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Factor Theorem

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  • Last Updated : 15 Jun, 2022

In mathematics, there are situations where we try to represent a problem in the form of mathematical symbols and mathematical expressions. For this purpose, we use variables like x, y, z, and mathematical operations like addition, subtraction, multiplication, and division to form a meaningful mathematical expression. Algebra is used in all the branches of mathematics such as trigonometry, calculus, coordinate geometry, and elementary geometry. These are further used for forming mathematical equations. One simple example of an expression in algebra is 2x + 4 = 8.

Algebra deals with symbols and these symbols are related to each other with the help of operators. The operators are used for describing a relation between the operands, this is used not only for applying and solving the equations but also for real-life situations; sometimes we use simple algebra in real life without even knowing it. Polynomials are algebraic expressions with variables and constants in them. All these terms are combined using mathematical operators such as addition, subtraction, multiplication, etc. The division is not used in the polynomials

x2 + 6x +32 is a polynomial.

\frac{1}{x^3}    is not a polynomial.

Degree of a polynomial

The highest exponential value of the variable in the given polynomial is called the degree of the polynomial. The following table illustrates this.

Type of polynomial Degree Example
Constant 0 23
Linear 1 x + 69
Quadratic 2 x2 + 45x + 3
Cubic 3 x3 + 4x2 + 54x + 2

Factor Theorem

The factor theorem has a unique purpose in the way that using it we can find the factors of the given polynomial and then find its roots. Suppose there exists a polynomial f(x) 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. The formula of the factor theorem is,

g(y) = (y – a)q(a)

Proof of Factor Theorem

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

The given polynomial can be written as,

Dividend = (Divisor × Quotient) + Remainder

By using the division algorithm,

⟹ p(x) = (x – b) q(x) + remainder. Here, 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.

Sample Problems

Question 1: Suppose a polynomial f(x) = x2 – 5x + 6. If x = 2 is a root show that x – 2 is a factor.

Solution:

f(x) = x2 – 5x + 6.

Now applying factor theorem.

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

f(2) = 4 – 10 + 6 = 0

x – 2 is a factor of f(x).

Question 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.

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

f(-3) = -27 + 27 + 18 – 18 = 0

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

Question 3: Suppose a polynomial f(x) = 2x2 + 5x – 7. If x = 1 is a root show that x – 1 is a factor.

Solution:

f(x) = 2x2 + 5x – 7.

Now applying factor theorem.

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

f(2) = 2 + 5 – 7 = 0

x – 2 is a factor of f(x).

Question 4: Show that x = 2 is a root of p(x) = x2 – 9x + 14. It is given that x – 2 is a factor.

Solution:

 p(x) = x2 – 9x + 14.

Now applying factor theorem.

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

p(2) = 4 – 18 + 14 = 0

x – 2 is a factor of f(x) [Given]

This means that if a is a root then x – a is factor of p(x).

Question 5: Suppose a polynomial f(x) = 2x2 + 5x – 7. If x = 1 is a root show that x – 1 is a factor.

Solution:

f(x) = 2x2 + 5x – 7.

Now applying factor theorem.

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

f(2) = 2 + 5 – 7 = 0

x – 2 is a factor of f(x).

Question 6: Given a polynomial f(x) = x2 + 5x + 4. If x = -1 is a root show that x + 1 is a factor.

Solution:

f(x) = x2 + 5x + 4.

Now applying factor theorem.

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

f(-1) = 1 – 5 + 4 = 0

x + 1 is a factor of f(x).

Question 7: Given a polynomial f(x) = x3 – 2x + 4. Factorize and find its roots.

Solution:

f(x) = x3 – 2x + 4.

Now applying factor theorem.

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

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

Using this to factorize f(x)

f(x) = (x + 2)(x2 – 2x + 2)

x + 2 is a factor of f(x).

And x = -2 is a root.

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