4x^{2} + x + 6* *

**Final Answer**

The polynomial of degree 3 that has zeros 2, -1, and 3 is x^{3} – 4x^{2} + x + 6.

## Example 4: Proving an Equation Is a Factor of a Quadratic Equation

Show that (x + 2) is a factor of *P(x) *= x^{2} + 5x + 6 using the Factor Theorem.

**Solution**

Substitute the value of c = -2 to the given quadratic equation. Prove that x + 2 is a factor of x^{2} + 5x + 6 using the Factor Theorem.

*P(-2) *= (-2)^{2} + 5(-2) + 6

*P(-2)* = 4 – 10 + 6

*P(-2)* = 0

**Final Answer**

Therefore, x + 2 is a factor of x^{2} + 5x + 6.

## Example 5: Identifying if the Statement Is True Using Factor Theorem

Use the Factor Theorem to decide whether each statement is true.

a. x – 1 is a factor of x^{7} – 1.

b. x – 1 is a factor of 3x^{5} + 4x^{2} – 7.

c. x – 1 is a factor of x^{1992} – x^{1860} + x^{1754} – x^{1636}.

d. x + 1 is a factor of x^{1992} – x^{1860} + x^{1754} – x^{1636}.

**Solution**

a. To determine whether the statement is true, utilize the factor theorem and substitute c = 1 to the equation x^{7} – 1.

*f(x) *= x^{7} - 1

*f(x) *= (1)^{7} - 1

*f(x) *= 0

**Final Answer**

Since the result is zero, x – 1 is a factor of x^{7} – 1.

b. To determine whether the statement is true, utilize the factor theorem and substitute c = 1 to the equation x^{7} – 1.

*f(x) *= 3x^{5} + 4x^{2} – 7

*f(x)* = 3(1)^{5} + 4(1)^{2} – 7

*f(x)* = 3 + 4 - 7

*f(x)* = 0

**Final Answer**

Since the result is zero, x – 1 is a factor of 3x^{5} + 4x^{2} – 7.

c. To determine whether the statement is true, use the factor theorem and substitute c = 1 to the equation x^{1992} – x^{1860} + x^{1754} – x^{1636}.

*f(x) *= x^{1992} – x^{1860} + x^{1754} – x^{1636}

*f(x) *= (1)^{1992} – (1)^{1860} + (1)^{1754} – (1)^{1636}

*f(x) *= 0

**Final Answer**

Since the result is zero, x – 1 is a factor of x^{1992} – x^{1860} + x^{1754} – x^{1636}.

d. To determine whether the statement is true, use the factor theorem and substitute c = -1 to the equation

*f(x) *= x^{1992} – x^{1860} + x^{1754} – x^{1636}

*f(x) *= (-1)^{1992} – (-1)^{1860} + (-1)^{1754} – (-1)^{1636}

*f(x) *= 0

**Final Answer**

Since the result is zero, x + 1 is a factor of x^{1992} – x^{1860} + x^{1754} – x^{1636}.

## Example 6: Proving X - C Is a Factor of a Function Given the Value of C

Use the Factor Theorem to show that x – c is a factor of *f(x).*

a. *f(x)* = x^{3} + x^{2} – 2x +12; c = -3

b. *f(x)* = x^{12} – 4096; c = -2

c. *f(x)* = x^{4} – 2x^{3} + 3x – 36; c = 3

d. *f(x)* = x^{3} + x^{2} – 11x + 10; c = 2

e. *f(x)* = x^{5} + 1024; c = -4

**Solution:**

a. Using the factor theorem, substitute the value of c = -3 to the given cubic function *f(x)* = x^{3} + x^{2} – 2x +12.

*f(x)* = (-3)^{3} + (-3)^{2} – 2(-3) +12

*f(x)* = -27 + 9 + 6 + 12

*f(x)* = 0

**Final Answer**

Therefore, x + 3 is a factor of x^{3} + x^{2} – 2x +12.

b. To show that x + 2 is a factor of *f(x)* = x^{12} – 4096, substitute the value of c to the given polynomial equation.

*f(x)* = x^{12} – 4096

*f(x)* = (-2)^{12} – 4096

*f(x)* = 0

**Final Answer**

Therefore, x + 2 is a factor of the polynomial equation x^{12} – 4096.

c. Using the factor theorem, substitute the value of c = 3 to the given cubic function *f(x)* = x^{4} – 2x^{3} + 3x – 36.

*f(x)* = x^{4} – 2x^{3} + 3x – 36

*f(3)* = (3)^{4} – 2(3)^{3} + 3(3) – 36

*f(3)* = 0

**Final Answer**

Since the resulting answer is zero, x – 3 is a factor of x^{4} – 2x^{3} + 3x – 36.

d. Using the factor theorem, substitute the value of c = 2 to the given cubic function *f(x)* = x^{3} + x^{2} – 11x + 10.

*f(x)* = x^{3} + x^{2} – 11x + 10

*f(2)* = (2)^{3} + (2)^{2} – 11(2) + 10

*f(2)* = 0

**Final Answer**

Since the resulting answer is zero, x – 2 is a factor of x^{3} + x^{2} – 11x + 10.

e. Using the factor theorem, substitute the value of c = -4 to the given cubic function *f(x)* = x^{5} + 1024.

*f(x)* = x^{5} + 1024

*f(-4)* = (-4)^{5} + 1024

*f(-4)* = 0

**Final Answer**

Since the resulting answer is zero, x + 4 is a factor of x^{5} + 1024.

## Example 7: Finding the Equation Given the Zeros with the Use of Factor Theorem

Use the Factor Theorem to find the polynomial equation of degree 4 given the zeros -2, -1, 1, and 4.

**Solution**

Given the zeros -2, -1, 1, and 4, you can use the Factor Theorem’s definition to get the factors. Also, given the degree of 4, there should be 4 factors.

(x + 2)(x + 1)(x – 1)(x – 4)

Use the FOIL Method and multiply (x + 1) and (x – 1) first.

(x + 1)(x – 1)

(x + 2)(x^{2} – 1)(x – 4)

Next, utilize the FOIL Method and multiply (x + 2) and (x – 4).

(x + 2) (x – 4)

(x^{2} – 1)(x^{2} – 2x – 8)

Simplify the equation by using the distributive property of multiplication.

x^{4} – 2x^{3} – 8x^{2} – x^{2} + 2x + 8

x^{4} – 2x^{3} – 9x^{2} + 2x + 8

**Final Answer**

The resulting polynomial equation in the 4^{th} degree is x^{4} – 2x^{3} – 9x^{2} + 2x + 8.

## Example 8: Determining if X - C Is a Factor of an Equation

Determine if x + 3 is a factor of the equation *f(x)* = x^{3} + x^{2} – 2x + 12.

**Solution**

Use the Factor Theorem and plug in the given value of c = -3.

*f(x)* = x^{3} + x^{2} – 2x + 12

*f(-3)* = (-3)^{3} + (-3)^{2} – 2(-3) + 12

*f(-3)* = 0

**Final Answer**

Therefore, *f(-3)* = 0 so x + 3 is a factor of *f(x)* = x^{3} + x^{2} – 2x + 12.

## Example 9: Determining if X – C Is a Factor of an Equation

Determine if x - 2 is a factor of the equation *f(x)* = x^{4} – 3x^{3} – 2x^{2} + 5x + 6.

**Solution**

Use the Factor Theorem and plug in the given value of c = 2.

*f(x)* = x^{4} – 3x^{3} – 2x^{2} + 5x + 6

*f(2)* = (2)^{4} – 3(2)^{3} – 2(2)^{2} + 5(2) + 6

*f(2)* = 16 – 3(8) – 2(4) + 10 + 6

*f(2)* = 16 – 24 – 8 + 10 + 6

*f(2)* = 0

**Final Answer**

Therefore, *f(2)* = 0 so x - 2 is a factor of *f(x)* = x^{4} – 3x^{3} – 2x^{2} + 5x + 6.

## Example 10: Finding the Polynomial Equation Given the Zeros

Use the Factor Theorem to find the polynomial equation of degree 3 given the zeros -2, 0, and 5.

**Solution**

Given the zeros -2, 0, and 5, you can use the Factor Theorem’s definition to get the factors. Also, given the degree of 3, there should be 3 factors.

(x + 2)(x)(x – 5)

Distribute x to x – 5.

(x + 2)(x^{2} – 5x)

Next, utilize the FOIL Method and foil the polynomials (x + 2) and (x^{2} – 5x).

(x + 2) (x^{2} – 5x)

(x^{3} – 5x^{2} + 2x^{2} – 10x)

Simplify the equation by adding or subtracting like terms.

x^{3} – 5x^{2} + 2x^{2} – 10x

x^{3} – 3x^{2} – 10x

**Final Answer**

The resulting polynomial equation in the 3^{rd} degree is x^{3} – 3x^{2} – 10x.

## Factor Theorem and Synthetic Division of Polynomials

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**© 2020 Ray**