# 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.*

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

## Explore Other Math Articles

- How to Solve for the Surface Area and Volume of Prisms and Pyramids

This guide teaches you how to solve the surface area and volume of different polyhedrons such as prisms, pyramids. There are examples to show you how to solve these problems step-by-step. - Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition

A guide to solving for centroids and centers of gravity of different compound shapes using the method of geometric decomposition. Learn how to obtain the centroid from different examples provided. - How to Graph a Parabola in a Cartesian Coordinate System

The graph and location of a parabola depend on its equation. This is a step-by-step guide on how to graph different forms of parabola in the Cartesian coordinate system. - How to Find the General Term of Sequences

This is a full guide in finding the general term of sequences. There are examples provided to show you the step-by-step procedure in finding the general term of a sequence. - Calculator Techniques for Polygons in Plane Geometry

Solving problems related to plane geometry especially polygons can be easily solved using a calculator. Here is a comprehensive set of problems about polygons solved using calculators. - Age and Mixture Problems and Solutions in Algebra

Age and mixture problems are tricky questions in Algebra. It requires deep analytical thinking skills and great knowledge in creating mathematical equations. Practice these age and mixture problems with solutions in Algebra. - AC Method: Factoring Quadratic Trinomials Using the AC Method

Find out how to perform AC method in determining if a trinomial is factorable. Once proven factorable, proceed with finding the factors of the trinomial using a 2 x 2 grid. - Methods of Depreciation: Formulas, Problems, and Solutions

Learn how to solve problems on different types of depreciation methods in Engineering Economics using the formulas and solutions provided. - Calculator Techniques for Circles and Triangles in Plane Geometry

Solving problems related to plane geometry especially circles and triangles can be easily solved using a calculator. Here is a comprehensive set of calculator techniques for circles and triangles in plane geometry. - How to Solve for the Moment of Inertia of Irregular or Compound Shapes

This is a complete guide in solving for the moment of inertia of compound or irregular shapes. Know the basic steps and formulas needed and master solving moment of inertia. - Calculator Techniques for Quadrilaterals in Plane Geometry

Learn how to solve problems involving Quadrilaterals in Plane Geometry. It contains formulas, calculator techniques, descriptions, and properties needed in order to interpret and solve Quadrilateral problems. - How to Graph an Ellipse Given an Equation

Learn how to graph an ellipse given the general form and standard form. Know the different elements, properties, and formulas necessary in solving problems about ellipse. - How to Graph a Circle Given a General or Standard Equation

Learn how to graph a circle given the general form and standard form. Familiarize with converting general form to standard form equation of a circle and know the formulas necessary in solving problems about circles. - How to Calculate the Approximate Area of Irregular Shapes Using Simpson’s 1/3 Rule

Learn how to approximate the area of irregularly shaped curve figures using Simpson’s 1/3 Rule. This article covers concepts, problems, and solutions about how to use Simpson’s 1/3 Rule in area approximation. - Finding the Surface Area and Volume of Frustums of a Pyramid and Cone

Learn how to calculate the surface area and volume of the frustums of the right circular cone and pyramid. This article talks about the concepts and formulas needed in solving for the surface area and volume of frustums of solids. - Finding the Surface Area and Volume of Truncated Cylinders and Prisms

Learn how to compute for the surface area and volume of truncated solids. This article covers concepts, formulas, problems, and solutions about truncated cylinders and prisms.

*This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.*

**© 2020 Ray**