Notes
Introduction :
Polynomial : Algebraic expression in which the power of the variable is a positive integer is called a polynomial.
Example :
(i) 4x3 + \(\sqrt{3}\)x2 - 8x + 13 is a polynomial. Here, 4, \(\sqrt{3}\), -8 and 13 are real numbers and the powers of x (the variable) are 3, 2, 1 and 0 which are positive integers. [13 means 13x°]
(ii) 4x3 + 5x + \(\frac{8}{x}\) is not a polynomial. Here, \(\frac{8}{x}\) means 8x-1. The power of x is a negative integer.
(iii) \(\sqrt{y}\) + 5 is not a polynomial, because \(\sqrt{y}\) + 5 = y1/2 + 5, here power of y is \(\frac{1}{2}\) which is not a whole number.
(iv) x3 - 7x2 + 5x + \(\sqrt{-2}\) is not a polynomial. Here, \(\sqrt{-2}\) is not a real number.
- A polynomial in x is denoted by p(x), that in y by p(y), etc. For example, p(x) = x3 + 2x2 + 5x - 3, q(m) = m2 + 12 m - 7, r(y) = y2 + 5
- 4, -3, 9, \(-\frac{1}{2}\), 0, (\sqrt{3}\) etc. are constant numbers. They can be called constant polynomials.
Types of Polynomials (based on the number of terms) :
Monomial : A polynomial having only one term is called a monomial. e.g. 5x2, - 5, 4xy.
Binomial : A polynomial having two terms is called a binomial. e.g. x3 + 9x, p2 + 6p, y3 - 3y2.
Trinomial : A polynomial having three terms is called a trinomial. e.g. x2 + 4x + 3, 2m3 + m2 - 7,
Degree of a polynomial in one variable :
(i) The degree of a polynomial in one variable is the highest power of the variable.
e.g. in the polynomial 2x3 - 7x5 + 4x2 + 7, the highest power of the variable x is 5.
∴ the degree of this polynomial is 5.
(ii) The degree of a non-zero constant polynomial is zero.
e.g. the degree of polynomial 12 is 0. Because 12 = 12 × 1 = 12 × x°.
(iii) The degree of zero polynomial is not defined.
Degree of a polynomial in more than one variable :
- The highest sum of the powers of variables in each term of the polynomial is the degree of the polynomial.
- g. the degree of the polynomial x3y6 + 3x2y + 3xy2 + y6 is 9. The sum of the powers of x and y in the term x3y6 is 3 + 6 = 9 which is the highest sum of the powers in the given polynomial.
Types of polynomials (based on degree) :
A polynomial in one variable :
(i) Linear polynomial : A polynomial of degree 1 is called a linear polynomial.
- g. 6m + 3, 25y, etc.
- The general form of a linear polynomial in one variable is ax + b, where a and b are constants and a ≠
(ii) Quadratic polynomial : A polynomial of degree 2 is called a quadratic polynomial.
- g. 4z2 - 5z + 7, 5m2 + 3m - 2, etc.
- The general form of a quadratic polynomial in one variable is ax2 + bx + c, where a, b, c are constants and a ≠
(iii) Cubic polynomial : A polynomial of degree 3 is called a cubic polynomial.
- g. y3 - 3y2 + 5y - 1, 5m3 + 9m2 + 11m + 7, etc.
- The general form of a cubic polynomial in one variable is ax3 + bx2 + cx + d, where a, b, c, d are constants and a ≠
Polynomial : anxn + an - 1xn-1 + ... + a2x2 + a1x + a0 is a polynomial in x with degree n.
- an an - 1, ... , a2, a1, a0 are the coefficients and an ≠
Standard form, Index form and Coefficient form of the polynomial :
p (x) = 3x2 + 2x4 - 8. Write it in the descending power of the variable x.
2x4 + 3x2 - 8. This is called the standard form of the polynomial.
In this polynomial, the terms x3 and x are missing. It means the coefficients corresponding to these terms are zeros.
Now, write the polynomial including the missing terms. 2x4 + 0x3 + 3x2 + 0x - 8. This is called the index form of the polynomial.
This polynomial is represented as (2, 0, 3, 0, -8). This form of the polynomial is called the coefficient form of the polynomial.
To write a polynomial in the coefficient form proceed as follows :
Ex. 4x5 - 9x + 7x3 - 5.
| Write the given polynomial in standard form. | 4x5 + 7x3 - 9x - 5. |
| Write the missing terms, if any, at their proper positions. Take the coefficients of each missing term to be 0. | 4x5 + 0x4 + 7x3 + 0x2 - 9x - 5.
This is the index form.
|
| For ensuring that no term is left out, use the formula : number of terms = degree of the polynomial + 1 | (6 terms = degree 5 + 1) |
| Write the coefficients of all terms in round brackets and separate each of them by a comma (,). | (4, 0, 7, 0, - 9, - 5).
This is the coefficient form. |
| While converting coefficient form into index form, the degree of the polynomial should be less by 1 than the number of coefficients. | (Here, the degree of the polynomial is 6-1 = 5.) |
Solved Examples :
Q 1. Write the polynomial x3 + 3x - 5 in coefficient form.
Solution :
Step -1 : Write polynomial in standard form = x3 + 3x - 5
Step -2 : Write it in the index form by considering all the missing terms with coefficient zero. = x3 + 0x2 + 3x - 5
Step -3 : Write it in a coefficient form
∴ given polynomial in coefficient form is (1, 0, 3, - 5)
Q 2. (2, -1, 0, 5, 6) is the coefficient form of the polynomial. Represent it in index form.
Solution :
Coefficient form of the polynomial is (2, - 1, 0, 5, 6)
[While converting coefficient form into index form, the degree of the polynomial should be less by 1 than the number of coefficients. Here, the degree of the polynomial is 5 - 1 = 4.]
∴ index form of the polynomial is 2x4 - x3 + 0x2 + 5x + 6 i.e. 2x4 - x3 + 5x + 6
Operations on polynomials :
Addition and Subtraction of Polynomials :
The method of addition and subtraction of polynomials is similar to the addition and subtraction of algebraic expressions.
Study the following examples :
(1) Add : 3x2 - 7x + 5 and 7x2 + 12x - 9.
(3x2 - 7x + 5) + (7x2 - 12x - 9)
= 3x2 - 7x + 5 + 7x2 - 12x – 9 ... (Removing brackets)
= 3x2+ 7x2 -7x -12x + 5 – 9 ... (Collecting like terms)
= 10x2 - 19x – 4 ... (Adding like terms)
(2) Subtract : 2x + 3x2 - 7 from 5x2 + 4x - 3.
(5x2 + 4x - 3) - (2x + 3x2 - 7)
= 5x2 + 4x – 3 - 2x - 3x2 + 7
= 5x2 - 3x2 + 4x - 2x – 3 + 7 ... (Collecting like terms)
= 2x2 + 2x + 4 ... (Adding like terms)
(3) Multiply :(m2 - 5) × (m3 + 2m - 2)
(m2 - 5) × (m3 + 2m - 2)
= m2 (m3 + 2m - 2) - 5 (m3 + 2m - 2) ….(Each term of second polynomials is multiplied by each term of first polynomial)
= m5 + 2m3 - 2m2 - 5m3 - 10m + 10
= m5 + 2m3 - 5m3 - 2m2 - 10m + 10 ... (Collecting like terms)
= m5 - 3m3 - 2m2 - 10m + 10 ... (Adding like terms)
Here the degree of the product is 5.
[In example the degree of the polynomial m2 – 5 is '2' and the degree of the other polynomial m3 + 2m - 2 is '3'.
∴ the degree of the product m5 - 3m3 - 2m2 - 10m + 10 is '5'. '5 = 2 + 3'.]
(4) Divide : 2x2 + 4x - 3 by x + 1.
Method 1 :
∴ s(x) = p(x) x q(x) + r(x)
r(x) = 0 or degree of r(x) < 0.
Method 2 : Linear method of division.
Divide 2x2 + 4x - 3 by x + 1.
To get the term 2x2 multiply (x + 1) by 2x and add 2x. (to get 4x)
2x(x + 1) + 2x - 3 = 2x2 + 2x + 2x - 3
To get the term 2x multiply (x + 1) by 2 and subtract 5.
2(x+ 1) - 5 = 2x + 2 - 5
∴ (2x2 + 4x - 3) = 2x(x + 1) + 2(x + 1) – 5 = (x + 1)(2x + 2) – 5
| Euclid's division lemma :
If s(x) and p(x) are two polynomials such that degree of s(x) is greater than or equal to the degree of p(x) and after dividing s(x) by p(x) the quotient is q(x) then s(x) = p(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < 0. |
Synthetic division :
In the earlier part we studied the method of dividing a polynomial. Now, we study the division of a polynomial by another method known as synthetic division.
Study the following example carefully.
Example : Divide the polynomial (3x3 + 2x2 - 1) by (x + 2).
Solution : Let us write the dividend polynomial in the coefficient form.
Index form of the dividend polynomial is 3x3 + 2x2 - 1 = 3x3 + 2x2 + 0x - 1
∴ coefficient form of the given polynomial = (3, 2, 0, - 1)
Divisor polynomial = x + 2
Let us use the following steps for synthetic division.
Step-1: Draw one horizontal and one vertical line as shown alongside.
Step-2: Divisor is x + 2. Hence take opposite number of 2 which is -2
Write -2 to the left of the vertical line as shown.
Write the coefficient form of the dividend polynomial in the first row.
Write the first coefficient as it is in the third row.
Step-3: The product of 3 in the third row with divisor -2 is -6.
Write this -6 in the second row below the coefficient 2.
Addition of 2 and -6 which is -4, is to be written in the third row.
Similarly by multiplying and adding, last addition is the remainder, which is (- 17) and coefficient form of the Quotient is (3, - 4, 8).
∴ 3x3 + 2x2 - 1 = (x + 2)(3x2 - 4x + 8) - 17
This method is called the method of synthetic division.
The same division can be done by linear method of division as shown below.
3x3 + 2x2 - 1 = 3x2(x + 2) - 6x2+ 2x2 -1
= 3x2(x + 2) - 4x2 - 1
= 3x2(x + 2) - 4x2 -8x + 8x - 1
= 3x2 (x + 2) - 4x(x + 2) + 8x - 1
= 3x2(x + 2) - 4x(x + 2) + 8x + 16 - 16 - 1
= 3x2(x + 2) - 4x(x + 2) + 8(x + 2) - 17
∴ 3x3 + 2x2 - 1 = (x + 2)(3x2 - 4x + 8) - 17
Solved Example :
Q 1. Divide (2y4 - 3y3 + 5y - 4) ÷ (y - 1)
Solution :
Synthetic division : Dividend = 2y4 - 3y3 + 5y -4 = 2y4 - 3y3 + 0y2 + 5y - 4
Divisor = y - 1 Opposite of -1 is 1.
[Steps :
- Write the coefficient form of the dividend polynomial in the first row.
- Write the first coefficient as it is in the third row.
- The product of 2 in the third row with divisor 1 is 2.
- Write this 2 in the second row below the coefficient -3.
- Addition of 2 and -3 which is -1, is to be written in the third row.
Similarly by multiplying and adding, last addition is the remainder, which is (0) and coefficient form of the Quotient is (2, - 1, -1, 4)]
Coefficient form of the quotient is (2, -1, -1, 4) .
∴ Quotient = 2y3 - y2 - y + 4 and Remainder = 0
Linear method : 2y4 - 3y3 + 5y - 4 = 2y3(y - 1)+ 2y3 - 3y3 + 5y - 4
= 2y3(y - 1) - y2(y - 1) - y2 + 5y - 4
= 2y3(y - 1) - y2(y - 1) - y(y - 1) + 4y - 4
= (2y3 - y2 - y + 4)(y - 1)
Value of a polynomial :
If p(x) is the polynomial in x, then the value of the polynomial for x = a is written as p(a) and is obtained by substituting the value of x as a in the given polynomial.
If p(x) = x3 - 9x2 + 11x - 8, then the corresponding value of p(x) when x = 1 is obtained by substituting 1 for x.
p(1) = (1)3 - 9(1)2 +11(1) - 8 = 1 – 9 + 11 – 8 = 12 - 17 = - 5.
∴ p(1) = -5.
Remember :
To find the value of a polynomial for a given value of the variable, put the value in place of the variable in each term of the polynomial.
Remainder Theorem :
There is a relation between the values of p(x) for x = (-a × 1) i.e. p(-a) and the remainder when p(x) is divided by (x + a).
Solved example :
Q 1. Divide p(x) = (x2 + 7x + 15) by (x + 4)
Solution :
| Dividend = x2 + 7x + 15
Divisor = x + 4
The quotient is x + 3 and remainder is 3. ... (1) |
Synthetic division :
The coefficient form of x2 + 7x + 15 is (1, 7, 15), Divisor = x + 4. Opposite of 4 is -4
The coefficient form of the quotient is (1, 3) ∴ the quotient = x + 3. Remainder is 3. |
Now, we find the relation between the remainder and the value of the polynomial :
Dividend = x2 + 7x + 15. Divisor x + 4.
Substitute x = - 4 in the dividend
p(x) = x2 + 7x + 15
∴ p(-4) = (-4)2 +7(-4) + 15 = 16 – 28 + 15 = 31 – 28 = 3
∴ the value of the polynomials p (x) for x = -4 is 3. ... (2)
From (1) and (2),
the remainder when p (x) = x2 + 7x + 15 is divided by (x + a) i.e. (x + 4) and the value of p (x) for x = - 4, i.e. p(-4), both are equal.
∴ we get the following property :
If the polynomial p (x) is divided by (x + a), then the remainder is p (-a) means it is the same as the value of the polynomial p(x) for x = -a.
This is known as the Remainder Theorem.
Remember : If a polynomial p(x) is divided by (x + a) then the remainder is p(-a) where ‘a’ is a real number.
Activity :
Verify the following examples.
(1) Divide p(x) = 3x2 + x + 7 by x + 2. Find the Remainder.
Solution :
Answer is : The remainder is 17.
(2) Find the value of p(x) = 3x2 + x + 7 when x = - 2.
Solution :
p(x) = 3x2 + x + 7
∴ p(-2) =3(-2)2 + (-2) + 7 = 3(4) – 2 + 7 = 12 – 2 + 7 = 17
Answer is : p(x) = 17.
(3) See whether remainder obtained by division is same as the value of p(-2). Take one more example and verify.
Solution :
Yes, The remainder obtained by division is the same as the value of p (-2).
[**Students should take one more example and verify.]
Factor Theorem :
If p(x) is a polynomial of degree n≥ 1 and a is any real number, then (x -a) is a factor of p(x), if the remainder p(a) = 0.
Conversely, if p(a) = 0, then (x - a) is a factor of p(x).
e.g. divide (x3 - 9x2 + 27x - 27) by (x - 3).
(a) By synthetic division :
Quotient = x2 - 6x + 9; Remainder = 0
∴ by the factor theorem,
(x - 3) is a factor of p(x).
∴ (x3 - 9x2 + 27x - 27) = (x - 3)(x2 - 6x + 9).
(b) By remainder theorem :
p (x) = x3 - 9x2 + 27x - 27
∴ p(3) = (3)3 - 9(3)2 + 27(3) – 27 = 27 – 81 + 81 – 27 = 0
∴ (x -3) is a factor of p (x).
Remember :
p (x) is a polynomial and a is any real number, and if p(a) = 0, then (x - a) is the factor of p (x).
Conversely, if (x - a) is the factor of the polynomial p(x), then p(a) = 0.
Activity :
Q 1. Verify that (x - 1) is a factor of the polynomial x3 + 4x - 5.
Solution :
p(x) = x3 + 4x - 5
∴ p(1) = (1)3 +4(1) – 5 =1 + 4 – 5 = 0
∴ p(1) = 0
i.e. the remainder is zero.
∴ (x - 1) is a factor of the polynomial x3 + 4x - 5.
Q 2. Govind has borrowed rupees one lakh twenty, five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a.
He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x2 was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 on cultivation and wages for cotton and tur crop.
His total income was rupees = 14000x2 + \(\frac{25000}{3}\)y2 + 16000y
By taking x = 2, y = 3 write the income-expenditure account of Govind's farming.
Solution :
| Credit (Income)
₹ 1,25,000 Bank loan |
Debit (Expenses)
₹ 1,37,000 loan paid with interest for seeds |
| Income from soyabean 16
14000x2 = 14000(2)2 = 14000 x 4 = ₹ 56,000 |
For seeds
₹ 10,000 |
| Income from cotton
\(\frac{25000}{3}\)y2 = \(\frac{25000}{3}\)(3)2 = ₹ 75,000 |
Fertilizers and pesticides for soyabean
2000x = 2000(2) = ₹ 4,000 |
| Income from tur
16000y = 16000(3) = ₹ 48,000 |
Wages and cultivation charges for soyabean
4000x2 = 4000(2)2 = ₹ 16,000 |
| Total income = ₹ 1,79,000 | Fertilizers and pesticides for cotton & tur
8000y = 8000(3) = ₹ 24,000 |
| Wages and cultivation charges for cotton & tur
9000y2 = 9000(3)2 = ₹ 81,000 |
|
| Total expenditure = ₹ 1,35,000 |
Factors of polynomials :
Sometimes polynomial can be written in the form ax2 + bx + c and hence it is easy to find its factors.
Solved Examples :
Q.1 Factorise : (y2 - 3y)2 - 5(y2 - 3y) - 50.
Solution :
Factorization of polynomials reducible in the form ax2 + bx + c.
Let (y2 - 3y) = x
∴ (y2 - 3y)2 - 5(y2 - 3y) - 50 = x2 - 5x - 50
= x2 - 10x + 5x - 50
= x(x - 10) + 5(x - 10)
= (x - 10) (x + 5)
= (y2 - 3y - 10) (y2 - 3y + 5)
= [y2 - 5y + 2y - 10] (y2 - 3y + 5)
= [y(y - 5) + 2(y - 5)] (y2 - 3y + 5)
= (y - 5)(y + 2) (y2 - 3y + 5)
Q.2 Factorise : 3x2 - 2x - 8
Solution :
| 3x2 - 2x - 8
= 3x2 - 6x + 4x – 8 =3x(x - 2) + 4(x - 2) = (x - 2)(3x + 4) |
 |
Q.3 Factorise : 6x2 - 13x + 6
Solution :
| 6x2 - 13x + 6
= 6x2 - 4x - 9x + 6 = 2x(3x - 2) - 3(3x - 2) = (3x - 2)(2x - 3) |
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