8th Grade > Mathematics
ALGEBRAIC EXPRESSIONS AND IDENTITIES MCQs
Total Questions : 88
| Page 1 of 9 pages
Answer: Option B. -> x2+8x+15
:
B
To simplify: (x+3)(x+5)
Using the identity [(x+a)(x+b)2=x2+(a+b)x+ab]
we get,
(x+3)(x+5)=x2+(3+5)x+3×5=x2+8x+15
:
B
To simplify: (x+3)(x+5)
Using the identity [(x+a)(x+b)2=x2+(a+b)x+ab]
we get,
(x+3)(x+5)=x2+(3+5)x+3×5=x2+8x+15
Answer: Option A. -> 2x + 7
:
A
A polynomial which involves twoterms is called a binomial. For example, 3y + 9, 4a - 10 etc.
:
A
A polynomial which involves twoterms is called a binomial. For example, 3y + 9, 4a - 10 etc.
Answer: Option C. -> 48pq2
:
C
(4pq+3q)2−(4pq−3q)2
We have:
(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2
So,(4pq+3q)2−(4pq−3q)2
=[(4pq)2+(3q)2+2(4pq)(3q)]−[(4pq)2+(3q)2−2(4pq)(3q)]
=24pq2+24pq2
=48pq2
:
C
(4pq+3q)2−(4pq−3q)2
We have:
(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2
So,(4pq+3q)2−(4pq−3q)2
=[(4pq)2+(3q)2+2(4pq)(3q)]−[(4pq)2+(3q)2−2(4pq)(3q)]
=24pq2+24pq2
=48pq2
:
Using the identity
(a)2−(b)2=(a+b)(a−b) ,
(1.05)2−(0.95)2
=(1.05+0.95)(1.05−0.95)
=(2)(0.1)=0.2
Answer: Option D. -> x4−y4
:
D
To simplify :(x+y)(x−y)(x2+y2)
Usingthe identity,
(a+b)(a−b)=a2−b2
we get,
(x+y)(x−y)(x2+y2)
=(x2−y2)(x2+y2)
Using the same identity again,
wherea=x2andb=y2,
we get,
(x2−y2)(x2+y2)
=(x2)2−(y2)2
=x4−y4
:
D
To simplify :(x+y)(x−y)(x2+y2)
Usingthe identity,
(a+b)(a−b)=a2−b2
we get,
(x+y)(x−y)(x2+y2)
=(x2−y2)(x2+y2)
Using the same identity again,
wherea=x2andb=y2,
we get,
(x2−y2)(x2+y2)
=(x2)2−(y2)2
=x4−y4
Answer: Option C. -> −x5+x4+x3+3x−7
:
C
x4−6x3+x2−3x+1x5−7x3+x2−6x+8(−)(+)(−)(+)(−)−x5+x4+x3+3x−7
=(−x5+x4+x3+3x−7)
:
C
x4−6x3+x2−3x+1x5−7x3+x2−6x+8(−)(+)(−)(+)(−)−x5+x4+x3+3x−7
=(−x5+x4+x3+3x−7)
Answer: Option B. -> False
:
B
Expressions that contain exactly three terms are called trinomials. The given expression has four terms. Hence,the given statement is false.
:
B
Expressions that contain exactly three terms are called trinomials. The given expression has four terms. Hence,the given statement is false.
:
Application: 1Mark each
(i)403×397
=(400+3)(400−3)
=[(400)2−(3)2]
=160000−9
=159991
(ii)8.22
=(8+0.2)2
=(8)2+(0.2)2+2(8)(0.2)
=64+0.04+3.2
=67.24
(iii)9.7×9.8
=(9+0.7)(9+0.8)
=(9)2+(0.7+0.8)9+(0.7)(0.8)
=81+13.5+0.56
=95.06
(iv)22.92−7.12
=(22.9+7.1)(22.9−7.1)
=(30)(15.8)
=474
:
Application: 1Mark each
(i)(1.5x−4y)(1.5x+4y+3)
2.25x2+6xy+4.5x−6xy−16y2−12y
⇒2.25x2−16y2+4.5x−12y
(ii)(x+y)(2x+y)+(x+2y)(x−y)
(2x2+xy+2xy+y2)+(x2+xy−2y2)
3x2+4xy−y2
(iii)(a2+5)(b3+3)+5
a2b3+3a2+5b3+15+5
⇒a2b3+3a2+5b3+20
:
Application: 1 Mark each
(i)1002×1003
=(1000+2)(1000+3)⇒10002+(2+3)1000+(3)(2)
=1000000+5000+6
=1005006
(ii)97×103
=(100−3)(100+3)
=[(100)2−(3)2]
=[10000−9]
=9991
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