Factor Theorem of Polynomials – Definition, Proof & Examples

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Polynomials which are special types of algebraic expressions are used to form polynomial equations, which encode a wide range of problems, from simple to complicated problems in mathematics; they are used to define polynomial functions, used in calculus and numerical analysis to approximate the other functions. 

You can perform the four mathematical operations – addition, subtraction, multiplication, and division on polynomials. Factor theorem is used in factoring a polynomial and finding the roots of the polynomial. Let’s understand what the factor theorem of polynomials is and how it is used to solve problems.

What is Factor Theorem of Polynomials?

Factor theorem is used to factor the polynomials and to find the roots of the polynomials. It makes the process easy by removing all the known zeros from a given polynomial equation and leaving all the unknown zeros. 

The factor theorem states that if $f(x)$ is a polynomial of degree $n$ greater than or equal to $1$, and $a$ is any real number, then $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$. 

In other words, we can say that $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$. 

This theorem is very helpful in the factorization of a polynomial.

Steps to Use Factor Theorem

Following are the steps to use the factor theorem of polynomials.

Step 1: Use the synthetic division of the polynomial method to divide the given polynomial $p(x)$ by the given binomial $(x − a)$

Step 2: After the completion of the division, confirm whether the remainder is 0. If the remainder is not zero, then it means that $(x – a)$ is not a factor of $p(x)$.

Step 3: Using the division algorithm, write the given polynomial as the product of $(x – a)$ and the quotient $q(x)$

Step 4: If it is possible, factor the quotient further.

Step 5: Express the given polynomial as the product of its factors.

Proof of Factor Theorem

Method 1

Let’s first consider a polynomial $p(x)$ that is being divided by $(x – a)$ only if $p(a) = 0$. By using the division algorithm, the given polynomial can be written as the product of its divisor and its quotient:

$ \text{Dividend } = ( \text{Divisor } \times \text{ Quotient} ) + \text{ Remainder}

$=>xp(x) = (x – a) q(x) + \text{ remainder}$. Here, $p(x)$ is the dividend, $(x – a)$ is the divisor, and $q(x)$ is the quotient.

From the remainder theorem, we get:

$p(x) = (x – a) q(x) + p(a)$

If we substitute $p(a) = 0$ then the remainder is $0$,

$=> p(x) = (x – a) q(x) + 0$

$=> p(x) = (x – a) q(x)$

Thus, we can say that $(x – a)$ 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 $p(x)$ has a factor $(x – a)$, if and only if, $a$ is a root i.e., $p(a) = 0$.

Method 2

Let’s first consider a polynomial $p(x)$ that is being divided by $(x – a)$ only if $p(a) = 0$. By using the division algorithm, the given polynomial can be written as the product of its divisor and its quotient:

$ \text{Dividend } = ( \text{Divisor } \times \text{ Quotient} ) + \text{ Remainder}

$p(x) = (x – a)q(x)+ p(a)$

If $(x – a)$ is a factor of $p(x)$, then the remainder must be zero.

$(x – a)$ exactly divides $p(x)$

Therefore, $p(a)=0$.

The following statements are equivalent for any polynomial $p(x)$

• The remainder is zero when $p(x)$ is exactly divided by $(x – a)$
• $(x – a)$ is a factor of $p(x)$
• $a$ is the solution to $p(x)$
• $a$ is a zero of the polynomial $p(x)$, or $p(a) = 0$

Factor Theorem Examples

Ex 1: Determine whether $(x + 1)$ is a factor of the polynomial $6x^4 + 7x^3 – 5x – 4$.

By factor theorem, if $p(a)  =  0$, then $(x – a)$ is a factor of $p(x)$

Let $p(x) = 6x^4 + 7x^3 – 5x – 4$ and $(x – a) = (x + 1)$.

Therefore, $a = -1$

Thus, $p(-1)  = 6(-1)^4 + 7(-1)^3 – 5(-1) – 4$

$= 6 – 7 + 5 – 4$

$=>p(-1) = 0$

Therefore, $(x + 1)$ is a factor of the polynomial $6x^4 + 7x^3 – 5x – 4$.

Ex 2: Determine whether $(x + 1)$ is a factor of the polynomial $2x^4 + 9x^3 + 2x^2 + 10x + 15$.

By factor theorem, if $p(a)  =  0$, then $(x – a)$ is a factor of $p(x)$

Let $p(x) = 2x^4 + 9x^3 + 2x^2 + 10x + 15$ and $(x – a) = (x + 1)$.

Therefore, $a = -1$

Thus, $p(-1)  = 2(-1)^4 + 9(-1)^3 + 2(-1)^2 + 10(-1) + 15$

$= 2 – 9 + 2 – 10 + 15$

$=>p(-1) = 0$

Therefore, $(x + 1)$ is a factor of the polynomial $2x^4 + 9x^3 + 2x^2 + 10x + 15$.

Ex 3: Determine whether $(2x + 1)$ is a factor of the polynomial $4x^3 + 4x^2 – x – 1$.

By factor theorem, if $p(a)  =  0$, then $(x – a)$ is a factor of $p(x)$

Let $p(x) = 4x^3 + 4x^2 – x – 1$ and $(x – a) = (2x + 1)$.

$2x + 1 = 0 => 2x = – 1 => x = -\frac{1}{2}$ 

Therefore, $a = -\frac{1}{2}$

Thus, $p(-\frac{1}{2})  = 4(-\frac{1}{2})^3 + 4(-\frac{1}{2})^2 – (-\frac{1}{2}) – 1$

$= 4(-\frac{1}{8}) + 4(\frac{1}{4})^2 +\frac{1}{2} – 1$

$ = -\frac{1}{2} + 1 + \frac{1}{2} – 1$

$=>p(-1) = 0$

Therefore, $(2x + 1)$ is a factor of the polynomial $4x^3 + 4x^2 – x – 1$.

Ex 4: Determine the value of $m$ if $(x + 3)$ is a factor of the polynomial $x^3 – 3x^2 – mx + 24$

By factor theorem, if $p(a)  =  0$, then $(x – a)$ is a factor of $p(x)$

Let $p(x) = x^3 – 3x^2 – mx + 24$ and $(x – a) = (x + 3)$.

Therefore, $a = -3$

Thus, $p(-3)  = (-3)^3 – 3(-3)^2 – m(-3) + 24$

$= -27 – 3(9) + 3m + 24$

$= -27 – 27 + 3m + 24$

$=>p(-1) = -30 + 3m$

Therefore, $p(-1) = 0 => -30 + 3m = 0 => 3m = 30 => m = \frac{30}{3} => m = 10$

Ex 5: Given that $f(x) = 2x^{3} – 7x^{2} – 10x + 32$. Use the factor theorem to show that $(x – 2)$ is a factor of $f(x)$. Factorize $f(x)$ completely.

By factor theorem, if $f(a)  =  0$, then $(x – a)$ is a factor of $f(x)$

$f(x) = 2x^{3} – 7x^{2} – 10x + 32$

Let $(x – a) = (x – 2) => a = 2$

Therefore, $f(2) = 2(2^3) – 7(2^2) – 10(2) + 32 = 16 – 28 – 20 + 32 = 0$

Dividing $f(x) = 2x^{3} – 7x^{2} – 10x + 32$ by $(x – 2)$ using Synthetic Division, we get

Therefore, $2x^3 – 7x^2 – 10x + 32 = (x – 2)(2x^2 – 3x – 16)$

Now, consider $2x^2 – 3x – 16$

Using completing the square method, we get $2x^2 – 3x – 16 = 2(x^2 – \frac{3}{2}x – 8)$

$ = 2(x^2 – 2 \times x \ times \frac{3}{4} – 8)$

$ = 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – (\frac{3}{4})^2 – 8)$

$ = 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – \frac{9}{16} – 8)$

$ = 2(x^2 – 2 \times x \ times \frac{3}{4} + (\frac{3}{4})^2 – \frac{137}{16})$

$ = 2(x – \frac{3}{4})^2 – \frac{137}{16}$

Therefore, $2(x – \frac{3}{4})^2 – \frac{137}{16} = 0$

$=> 2(x – \frac{3}{4})^2 – \frac{137}{16} = 0$

$=> 2(x – \frac{3}{4})^2 = \frac{137}{16}$ 

$=> (x – \frac{3}{4})^2 = \frac{137}{32}$

$=> x – \frac{3}{4} = \pm \frac{\sqrt{137}}{4\sqrt{2}}$

$=> x – \frac{3}{4} = \pm \frac{\sqrt{274}}{8}$

$=> x = \frac{3}{4} \pm \frac{\sqrt{274}}{8}$

$=> x = \frac{6 \pm \sqrt{274}}{8}$

$=> x = \frac{6 + \sqrt{274}}{8}$ and $x = \frac{6 – \sqrt{274}}{8}$

Therefore, $2x^2 – 3x – 16 = \left(x – \frac{6 + \sqrt{274}}{8} \right) \left(x – \frac{6 – \sqrt{274}}{8} \right)$

Thus, $2x^{3} – 7x^{2} – 10x + 32 = (x – 2)\left(x – \frac{6 + \sqrt{274}}{8} \right) \left(x – \frac{6 – \sqrt{274}}{8} \right)$

Difference Between the Remainder Theorem and Factor Theorem

Following are the differences between the remainder theorem and the factor theorem:

Points to Remember about the Factor Theorem

Practice Problems

1. Determine whether $(x +1)$ is a factor of the following polynomials.
   • $3x^3 + 8x^2 – 6x – 5$
   • $x^3 – 14x^2 + 3x + 12$
2. Given $f(x) = 2x^3 – 3x^2 – 39x + 20$. Use the factor theorem to show that $(x + 4)$ is a factor of $f(x)$. Factorize $f(x)$ completely.
3. Given $f(x) = 2x^{3} – 7x^{2} – 5x + 4$. Use the factor theorem to show that $(x + 1)$ is a factor of $f(x)$. Factorize $f(x)$ completely.

FAQs

Conclusion

The factor theorem states that if $f(x)$ is a polynomial of degree $n$ greater than or equal to $1$, and $a$ is any real number, then $(x – a)$ is a factor of $f(x)$ if $f(a) = 0$. Factor theorem is used to factor the polynomials and to find the roots of the polynomials. It makes the process easy by removing all the known zeros from a given polynomial equation and leaving all the unknown zeros.

Recommended Reading

• Remainder Theorem of Polynomials – Definition, Proof & Examples
• What are Polynomials? (Definition, Types & Examples)
• Nature of Roots of Quadratic Equation(With Methods & Examples)
• Solving Quadratic Equations – Formulas, Tricks & Examples
• Quadratic Equation Definition (With Different Forms & Examples)
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• Linear Equations in Two Variables – Definition, Types, and Graphs
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• What are Algebraic Identities(With Definition, Types & Derivations)
• What is the Meaning of Equation – Definition, Types & Examples
• Division of Algebraic Expressions(With Methods & Examples)
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• What is Algebra – Definition, Basics & Examples
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