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Published on: 18/09/2019
Factorisation
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Questions + Answers key
Take MCQ Mathematics Test
1.
Simplify: 4x2y2(3z - 24) \(\div \) 36xy(z - 8)
2.
Factorise: 81a4 - 16b4
3.
Factorise: 54x2 - 96y2
4.
Factorise the following expressions: a4 + 2a2b2 + b4
5.
Factorise the following expressions:(l + m)2 - 4lm
6.
Factorise the following expressions. 5x2y - 15xy2
7.
Factorize 70 in the prime factors.
8.
Factorise the following expressions and divide them (p2- 14p - 32)\(\div\) (p +2)
9.
Factorise the expressions and divide them as directed.39y3 (50y2- 98) \(\div\)26y2 (5y+7)
10.
Solve the following algebraic expressions by factorising. [(x-y)2 + (x -y) \(\div\) (x - y + 1)
11.
Solve the following expression:
(pr yz - r xyz - q r xy) ÷ x y z
12.
Find and correct the errors in the following mathematical statements. Substituting x =-3 in
x2- 5x+ 4 gives (-3)2- 5 (-3)+ 4= 9- 15 + 4=-2
13.
Find and correct the errors in the following mathematical statements. Substituting x =- 3 in
x2+ 5x+ 4 gives (-3)2 + 5 (-3) + 4 = 9 + 2 + 4 = 15
14.
Factorise 15xy
15.
if a2+b2 = 74 and ab = 35,then the value of a+b.
1.
We have 3z - 24 = 3(z - 8)
∴ 4x2y2(3z - 24) \(\div \) 36xy(z - 8)
= \(\frac { 4{ x }^{ 2 }y^{ 2 }(3z-24) }{ 36xy(z-8) }\)
\(=\frac { { 4x }^{ 2 }y^{ 2 }(3)(z-8) }{ 4xy\times 9(z-8) }\)
\(=\frac { xy }{ 3 } \)
2.
we have 81a4 - 16b4=(9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
[Using A2 - B2 = (A + B)(A - B)]
= (9a2 + 4b2)[(3a)2 - (2b)2]
= (9a2 + 4b2)[(3a + 2b)(3a - 2b)]
= (9a2 + 4b2)(3a + 2b)(3a - 2b)
3.
we have 54x2 - 96y2 =6[9x2 - 16y2]
=6[(3x)2 - (4y)2]
= 6[(3x + 4y)(3x - 4y)]
[Using a2 - b2 = (a + b)(a - b)]
Thus, 54x2- 96y2 = 6 (3x + 4y)(3x - 4y)
4.
we have a4 + 2a2b2 + b4
= (a2)2 + 2(a2)(b2) + b2)2
= (a2+b2)2 = (a2+b2)(a2 + b2)
∴ a4 + 2a2b2+b2
=(a2 + b2)2 = (a2 + b2)(a2 + b2)
5.
We have (l + m)2 - 4lm
=(l2 + 2lm + m2)-4lm
[Collecting the like terms 2lm and -4lm]
= l2+ (2lm - 4lm) + m2
= l2 + 2lm + m2
= (l2)+2(l)(m) + m2
= (l-m)2 = (l-m)(l-m)
∴ (l+m)2-4lm = (l-m)2 = (l-m)(l-m)
6.
∵ 5x2y=\(5\times x\times x\times y\)
\(=(5\times x\times y)[x]\)
\({ 15xy }^{ 2 }=5\times 3\times x\times y\times y\)
\(=(5\times x\times y)[3\times y]\)
\(\therefore 5{ x }^{ 2 }y-15x{ y }^{ 2 }=(5\times x\times y)[x-3\times y]\)
= 5xy(x - 3y)
7.
| 2 | 70 |
| 5 | 35 |
| 7 | 7 |
| 1 |
Hence, 2 X 5 X 7 = 70
8.
(p - 16)
9.
50y2 - 98 = 2(25y2 - 49)
= 2[(5y)2 - (7)2]
=2[(5y - 7)(5y + 7)]
\(\therefore \ \frac{39 y^{3}\left(50 y^{2}-98\right)}{26 y^{2}(5 y+7)}=\frac{3 \times 13 \times y \times y \times y \times 2 \times(5 y-7) \times(5 y+7)}{2 \times 13 \times y \times y \times(5 y+7)} \)
\(=\frac{3 \times y \times(5 y-7)}{1}\)
Thus, 39y3(50y2 - 98) / 26y2(5y + 7) = 3y(5y - 7)
10.
(x - y)
11.
\(\frac{pr}{x}-r-\frac{qr}{z}\)
12.
Given mathematical statement is incorrect because when we multiply two negative numbers, then their
product will be positive. Hence, correct statement is given below:
Substituting x = - 3 in x2 - 5x + 4 gives
(-3)2 -5(-3) + 4 = 9 + 15 + 4 = 28
13.
Given mathematical statement is incorrect because while substituting a negative value, we have to remember use of brackets.
Hence, correct statement is given below:
Substituting x = - 3 in x2 + 5x + 4 gives
(-3)2 + 5(-3) + 4
= 9 -15 + 4 = 13 -15 = - 2
14.
15xy = 3 x 5 \(\times \) x \(\times \) y
15.
12
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