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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.
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
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
The degree of a polynomial is determined by the highest exponent present in the equation.
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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) = a2 - 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)
Read More: Polynomials important questions
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 ]
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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).
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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).
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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)
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