PROBABILITY CLASS 10 maths chapter 15 NCERT full solution

NCERT class 10 Mathematics
Chapter15– PROBABILITY  full solution 

Probability provides an introduction to the theoretical probability of an event. Probability are undoubtedly an essential study material for the students in CBSE class 10. We, brainy trainy provided here the meaning of probability, important basic tips regarding probability which can help the students to prepare for their exams. This chapter is a continuation of every class from 7th grade. We here, discuss the problems in easy language. NCERT is based on the CBSE syllabus. Board exam question paper contains total 6 marks out of 80 from probability. Students of 10th class can easily scored 6marks once the method or idea gets familiar. 

                 In class 9th , you all have studied about experimental probabilities of events which were based on the results of actual experiments. Example: Tossing a coin 500 times in which frequencies of the outcomes were as, Head : 200 times ; and Tail : 300 times.Based on this experiment, the experimental (empirical) probability of getting a head is 200/500=2/5 and  probability of getting a tail is 300/500 = 3/5.They are called experimental or empirical probabilities.We can written it as ; P(E) = number of trials in which the event happened/ total number of trials                      

The probability that we may fail in the struggle ought not to deter us from the support of a cause we believe to be just.

                     By Abraham Lincoln.         

         Probability chapter - 15 class 10

Probability—A theoretical probability  which is also called classical probability.

The theoretical probability of an event E, written as P( E ), is defined asP( E) = Number of favourable outcomes/total numbers of outcomes                     

Important points:

1) The sum of the probabilities of all the elementary events of an experiment is 1.
2) Probability never be negative.
3) The probability of an event E is a number P(E) such that 
 0 ≤ P(E) ≤ 1.
4) If P(E) = 1, then it is called a ‘Certain Event’ or ‘sure event’.
5) If P(E) = 0, then it is called an ‘Impossible Event’.
6) Complementary events : The event E representing ‘ not E ‘ , is called complement of the event E. So, we can say that E and  not E are complementary events .                 

   Sample space: A collection of all possible outcomes of an experiment is known as sample space. It is denoted by ‘S’ and represented in curly brackets.

Example of sample space: 
A coin is tossed once ;
  Event , E1 = getting a head outcome                                                 
  Event , E2 = getting a tail outcome 
Sample space ( S ) = { H , T }
Total number of outcomes = 2.

Important tips related to:

Coin: A fair coin which have two faces, one is head and other is tail.

Dice: A fair dice which have six faces, i.e. 0,1,2,3,4,5,6 which is used in games like Ludo , snake and ladder etc.

Playing cards: A pack of 52 cards which are divided into 4 suits of 13 cards each— spades, hearts, diamonds, and clubs.

1• Clubs and spades are of black colour.

2.Hearts and diamonds are of red colour.

3• Each suit are ace, king, queen, jack, 10, 9,8,7,6,5,4,3,2 . There is 13 cards of each suit = 13 *4=52 cards.

4• Kings, queens, jacks are called FACE CARDS.

5• First card - termed as ‘ACE ‘denoted as ‘A ‘ has 4 cards - each suit.

6• Jack card denoted as ‘J‘ , queen card denoted as‘Q’ , king card denoted as ‘K’ have  4 cards of each shape .i.e spades, hearts, diamonds, and clubs.

   NCERT full  solution chapter 15    

                          Exercise 15.1 

1. Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .

(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .

(iii) The probability of an event that is certain to happen is _________ . Such an event is called _________ .

(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .

(v) The probability of an event is greater than or equal to and less than or equal to __________.

Solution:

(i) Probability of an event E + Probability of the event ‘not E’ = 1.

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to Answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Solution:

(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

(ii) Even this statement does not have equally likely outcomes as the player may shoot or miss the shot.

(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.

(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be a boy or a girl.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution:

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

4. Which of the following cannot be the probability of an event?

(A) 2/3 

(B) -1.5 

(C) 15%

 (D) 0.7

Solution:

The probability of any event (E) always lies between 0 and 1 i.e. 0 ≤ P(E) ≤ 1. 

So, from the above options,  option (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Solution:   

 As we know that,

P(E) + P(not E) = 1    

  P(E) = 0.05 ( given )

So,   P(not E) = 1 – P(E)

Or,   P(not E) = 1 – 0.05

 Therefore,  P(not E) = 0.95

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

Solution:

(i) We know that the bag only contains lemon-flavoured candies.So, The no. of orange flavoured candies = 0

The probability of taking out orange flavoured candies = 0/1 = 0

(ii) As there are only lemon flavoured candies, 

 Therefore,   P(lemon flavoured candies) = 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

 Let the event where 2 students having the same birthday be E                Given, P(E) = 0.992    

  We know,       P(E) + P(not E) = 1                              

 Or, P(not E) = 1 – 0.992 = 0.008

The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red?   (ii) not red?

Solution:  

 The total number of balls = No. of red balls + No. of black balls  

   P So, the total no. of balls = 5 + 3 = 8    

    =>  P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8

(ii) Probability of drawing black balls =  P(black balls) = (no. of black balls/total no. of balls) = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be  

  (i) red? 

 (ii) white?  

  (iii) not green?

Solution:

The Total no. of balls = 5 + 8 + 4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red balls = 5

P (red ball) = 5/17 = 0.29

(ii) Total number of white balls = 8

P (white ball) = 8/17 = 0.47

(iii) Total number of green balls = 4

P (green ball) = 4/17 = 0.23

∴ P (not green) = 1 – P (green ball) = 1 – (4/7) = 0.77

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a 50 p coin?

(ii) will not be a ₹5 coin?

Solution: 

  Total no. of coins = 100 + 50 + 20 + 10 = 180

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

(ii) There are 100+50+20=170 coins other than ₹5 coin.

So, Favourable number of outcomes =170

∴ P (falling out of a coin other than ₹5) = 170/180 = 17/18

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

Solution:

The total number of fish in the tank = 5 + 8 = 13

Total number of male fish = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (male fish) = 5/13 = 0.38

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Solution:

Total number of possible outcomes = 8

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of favourable events (i.e. 8) = 1

∴ P (pointing at 8) = ⅛ = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = ½ = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

13. A die is thrown once. Find the probability of getting
(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Solution:

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Solution:

Total number of possible outcomes = 52

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total numbers of king of red colour = 2

P (getting a king of red colour) = 2/52 = 1/26 = 0.038

(ii) Total numbers of face cards = 12

P (getting a face card) = 12/52 = 3/13 = 0.23

(iii) Total numbers of red face cards = 6

P (getting a king of red colour) = 6/52 = 3/26 = 0.11

(iv) Total numbers of jack of hearts = 1

P (getting a king of red colour) = 1/52 = 0.019

(v) Total numbers of king of spade = 13

P (getting a king of red colour) = 13/52 = ¼ = 0.25

(vi) Total numbers of queen of diamonds = 1

P (getting a king of red colour) = 1/52 = 0.019

15.  Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is 

(a) an ace? 

(b) a queen?

Solution:

Total numbers of cards = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Numbers of queen = 1

P (picking a queen) = ⅕ = 0.2

(ii) If the queen is drawn and put aside, the total numbers of cards left is (5 – 4) = 4

(a) Total numbers of ace = 1

P (picking an ace) = ¼ = 0.25

(b) Total numbers of queen = 0

P (picking a queen) = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution:

Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132 + 12 = 144 pens

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P(picking a good pen) = 132/144 = 11/12 = 0.916

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

(i) Numbers of defective bulbs = 4

The total numbers of bulbs = 20

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2

(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19

So, the total numbers of events (or outcomes) = 19

Numbers of defective bulbs = 19 – 4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

Solution:    The total numbers of discs = 90

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of discs having two digit numbers = 81

(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 – 9 = 81)

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2

19. A child has a die whose six faces show the letters as given below: The die is thrown once. What is the probability of getting

  (i) A?    

 (ii) D?

Solution:

The total number of possible outcomes (or events) = 6

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) The total number of faces having A on it = 2

P (getting A) = 2/6 = ⅓ = 0.33

(ii) The total number of faces having D on it = 1

P (getting D) = ⅙ = 0.166

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

Solution:

First, calculate the area of the rectangle and the area of the circle. 

Here, the area of the rectangle is the possible outcome , 

and the area of the circle will be the favourable outcome.

So, the area of the rectangle = (3 × 2)sq m = 6 sq m

and,

The area of the circle = πr2 = π(½)2 m2= π/4 m2 = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 

or, 0.78/6 = 0.13

21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

Solution:

The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144 – 20 = 124

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total numbers events in which she will buy them = 124

So, P (buying) = 124/144 = 31/36 = 0.86

(ii) Total numbers events in which she will not buy them = 20

So, P (not buying) = 20/144 = 5/36 = 0.138

22. Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

Solution:

If 2 dices are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total numbers of events: 6 × 6 = 36

(i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) = 2/36

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution:

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = ¾ = 0.75

24. A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution:

Outcomes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total number of outcome = 6 × 6 = 36

    Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36 – (5 + 6)] = 25

∴ P(E) = 25/36

(ii) Number of events when 5 comes at least once = 11 (5 + 6)

∴ The required probability = 11/36

25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution:.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Solution:

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P (getting two heads) = ¼

and, P (getting one of the each) = 2/4 = ½

∴ This statement is incorrect.

(ii) Since the two outcomes are equally likely, this statement is correct.

Class 10 Maths Chapter 15 Exercise: 15.2 

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

Solution:

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5 × 5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = ⅕

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) = 8/25

(iii) P (both visiting on the different days) = 1 – P (both visiting on the same day)

So, P (both visiting on the different days) = 1 – ⅕ = ⅘

2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

Solution: 

         Complete table is as under:

Clearly total no of outcomes = 6*6 = 36

(i) E (Even) = 18

P (Even) = 18/36 = ½

(ii) E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

(iii) E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Solution:

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x + 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ P (drawing a blue ball) = [x/(x + 5)] ——–(i)

Similarly,

P (drawing a red ball) = [5/(x + 5)] ——–(i)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

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4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x

Solution:

Total number of black balls = x

Total number of balls = 12

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (getting black balls) = x / 12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x + 6

Now, P (getting black balls) = (x + 6)/18 ——————-(i)

Solving equation (i) and (ii)

x = 3

---

5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is ⅔. Find the number of blue balls in the jar.

Solution:

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24 – x

P(getting green marble) = x/24

From the question, x/24 = ⅔

So, the total green marbles = 16

And, the total blue marbles = 24 – x = 24 - 16 = 8 .

Polynomials class 10 / NCERT solutions for class 10 maths chapter 2 polynomials

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

Chapter with full detail information 

https://brainytrainy0183.blogspot.com gives you detail information regarding chapter. Here we provide you each topic with full detail ( with definitions, and examples). In this , you will  get definitions of each topic with detail examples and exercises 2.1 , 2.2 , 2.3 , and 2.4 (optional) solved. You can also comment us any further query. 

 Let’s start: 

                      Polynomials

What is polynomial: An algebraic expression p(x) is in the  form of p(x)=a₀+a₁x+a₂x²+………+a_nxⁿ , where the all  a’s are real numbers (sometimes called the coefficients of the polynomials) and all indices of x are non-negative integers, is called a polynomial in x.

Note: A polynomial in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of x in various terms.

Degree of polynomial: If the p(x) is a polynomial in x , the highest power of x in p(x) is called the degree of polynomial. For example: 5x-4 is a polynomial in the variable x of degree 1 , 4x²+3x+2 is a polynomial in the variable x of degree 2 .

Note: Expressions like 1/x-1 , √x+5 , etc. are not polynomials.

Different types of polynomials: 

1)                 Linear polynomial: A polynomial of highest degree 1 is called linear polynomial. It is in the form ax+b , where a,b are real numbers and a is non zero. For example: 4x-7 , 9x +4 , etc.

2)                 Quadratic polynomial: A polynomial of highest degree 2 is called quadratic polynomial. It is in the form ax²+bx+c , where a,b,c are real numbers and a is non- zero. For example: 2x²+5x-2=0 , 4x²+7x+4=0,etc.

3)  Cubic polynomial: A polynomial of highest degree 3 is called cubic polynomial. It  is in the form of ax³+bx²+cx+d , where a,b,c,d are real numbers and a is non- zero. For example: 4x³+3x²+5x+7=0 etc.

Zero of a polynomial: A real number k is called a zero of a polynomial p(x) , if p(k)=0.

i.e., a zero of a polynomial is the value of the variable for which the value of the polynomial becomes zero.

For example: 5x -10 = 0  or 5x=10 or x=2 , here we find value of x is 2 , so 2 is the zero of polynomial.

Note: A polynomial of degree n can have at the most n zeroes . So, a quadratic polynomial can have at the most 2 zeroes and a cubic polynomial can have at the most 3 zeroes.

How we locate Zeroes of a polynomial in graphically representation: The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y=p(x) intersects the x-axis.

Here, the graph cuts x-axis at two distinct points i.e A and A’ . Therefore, we say that above graph have two zeroes . 

                                           OR

 The number of zeroes is 2 as the graph intersects the x-axis at two points.

Relationship between Zeroes and coefficients of a polynomial: 

Ø Let’s say that α and β are the zeroes of a quadratic polynomial p(x)=ax²+bx+c, then 

Sum of zeroes = α+β= -b/a= -( coefficient of x)/ coefficient of x²

 Product of zeroes = αβ= c/a= constant term/coefficient of x²

Let’s say that α, βand γ are the zeroes of a cubic polynomial ax³+bx²+cx+d=0 , then 

Sum of zeroes = α+β+ γ = -b/a = -(coefficient of x²)/coefficient of x³

Sum and product of zeroes = αβ+ β γ+ γ α = c/a = coefficient of x / coefficient of x³

Product of zeroes = αβ γ = -d/a = -(constant term)/ coefficient of x³

Division Algorithm for polynomials: The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that 

P(x) = g(x)q(x) + r(x), where r(x) = 0 or degree r(x) <degree q(x).

     

                            Exercise 2.1

Question 1.) 

Solution:  

    i.         The number of zeroes is 0 the graph line is above x-axis and does not intersect the x- axis.

  ii.         The number of zeroes is one as the graph intersects the x-axis at one point only.

iii.         The number of zeroes is three as the graph intersects the x-axis at three points .

iv.         The number of zeroes is two as the graph intersects the x-axis at two points .

  v.         The number of zeroes is four as the graph intersects the

 x-axis at four points.

vi.         The number of zeroes is three as the graph intersects the x-axis at three points.

                  Exercise 2.2

Question 1.)  Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solutions:

(i)x²–2x –8

⇒x²– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x²–2x–8 are (4, -2)

Relationship:

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x²)

(ii)4s²–4s+1

⇒4s²–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s²–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s²)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s² )

(iii) 6x²–3–7x

⇒6x²–7x–3 = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x²–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x²)

(iv)4u²+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u² + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u²)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u²)

(v) t²–15

⇒ t² = 15 or t = ±√15

Therefore, zeroes of polynomial equation t² –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t²)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t² )

(vi) 3x²–x–4

⇒ 3x²–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x² – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x² )

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1

Solution:

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

X²–(α+β)x +αβ = 0

X²–(1/4)x +(-1) = 0

4x²–x-4 = 0

Thus,4x²–x–4 is the quadratic polynomial.

(ii)√2, 1/3

Solution:             Sum of zeroes = α + β =√2

                             Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

  X²–(α+β)x +αβ = 0

X² –(√2)x + (1/3) = 0

3x²-3√2x+1 = 0

Thus, 3x²-3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Solution:           Sum of zeroes = α+β = 0

                           Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

x²–(α+β)x +αβ = 0

x²–(0)x +√5= 0

 Thus ,x²+√5 is the quadratic polynomial.

(iv)1, 1

Solution.        Sum of zeroes = α+β = 1

                        Product of zeroes = α β = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

X²–(α+β)x +αβ = 0

X²–x+1 = 0

Thus , x²–x+1is the quadratic polynomial.

(v) -1/4, 1/4

Solution:            Sum of zeroes = α+β = -1/4

                             Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

X²–(α+β)x +αβ = 0

X²–(-1/4)x +(1/4) = 0

4x²+x+1 = 0

Thus,4x²+x+1 is the quadratic polynomial.

(vi) 4, 1

Solution:           Sum of zeroes = α+β =4

                           Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written as:

X²–(α+β)x+αβ = 0

X²–4x+1 = 0

Thus , x²–4x+1 is the quadratic polynomial.

Exercise 2.3 

Question 1.)   Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i)         p(x) = x³-3x²+5x–3 , g(x) = x²–2

  

Solution:         Dividend = p(x) = x³-3x²+5x–3

                       Divisor = g(x) = x²– 2

  Quotient = x–3

Remainder = 7x–9

ii) p(x) = x⁴-3x²+4x+5 , g(x) = x²+1-x

Solution: Given,

Dividend = p(x) = x⁴ – 3x² + 4x +5

Divisor = g(x) = x² +1-x= x²-x+1

Quotient = x² +x–3

Remainder = 8

(iii) p(x) =x⁴–5x+6, g(x) = 2–x²

Solution: Given,

Dividend = p(x) =x⁴ – 5x + 6 = x⁴ +0x²–5x+6

Divisor = g(x) = 2–x² = –x²+2

Quotient = -x²-2

Remainder = -5x + 10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t²-3, 2t⁴ +3t³-2t²-9t-12

Solutions: Given,

First polynomial = t²-3

Second polynomial = 2t⁴ +3t³-2t² -9t-12

As, the remainder is left as 0. Therefore, we say that, t²-3 is a factor of 2t⁴+3t³-2t²-9t-12.

(ii)x²+3x+1 , 3x⁴+5x³-7x²+2x+2

Solutions: Given,

First polynomial = x²+3x+1

Second polynomial = 3x⁴+5x³-7x²+2x+2

As , the remainder is left as 0. Therefore, we say that, x² + 3x + 1 is a factor of 3x⁴+5x³-7x²+2x+2.

(iii) x³-3x+1, x⁵-4x³+x²+3x+1

Solutions: Given,

First polynomial = x³-3x+1

Second polynomial = x⁵-4x³+x²+3x+1

As, the remainder is not equal to 0. Therefore, we say that, x³-3x+1 is not a factor of x⁵-4x³+x²+3x+1 .

3. Obtain all other zeroes of 3x⁴+6x³-2x²-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solutions:         Since two zeroes are √(5/3) and -√(5/3) are the factors of polynomial f(x).

∴ (x –√(5/3)) (x+√(5/3) = x²-(5/3) = 0

(3x²−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x²−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x⁴ +6x³ −2x² −10x–5 = (3x²-5)(x²+2x+1)

Now, on further factorizing (x²+2x+1) we get,

X²+2x+1 = x²+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.(Answer)

4. On dividing x³-3x²+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively. Find g(x).

Solutions:Given,Dividend, p(x) = x³-3x²+x+2

                              Quotient = x-2

                              Remainder = –2x+4

 To find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor×Quotient + Remainder

∴ x³-3x²+x+2 = g(x)×(x-2)+(-2x+4)

X³-3x²+x+2-(-2x+4) = g(x)×(x-2)

 X³-3x²+x+2+2x-4 = g(x) (x-2)

X³-3x²+3x-2=g(x) (x-2)

Now, for finding g(x) we will divide x³-3x²+3x-2 with (x-2)

Therefore, g(x) = (x2–x+1)

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solutions: According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor×Quotient+Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

(i):deg p(x) = deg q(x)

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, 3x²+3x+3 is a polynomial to be divided by 3.

So, (3x²+3x+3)/3 = x²+x+1 = q(x)

Thus, you can see, the degree of quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied here.

(ii)deg q(x) = deg r(x)

Let us take an example , p(x)=x²+x is a polynomial to be divided by g(x)=x.

So, (x²+x)/x = x+1 = q(x)

Also, remainder, r(x) = 0

Thus, you can see, the degree of quotient is equal to the degree of remainder.

Hence, division algorithm is satisfied here.

(iii)deg r(x) = 0

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x²+1 is a polynomial to be divided by g(x)=x.

So,( x²+1)/x= x=q(x)

And r(x)=1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.

Exercise 2.4

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³+x²-5x+2; -1/2, 1, -2

Solutions: Given, p(x) = 2x³+x²-5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = 2x³+x²-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³-4x²+5x-2 ;2, 1, 1

Solutions: Given, p(x) = x³-4x²+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2³-4(2)²+5(2)-2 = 0

p(1) = 1³-(4×1² )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = x³-4x²+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solutions: Let us consider the cubic polynomial is ax³+bx²+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x³-2x²-7x+14

3. If the zeroes of the polynomial x³-3x²+x+1 are a – b, a, a + b, find a and b.

Solutions: Given, p(x) = x³-3x²+x+1

                     zeroes are given as a – b, a, a + b 

Sum of zeroes = a – b + a + a + b

-b/a = 3a

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-1/1 = 1-b²

 Or , b² = 1+1 = 2

b = √2

Hence, a=1 and b= √2 (answer).

4. If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2 ±√3, find other zeroes.

Solutions: Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x⁴-6x³-26x²+138x-35

Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

 Let x= 2 ±√3

        X-2 =±√3

        Squaring on both sides,

       X²-4x+4 = 3

       X²-4x+1 = 0

X²-4x+1 is a factor of a given polynomial f(x).

Let us divide p(x) by X²-4x+1 to obtain other zeroes.

So, x⁴-6x³-26x²+138x-35 = (x²-4x+1)(x² –2x−35)

Now, on further factorizing (x²–2x−35) we get,

X²–(7−5)x −35 = x²– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:              

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

5. If the polynomial x⁴-6x³+16x²- 25x+10 is divided by another polynomial x²-2x+k , the remainder comes out to be x+a , find k and a.

Solution: let us divide x⁴-6x³+16x²-25x+10 by 

x²-2x+k. 

Now, remainder = (2k-9)x-(8-k)k+10………..(1)

But remainder = x+a ( given )……………….(2)

From (1) and (2) , we get;

(2k-9)x-(8-k)k+10 = x+a

On comparing their coefficients, we have 

2k-9 = 1 

K= 5

And  -(8-k)k+10 = a

            a= -(8-5)5+10

      Or,   a = -3*5+10

      Or,   a  = -15+10

      Or,   a = -5

Hence,     K = 5 and a = -5 ( answer).