ML AGGARWAL CLASS 8 CHAPTER 10 Algebraic Expressions and Identities Exercise 10.1
Exercise 10.1
Question 1
Sol :
(i) Given expression 12x2yz−4xy2
Terms
12xNyz
−4xy2
Numerical coefficient
12 , -4
Literal Coefficient
x2yz
xy2
(ii) Given expression 8+mn+nl−1m
Terms
8mn , nl ,-lm
Numerical cofficient
8 ,1 , 1 ,-1
Literal Coefficient
- ,mn ,nl ,lm
(iii) Given expression x23+46−xy2
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) 7x5−3xy→ Binomiol
(vi) 5p÷3q−3p2x q→ Binomial
Question 3
(i) 25x4−√3x2+5x−1
Itis polynomial of degree 4
(ii) 7x3−3x2+√5
due to −3x−2 term, It is not called as polynomial
∴ It is Not Polynomial
(iii) 4a3b2−3ab4+5ab+23
It is a polynomial of degree 5
(iv) 2x2y−3xy+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
ab−bc0+bc−ca−ab0+ca0+0+0
∴ab−bc+bc−ca+ca−ab=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
4x3−4x2+0x+9
+3x2+5x+4
7x3+0−11x+1
0+6x2−13x+0
-------------------------------------
10x3+2x2−29x+14
Question 5
Sol :
(i) 14a−5ab+7b−58a+3ab−2b+7(−)(−)(+)(−)6a−8ab+9b−12
(ii) 12xy−3yz−4zx+5xyz
8xy+4yz+5zx+0
(−)(−)(−)(−)
--------------------------------------
4xy−7yz−9zx+5xyz
(iii) 5p2q−2pq2+5pq−11q−3p+18
4p2q+3pq2−3pq+7q−8p−10
(−)(−)(+)(−)(+)(+)
-----------------------------------------------------------
pγq−7pq2+8pq−18q+5p+28
Question 6
Sol :
Horizontal method
3x2+5xy+7y2+3→1
2x2−4xy−3y2+7→(2)
9x2−8xy+11y2→(3)
(3) −[(1)+(2)]
9x2−8xy+11y2−[3x2+5xy+4y2+3+2x2−4xy−3y2+7]
9x2−8xy+11y2−[5x2+xy+4y2+10]
9x2−8xy+11y2−5x2−xy−4y2−10
4x2−9xy+7y2−10
Question 7
Sol :
Let 3a2−5ab−2b2−3−−−(1)
5a2−7ab−3b2+3a−−−(2)
do (2) - (1)
5av−7ab−3b2+3a
3av−5ab−2b2+0−3
(−)(+)(+)(−)(+)
---------------------------------------------------
2a2−2ab−b2+3a+3
Question 8
Sol :
Perimeter of triangle (p) = 7p2−5p+11→(1)
sides
s1=p2+2p−1→(2)
s2=3p2−6p+3→(3)
S3=?
P=S1+S2+S3S3=P−(S1+S2)
−7p2−5p+11−[p2+2p−1+3p2−6p+3]
=7p2−5p+11−[4p2−4p+2]
=7p2−5p+11−4p2+4p−2
S3=3p2−p+9
hence third side of triangle
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