ML AGGARWAL CLASS 8 CHAPTER 10 Algebraic Expressions and Identities Exercise 10.1

 Exercise 10.1

Question 1

Sol :
(i)  Given expression 12x2yz4xy2

Terms         
         
12xNyz

4xy2

Numerical coefficient 
12 , -4

Literal Coefficient 

x2yz

xy2


(ii) Given expression 8+mn+nl1m

Terms
8mn , nl  ,-lm


Numerical cofficient 
8 ,1 , 1 ,-1


Literal Coefficient 
- ,mn  ,nl ,lm


(iii) Given expression x23+46xy2

Terms

x23

y6

xy2


Numerical co-efficient 

13

16

-1


Literal co - efificient

x2

y

xy2


(iv) Given expression 4p+2.3q+1.7r

Terms
4P ,2.3q ,1.7r


Numerical co-efficient
-4 ,2.3 ,1.7

Literal co-efficient
P ,q ,r

Question 2

Sol :

(i) 5p×q×r2 Monomial

(ii) 3x2×y÷2z Monomial

(iii) 3+7x2 Binomial

(iv) 5a2+3b2+c2 Trinomial

(v) 7x53xy Binomiol

(vi) 5p÷3q3p2x q Binomial



Question 3


(i) 25x43x2+5x1

Itis polynomial of degree 4


(ii)  7x33x2+5

due to 3x2 term, It is not called as polynomial

It is Not Polynomial


(iii) 4a3b23ab4+5ab+23

It is a polynomial of degree 5


(iv) 2x2y3xy+5y3+3

due to negative power in the 3(xy)1

It is not a Polynomial


Question 4

Sol :

(i) Arrange terms for column method 

abbc0+bccaab0+ca0+0+0

abbc+bcca+caab=0



(ii) Arrange terms in columns 

5p2qv+4pq+7

2p2q2+9pq+3
---------------------------------
3pvq2+13pq+10



(iii) Arrange terms in columns 

l2+m2+n2+0+0+0

0+0+0+lm+mn+0

0+0+0+0+m n+nl

0+0+0+1 m+0+n l
---------------------------------------------------
l2+m2+n2+2lm+2mn+2nl



(iv) Arrange terms in columns 

4x34x2+0x+9

+3x2+5x+4

7x3+011x+1

0+6x213x+0
-------------------------------------
10x3+2x229x+14


Question 5

Sol :

 (i)  14a5ab+7b58a+3ab2b+7()()(+)()6a8ab+9b12

(ii) 12xy3yz4zx+5xyz

8xy+4yz+5zx+0

()()()()
--------------------------------------
4xy7yz9zx+5xyz



(iii) 5p2q2pq2+5pq11q3p+18


4p2q+3pq23pq+7q8p10

()()(+)()(+)(+)
-----------------------------------------------------------
pγq7pq2+8pq18q+5p+28

Question 6

Sol :
Horizontal method 

3x2+5xy+7y2+31

2x24xy3y2+7(2)

9x28xy+11y2(3)


(3) [(1)+(2)]

9x28xy+11y2[3x2+5xy+4y2+3+2x24xy3y2+7]

9x28xy+11y2[5x2+xy+4y2+10]

9x28xy+11y25x2xy4y210

4x29xy+7y210

Question 7

Sol :
Let 3a25ab2b23(1)

5a27ab3b2+3a(2)

do (2) - (1)

5av7ab3b2+3a

3av5ab2b2+03

()(+)(+)()(+)
---------------------------------------------------
2a22abb2+3a+3


Question 8

Sol :

Perimeter of triangle (p) = 7p25p+11(1)

sides 

s1=p2+2p1(2)

s2=3p26p+3(3)

S3=?

P=S1+S2+S3S3=P(S1+S2)

7p25p+11[p2+2p1+3p26p+3]

=7p25p+11[4p24p+2]

=7p25p+114p2+4p2

S3=3p2p+9

hence third side of triangle 

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