8th Standard Syllabus & Materials
8th Standard
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
Factorise 49x2 - 8xxy + 36y2
2.
Factorise : 49x2 - 64y2
3.
Factoris 7c2 + 2c - 5
4.
Factorize: 8p3+ q3
5.
Factorise: 81x3-3y3
1.
Now, 49x2 - 84xy + 36y2 = 72x2 - 84xy + 62y2
= (7x)2 - 2(7x)(6y) + (6y)2
Comparing this with a2 - 2ab + b2 = (a - b)2
(7x)2- 2(7x)(6y) + (6y)2 = (7x - 6y)2
\(\therefore\) 49x2- 84xy + 36y2 = (7x-6y)2
2.
Now, 49x2- 64y2 = 72x2 - 82y2
= (7x)2 - (8y)2
Comparing this with a2-b2 = (a + b)(a - b)
we get a = 7x, b = 8y
(7x)2- (8y)2 = (7x + 8y)(7x - 8y)
3.
This is in the form of ax2 + bx + c
We get a = 7,b = 2,c = −5
Now, the product = a x c = 7 x (−5) = –35 and sum b = 2
= 7c2 + 2c − 5
= (7c2 − 5c) + (7c − 5) (the middle term 2c can be written as – 5c + 7c)
= c(7c − 5) + 1(7c−5) (taking out the common factor 7c – 5 )
= (7c − 5)(c + 1)
Therefore, (7c–5), (c+1) are the two factors.
4.
We have 8p3+ q3
This can be written as = 23p3+ q3
= (2p)3+ q3
Comparing this with a3+ b3, we get a = 2p,b = q
We know a3+ b3 = (a+b) (a2−ab+b2)
(2p)3+ q3 = (2p+q)[(2p)2−(2p)(q)+q2]
8p3+ q3 = (2p+q) [4p2−2pq+q2]
5.
We have 81x3-3y3 = 3(27x3-y3)
= 3(33x3-y3)
= 3[(3x)3-y3]
Comparing this a3-b3,we get a = 3x, b = y
= 3{(3x-y)[(3x)2+(3x)(y)+y2]}
Therefore, 81x3-3y3 = 3{(3x-y)[9x2+3xy+y2]}
8th Standard Syllabus & Materials
8th Standard
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards
