- Constants and Variable
- |
- Polynomial expression
- |
- how to find the degree of a polynomial
- |
- Value of the polynomial
- |
- Zeros or roots of the polynomial
- |
- Adding Polynomials
- |
- subtracing Polynomials
- |
- Multiplying Polynomials
- |
- Dividing Polynomails
- |
- How to factor polynomials
- |
- Solved Examples Polynomials
- |

Multiplication of polynomial is simple task and we can follow below steps to multiply polynomials

1)Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e

2) We have to distribute each term of the first polynomial to every term of the second polynomial.

a) when you multiply two terms together you must multiply the coefficient (numbers) and add the exponents

b) Also as we already know ++ equals =, +- or -+ equals - and -- equals +

3) group like terms

So multiplication may involve

a) Multiplication of monomial to monomial

b) Multiplication of monomial to binomial, trinomial or more terms polynomials

c) Multiplication of binomial, trinomial or more terms polynomials to monomial

d) Multiplication of binomial to binomial, trinomial or more terms polynomials

e) Multiplication of trinomial to trinomial or more terms polynomials

**Multiple the Monomials**

- x
^{2}x (2x^{22}) x (4x^{26})

2 (-10xy^{3}/3) ×(6x^{3y}/5)

3. ( x) x (x^{2}) x (x^{3}) x (x^{8})

**Answer:**

We will use the below property extensively in above questions

x^{m} x x^{n} x x^{o} = x^{m+n+o}

1. As you know

So, we get

a^{2} x (2a^{22}) x (4a^{26}) = 8a^{48}

2. (2xy/3) ×(-9x^{2}y^{2}/10)

=(-3x^{3}y^{3}/5)

3. ( x) x (x^{2}) x (x^{3}) x (x^{8})

= x^{14}

**Multiply the binomials.**

i) (2x + 5) and (4x – 3)

(ii) (x – 8) and (3x – 4)

(iii) (2.5x – 0.5) and (2.5x + 0.5)

**Answer:**

Let ( a+b) (c+d) to be done

then

( a+b) (c+d)= a(c+d) + b( c+d)

=(a x c)+(a x d)+(b x c)+(b x d)

We will use the same concept in all the question below

1) (2x + 5)(4x - 3)

= 2x x 4x - 2x x 3 + 5 x 4x - 5 x 3

= 8x² - 6x + 20x -15

= 8x² + 14x -15

2) ( y - 8)(3y - 4)

= y **x** 3y - 4y - 8 **x** 3y + 32

= 3y^{2} - 4y - 24y + 32

= 3y^{2} - 28y + 32

3) (2.5l - 0.5m)(2.5l + 0.5)

Using (a+b)(a-b) = a^{2} - b^{2}

We get = 6.25l^{2} - 0.25m^{2}

**Multiply Binomial by Trinomial**

(3x + 2)(4x^{2} – 7x + 5)

**Answer**

=12x^{3} -21x^{2} +15x +8x^{2} -14x+10

=12x^{3} -13x^{2} +x+10

When a polynomial p(x) is divided by the polynomial g(x), we get quotient q(x) and remainder r(x)

p(x)=g(x).q(x)+r(x)

Notes

1) The degree of the reminder r(x) is always less then divisor g(x)

Now Let us see how to divide the polynomial by another non-zero polynomial

Steps to divide a polynomial by another polynomial. This is also called the long division method of polynomials

1) Arrange the term in decreasing order in both the polynomial

2) Divide the highest degree term of the dividend by the highest degree term of the divisor to obtain the first term

3) Now We multiply the divisor by the first term of the quotient, and subtract this product from the dividend

3) Similar steps are followed till we get the reminder whose degree is less than of divisor

Example Divide p(x) by g(x), where p(x) = x + 4x^{2} -1 and g(x) = 1 + x.

Solution

We carry out the process of division by means of the following steps:

Step 1 : We write the dividend x + 4x^{2} - 1 and the divisor 1 + x. in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 4x^{2} + x –1 and divisor is x + 1.

Step 2 :We divide the first term of the dividend by the first term of the divisor, i.e., we divide 4x^{2} by x, and get 4x. This gives us the first term of the quotient.

Step 3 :We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 4x^{2} + 4x from the dividend 4x^{2} + x - 1. This gives us the remainder as
-3x - 1.

Step 4 : We treat the remainder -3x - 1 as the new dividend. The divisor remains the same. We repeat Step 2 to get the next term of the quotient, i.e., we divide the first term -3x of the (new) dividend by the first term x of the divisor and obtain –3. Thus, –3 is the second term in the quotient.

Step 5 We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by - 3 and subtract the product -3x - 3 from the dividend - 3x - 1. This gives us 2 as the remainder

Now the degree of the reminder is less than degree of the divisor so process stops here

Complete division is illustrated below

Quotient is 3x -3 and Reminder is 2

Now 4x^{2} + x -1= (x+1)(4x-3) +2

Solution

We carry out the process of division by means of the following steps:

Step 1 : We write the dividend x + 4x

Step 2 :We divide the first term of the dividend by the first term of the divisor, i.e., we divide 4x

Step 3 :We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 4x

Step 4 : We treat the remainder -3x - 1 as the new dividend. The divisor remains the same. We repeat Step 2 to get the next term of the quotient, i.e., we divide the first term -3x of the (new) dividend by the first term x of the divisor and obtain –3. Thus, –3 is the second term in the quotient.

Step 5 We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by - 3 and subtract the product -3x - 3 from the dividend - 3x - 1. This gives us 2 as the remainder

Now the degree of the reminder is less than degree of the divisor so process stops here

Complete division is illustrated below

Quotient is 3x -3 and Reminder is 2

Now 4x

If p(x) is an polynomial of degree greater than or equal to 1 and p(x) is divided by the expression (x-a),then the reminder will be p(a)

Important notes

1) for (x-a) then remainder P(a)

2) for (x+a) => x -(-a),then remainder will be P(-a)

3) for (ax-b) => a(x-b/a) ,the remainder will be P(b/a)

4) for (ax+b) => a(x+b/a),the remainder will be P(-b/a)

5) for (b-ax)=> -a(x-b/a),the remainder will be P(b/a)

Class 9 Maths Home page Class 9 Science Home page

- NCERT Exemplar Problems: Solutions Mathematics Class 9
- IIT Foundation & Olympiad Explorer - Class 9 (Maths)
- Mathematics - Class 9 RD Sharma
- NCERT Solutions - Mathematics for Class IX
- Olympiad Excellence Guide for Mathematics (Class-9)
- MTG Foundation Course for JEE/Olympiads - Class 9 Maths
- Mathematics foundation course for Boards /JEE/PETs/ NTSE