

2 :

Algebraic long division Divide 2x³ + 3x²  x + 1 by x + 2 x + 2 is the divisor The quotient will be here. 2x³ + 3x²  x + 1 is the dividend 

3 :

Algebraic long division First divide the first term of the dividend, 2x³, by x (the first term of the divisor). This gives 2x². This will be the first term of the quotient. 

4 :

Algebraic long division Now multiply 2x² by x + 2 and subtract 

5 :

Algebraic long division Bring down the next term, x. 

6 :

Algebraic long division Now divide –x², the first term of –x²  x, by x, the first term of the divisor which gives –x. 

7 :

Algebraic long division Multiply –x by x + 2 and subtract 

8 :

Algebraic long division Bring down the next term, 1 

9 :

Algebraic long division Divide x, the first term of x + 1, by x, the first term of the divisor which gives 1 

10 :

Algebraic long division Multiply x + 2 by 1 and subtract 

11 :

Algebraic long division The remainder is –1. The quotient is 2x²  x + 1 

12 :

Dividing in your head Divide 2x³ + 3x²  x + 1 by x + 2 When a cubic is divided by a linear expression, the quotient is a quadratic and the remainder, if any, is a constant. Let the remainder be d. Let the quotient by ax² + bx + c 2x³ + 3x²  x + 1 = (x + 2)(ax² + bx + c) + d 

13 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(ax² + bx + c) + d The first terms in each bracket give the term in x³ x multiplied by ax² gives ax³ so a must be 2. 

14 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x² + bx + c) + d The first terms in each bracket give the term in x³ x multiplied by ax² gives ax³ so a must be 2. 

15 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x² + bx + c) + d Now look for pairs of terms that multiply to give terms in x² x multiplied by bx gives bx² bx² + 4x² must be 3x² 2 multiplied by 2x² gives 4x² so b must be 1. 

16 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x² + 1x + c) + d Now look for pairs of terms that multiply to give terms in x² x multiplied by bx gives bx² bx² + 4x² must be 3x² 2 multiplied by 2x² gives 4x² so b must be 1. 

17 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x²  x + c) + d Now look for pairs of terms that multiply to give terms in x x multiplied by c gives cx cx  2x must be x 2 multiplied by x gives 2x so c must be 1. 

18 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x²  x + 1) + d Now look for pairs of terms that multiply to give terms in x x multiplied by c gives cx cx  2x must be x 2 multiplied by x gives 2x so c must be 1. 

19 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x²  x + 1) + d Now look at the constant term 2 multiplied by 1 gives 2 2 + d must be 1 then add d so d must be 1. 

20 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x²  x + 1)  1 Now look at the constant term 2 multiplied by 1 gives 2 2 + d must be 1 then add d so d must be 1. 

21 :

Dividing in your head 2x³ + 3x²  x + 1 = (x + 2)(2x²  x + 1)  1 The quotient is 2x²  x + 1 and the remainder is –1. 

22 :

DEGREE OF THE POLYNOMIALS The degree of a polynomial is the highest degree for a term with nonzero coefficient in a polynomial expressed in canonical form (i.e. as a sum or difference of terms). The degree of a term is the sum of the powers of each variable in the term. The word degree has for some decades been favored in standard textbooks. In some older books, the word order is used.


23 :

For example, the polynomial 7x2y3 + 4x  9 has three terms. (Notice, this polynomial can also be expressed as 7x2y3 + 4x1y0  9x0y0.) The first term has a degree of 5 (the sum of 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5 which is the highest degree of any term 

24 :

QUESTIONS Question: What is the degree of the polynomial 2 x9 + 7 x3 + 191? Answer: 2 x9 Question: What is the largest number of real roots that a 7th degree polynomial could have? What is the smallest?
Answer: A seventh degree polynomial has at least one and at most 7 real roots. 

25 :

Question: What is the large scale behavior of f(x) = x6+ ax5 + bx4+ cx3+ dx2+ ex + g?
Answer: As x becomes very large positive or very large negative f(x) becomes more and more negative.
Question: What are the roots of g(x) = 3x2  7x  20?
Answer: 5/3 and 4 

26 :

Evaluate the following using suitable identities:
(99)3
= (100  1)3
= (100)3  (1)3  3(100) (1) (100  1)
= 1000000  1  300(99)
= 1000000  1  29700
= 970299
(ii) (102)3
= (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600 (102)
= 1000000 + 8 + 61200
= 1061208 

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This presentation is presented by Roll no. 3639. 36SOURAV DUBEY
37SHAIKH MUNNIRUDDIN
38UJJWAL MAHATO
39YASH AGARWAL 
