Factorisation of Polynomials: Methods and Factor Theorem

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Factorisation of polynomials refers to the process of splitting a polynomial into its constituent factors. The product of these factors forms an algebraic expression. An algebraic expression consists of constants and variables. Depending upon the number of terms present in an algebraic expression, it is classified as monomial, binomial, trinomial, etc.

Table of Content

Polynomial
Methods for Factorisation of a Polynomial
Factor Theorem
Things to Remember
Sample Questions

Key Takeaways: Polynomials, Algebraic Expressions, Constant, Variable, Degree of Polynomial, Factor Theorem

Polynomial

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Polynomial is an algebraic expression containing one or more terms. These terms consist of constants and variables and perform addition, subtraction, multiplication, and division. A polynomial is denoted by P(x), where x means the variable.

Terms of a Polynomial

Terms of a Polynomial

Depending upon the number of terms present in an equation, a polynomial is mainly of three types:

Monomial- If an equation contains only one term, it is called a monomial.
Binomial- If an equation contains two terms, along with a ‘+’ or ‘-’ sign between them, it is called a binomial.
Trinomial- If an equation contains three terms, it is called a trinomial.

Types of Polynomial

Types of Polynomial

The degree of a polynomial is determined by the highest exponent present in the equation.

The video below explains this:

Polynomials Detailed Video Explanation:

Methods for Factorisation of a Polynomial

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Factorisation of polynomials can be carried out in the following ways:

Using Highest Common Factor (HCF)

In this method, the HCF of all the terms of the polynomial is determined. Then, each term is divided by the HCF. The quotient is kept in brackets and the common factors are written outside the bracket.

For example:

24ax - 40ay + 8a

= (8 × 3 × a × x) - (8 × 5 × a × y) + (8 × a)

= 8a (3x) - 8a (5y) + 8a

= 8a (3x - 5y + 1)

Regrouping

Sometimes, all the terms in an equation do not have a common factor. In such cases, the terms are grouped in a way such that each group has its common factor. Then, factorisation of each group is done and a factor common to each group is taken out.

For example:

5xy - 5x + 6 - 6y

= (5xy - 5x) + (6 - 6y)

= 5x (y-1) + 6 (1-y)

= 5x (y-1) - 6 (y-1)

= (5x - 6) (y - 1)

Read More: Polynomials Revision Notes

Using Identities

Some algebraic expressions form identities that can be solved to find out the factors of the polynomial. These identities are:

(a + b)²= a² + 2ab + b²
(a - b)²= a² - 2ab + b²
(a - b) (a + b) = a^₂- b²
(x + a) (x + b) = x² + (a +b)x + ab

For example:

49x²- 9y²

= (7x)² - (3y)²

= (7x + 3y) (7x - 3y)

Middle Term Splitting

In this method, factorisation is done by splitting the middle term of an equation into two different terms. These terms should be in a way that when multiplied together, it becomes equal to the product of the remaining two terms. Also, their addition or subtraction must give the original middle term.

For example:

x² - 8x + 12

= x² - 2x - 6x + 12

= x (x - 2) - 6 (x - 2)

= (x - 6) (x - 2)

By the Sum or Difference of Two Cubes

This is done with the help of following identities:

a³ + b³ = (a + b) (a² - ab + b²)
a³ - b³ = (a - b) (a² + ab + b²)

For example:

27x³ - 343/x³

(3x)³ - (7/x)³

(3x - 7/x) [ (3x)2 + 3x × 7/x + (7/x)² ]

(3x - 7/x) [ 9x2 + 21 + 49/x2 ]

Read Also:

Remainder Theorem	HCF and LCM
Quadratic Equations	Zeros of Polynomial

Factor Theorem

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According to the theorem, if f(x) is a polynomial with degree n, where n ≥ 1 and ‘a’ is any real number, then

(x-a) is a factor of f(x), if f(a) = 0
If (x-a) is a factor of a polynomial f(x), then f(a) = 0

Proof: Given that f(x) is a polynomial of degree n ≥ 1 by remainder theorem,

f(x) = (x - a) . q(x) + f(a) ……. eq.1

Let f(a) = 0

then, eq.1 becomes f(x) = (x - a) . q(x)

Thus, (x - a) is a factor of f(x).

Hence proved.

Now, let (x - a) is a factor of f(x).

So, f(x) = (x - a) . q(x) for a polynomial q(x).

∴, f(a) = (a - a) . q(a) = 0

Hence, f(a) = 0 when (x - a) is a factor of f(x).

In other words, the factor theorem states that when a polynomial f(x) is divided by p(x) and leaves remainder 0 then, we can say that p(x) is a factor of f(x).

Also Read:

Polynomials	Polynomials Formulas
Roots of polynomials	Degree of polynomials

Things to Remember

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The process of splitting a polynomial into its factors is called Factorisation of Polynomials.
Polynomials are denoted by P(x), where x is any variable.
Depending upon the number of terms present in a polynomial, it can be classified as: Monomial, Binomial and Trinomial.
Factorisation can be done by using different methods like regrouping, middle term splitting, using HCF, etc.
Factor theorem states that if a polynomial f(x) leaves remainder zero upon division with p(x), then p(x) is a factor of f(x).

Also Read:

Zeros of Polynomial	Roots of Polynomial	Degree of Polynomial

Sample Questions

Ques. Show that (x - 3) is a factor of the polynomial x³ - 3x²+ 4x - 12. (3 marks)

Ans. Let p(x) = x³- 3x² + 4x - 12

By factor theorem, (x-a) is a factor of p(x) if p(a) = 0.

So, p(3) = x³ - 3x² + 4x - 12

p(3) = 33 - 3(3)² + 4(3) - 12

p(3) = 27 - 27 + 12 - 12 = 0

Hence, (x - 3) is a factor of the polynomial x³ - 3x² + 4x - 12

Ques. Find the value of k, if x-1 is a factor of 4x³ + 3x² - 4x + k. (2 marks)

Ans. Since x-1 is a factor 4x³ + 3x²- 4x + k, so p(1) = 0

p(1) = 4(1)³ + 3(1)² - 4(1) + k

0 = 4 + 3 - 4 + k

k = -3

Ques. What must be added to x³– 3x² + 4x – 15 to obtain a polynomial which is exactly divisible by x – 3? (3 marks)

Ans. Let f (x) = x3 – 3x² + 4x – 15 & g(x) = x -3

Since the degree of the g(x) is 1. So the degree of the remainder is 0

Let “b” be added to f(x), so that it may be exactly divisible by g(x)

i.e f(x) = (x³ – 3x² + 4x -15) + b

Since f(x) is exactly divisible by (x-3)

So f(3) = 0

27 – 27 + 12 – 15 + b = 0

b = 3

i.e. g(x) is a factor of f(x) while add constant of 3 to f(x)

Ques. Find the value of “n” and ‘m” such that the polynomial x³ – nx² – 13x + m has (x + 1) and ( x – 2) as factors. (5 marks)

Ans. Let f(x) =x³ – nx² – 13x + m

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

i.e f(-1) = 0

So f ( -1) = – 1 -n +13 +m = 0

m -n = -12 – – – – – – – – ( 1)

Since (x -2) is a factor of f(x)

i.e f(2) = 0

So f ( 3) = 8 -4n – 26 + m = 0

m -4n = 18 – – – – – – – – – (2)

Now from equations (1) and (2)

n = -10 & m = -22

Ques. Find the value of the polynomial 5x – 4x² + 3 at (3 marks) [NCERT]
(i) x = 0
(ii) x = – 1
(iii) x = 2

Ans. let p(x) = 5x – 4x2 + 3

(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3

Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)2 + 3

= – 5x – 4x2 + 3 = -9 + 3 = -6

Thus, the value of 5x – 4x2 + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3

= 10 – 16 + 3 = -3

Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3.

Ques. Factorise 27x³ +y³ +z³-9xyz. (3 marks) [NCERT]

Ans. We have,

27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)

Using the identity,

x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

We have, (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)

= (3x + y + z) [(3x)³+ y³ + z³ – (3x × y) – (y × 2) – (z × 3x)]

= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx)

Ques. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz. (3 marks) [NCERT]

Ans. Since, x + y + z = 0

⇒ x + y = -z (x + y)³ = (-z)³

⇒ x³ + y³ + 3xy(x + y) = – z³

⇒ x³ + y³ + 3xy(-z) = -z³ [? x + y = -z]

⇒ x³ + y³ – 3xyz = – z³

⇒ x³ + y³ + z³ = 3xyz

Hence, if x + y + z = 0, then

x³ + y³ + z³ = 3xyz

Ques. Find the common factors of the following terms. (3 marks)
(1) 25x²y, 30xy²
(2) 63m³n, 54mn?

Ans.

25x²y, 30xy²

25x²y = 5 × 5 × x × x × y

30xy² = 2 × 3 × 5 × x × y × y

Common factors are 5× x × y = 5 xy

63m³n, 54mn?

63m³n = 3 × 3 × 7 × m × m × m × n

54mn? = 2 × 3 × 3 × 3 × m × n × n × n × n

Common factors are 3 × 3 × m × n = 9mn

Ques. Factorise the following polynomials. (2 marks)
(1) 6p(p – 3) + 1 (p – 3)
(2) 14(3y – 5z)3³ + 7(3y – 5z)²

Ans.

6p(p – 3) + 1 (p – 3) = (p – 3) (6p + 1)
14(3y – 5z)³ + 7(3y – 5z)²

= 7(3y – 5z)²[2(3y – 5z) +1]

= 7(3y – 5z)²(6y – 10z + 1)

Ques. Factorise the following polynomials. (3 marks)
(1) xy(z² + 1) + z(x² + y²)
(2) 2axy² + 10x + 3ay² + 15

Ans.

xy(z² + 1) + z(x² + y²)

= xyz² + xy + 2x² + zy²

= (xyz² + zx²) + (xy + zy²)

= zx(yz + x) + y(x + yz)

= zx(x + yz) + y(x + yz)

= (x + yz) (zx + y)

2axy² + 10x + 3ay² + 15

= (2axy² + 3ay²) + (10x + 15)

= ay²(2x + 3) +5(2x + 3)

= (2x + 3) (ay² + 5)

Also Read:

Real-Valued Function	Trigonometric Functions	Complex numbers and Quadratic Equations

Factorisation of Polynomials: Methods and Factor Theorem