In this page we have Important questions for Factorization Class 8 Maths Chapter 14 . Hope you like them and do not forget to like , social share
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Question 1
User the below expression to solve the below problems
a
2 –b
2 = (a - b )(a + b)
(a) x
2 - 81
(b) 49a
2 - 36
(c) 121 – 49z
2
(d) 4x
2 - 9y
2
(e) 16x
2 - 225y
2
(f) 9x
2y
2 - 25
(g) 16x
2 - 81
(h) (a + b)
2 - 9c
2
(i) a
2 - (x – y )
2
(j) 4(x + y)
2 - x
2
(k) 36(a + b)
2 - 16(a - b)
2
(l) 20x
2 – 45y
2
(m) z
3 – 81z
(n) 12p
2 - 27
(o) 3z
5 – 27z
3
(p) 36a
2b
2 - 8
(q) z
2 - x
2 – 2x - 1
(r) 9x
2 - y
2 + 4y - 4
(s) p
2 – 2pq + q
2 - r
2
Solution
(a) x2 - 81 = x2 - 92= (x-9)(x+9)
(b) 49a2 - 36 =(7a)2 - 62= (7a-6)(7a+6)
(c) 121 – 49z2=(11-7z)(11+7z)
(d) 4x2 - 9y2=(2x)2-(3y)2=(2x-3y)(2x+3y)
(e) 16x2 - 225y2=(4x-15y)(4x+15y)
(f) 9x2y2 - 25=(3x-5)(3x+5)
(g) 16x2 - 81=(4x-9)(4x+9)
(h) (a + b)2 - 9c2=(a + b)2 - (3c)2 =(a+b-3c)(a+b+3c)
(i) a2 - (x – y )2=(a-x+y)(a+x-y)
(j) 4(x + y)2 - x2= [2(x+y)]2 - x2=[2(x+y) -x][2(x+y)+x]=(x+2y)(3x+2y)
(k) 36(a + b)2 - 16(a - b)2 =[6(a+b)]2 - [4(a-b)]2 =[6(a+b) -4(a-b)][6(a+b) +4(a-b)]=(2a+10b)(10a+2b)=4(a+5b)(5a+b)
(l) 20x2 – 45y2= 5(4x2 - 9y2) =5(2x-3y)(2x+3y)
(m) z3 – 81z=z(z2 - 81)=z(z-9)(z+9)
(n) 12p2 - 27= 3(4p2 - 9)=3(2p-3)(2p+3)
(o) 3z5 – 27z3=3z3(z2 -9) =3z3(z-3)(z+3)
(p) 36a2b2 - 8= 8( 4a2b2 -1) =8(2ab+1)(2ab-1)
(q) z2 - x2 – 2x - 1= z2 -( x2 +2x + 1)=z2 -(x+1)2=(z-x-1)(z+x+1)
(r) 9x2 - y2 + 4y - 4 =9x2 -(y2 - 4y + 4)=9x2 -(y-2)2= (3x-y+2)(3x+y-2)
(s) p2 – 2pq + q2 - r2=(p-q)2 - r2=(p-q-r)(p-q+r)
Question 2
User the below expression to solve the below problems
(a+b)
2 = (a
2 + b
2 +2ab)
(a-b)
2 = (a
2 + b
2 -2ab)
(a) p
2 + 4p + 4
(b) x
2 + 18x + 81
(c) 9 + 6z + z
2
(d) 9 + 6x + x
2
(e) z
2 + 6xz + 9x
2
(f) 9x
2 + 30x +25
(g) 36z
2 + 12z + 1
(h) 9x
2 + 30x + 25
(i) z
2 + z + 1/4
(j) 4z
2 – 6z + 9
(k) 4p
2 – 20p + 25
(l) 9x
2 - 30x + 25
(m) 1 – 2y + y
2
(n) 4 – 12y+ 9y
2
(o) x
2y
2 - 6xyz + 9z
2
(p) p
2 – 6pq + 9p
2
Solution
(a) p2 + 4p + 4 =(p+2)2
(b) x2 + 18x + 81= x2 + 2*9*x + 92= (x+9)2
(c) 9 + 6z + z2=(z+3)2
(d) 9 + 6x + x2=(x+3)2
(e) z2 + 6xz + 9x2=(z+3x)2
(f) 9x2 + 30x +25=(3x+5)2
(g) 36z2 + 12z + 1=(6z+1)2
(h) 9x2 + 30x + 25= (3x+5)2
(i) z2 + z + 1/4= (z+1/2)2
(j) 4z2 – 6z + 9= (2z-3)2
(k) 4p2 – 20p + 25= (2p+5)2
(l) 9x2 - 30x + 25=(3x-5)2
(m) 1 – 2y + y2= (y-1)2
(n) 4 – 12y+ 9y2= (3y-2)2
(o) x2y2 - 6xyz + 9z2=(xy-3z)2
(p) p2 – 6pq + 9p2=(p-3q)2
Question 3
Find and correct the errors in the statement: (3x)
2 + 5x = 9x + 5x = 14x
Solution
LHS = (2x)2 + 5x =4x2+5x
RHS = 9x + 5x = 14x
Third = 14x
LHS≠ RHS =third part
Correct statement would be
(3x)2 + 5x = 9x2+5x
Question 4
Find and correct the errors in the statement: (4x + 2)
2 = 2x
2 + 16x + 4
Solution
LHS = (4x + 2)2 =16x2 +16x+4
RHS = 2x2 + 16x + 4
LHS≠ RHS
Correct statement would be
(4x + 2)2 =16x2 +16x+4
Question 5
Find and correct the errors in the statement: (z - 1)
2 = z
2 - 1
Solution
LHS = (z - 1)2 =z2 -2z+1
RHS = z2 - 1
LHS≠ RHS
Correct statement would be
(z - 1)2 =z2 -2z+1
Question 6
Find and correct the errors in the statement: (y + 6)
2 = y
2 + 36
Solution
LHS=( y + 6)2 =y2 +12z+36
RHS = y2 + 36
LHS≠ RHS
Correct statement would be
( y + 6)2 =y2 +12z+36
Question 7
Verify that
(11m + 4n)
2 – (11mn – 4n)
2 = 176mn
2
Solution
LHS=(11m + 4n)2 – (11mn – 4n)2 = $121m^2 + 16n^2 + 88mn -(121m^2n^2 + 16n^2 -88mn^2)=121m^2 -121m^2n^2 + 88mn + 88mn^2$
RHS = 176mn2
LHS≠ RHS
.
So this is incorrect
Correct statement would be
$(11m + 4n)^2 -(11mn -4n)^2 =121m^2 -121m^2n^2 + 88mn + 88mn^2$
Question 8
True and false
(a) The value of (a + 1) (a – 1) (a
2 + 1) is a
4 + 1.
(b) x + y+ z is a monomial
(c) Some of the factors of $\frac {p^2}{2} + \frac {p}{2}$ are p,1/2 and (p+1)
(d)if xy=20 and (x- y) =1, the value (x+y) =9
Solution
(a) False
(b) False
(c) True
(d) True
Question 9
Factorise
$x^2 + \frac {1}{x^2} + 2 -3x - \frac{3}{x}$
Solution
$x^2 + \frac {1}{x^2} + 2 -3x - \frac{3}{x}= (x + \frac {1}{x})^2 - 3(x + \frac {1}{x})= (x+ \frac {1}{x}) (x+ \frac {1}{x} -3)$
Question 10
The Length and breadth of the rectangle is (x+y) and (x-y) respectively. What is the area of the rectangle
Solution
Area= Length x Breath
Therefore
Area is =$(x+y)(x-y)= x^2 -y^2$
Question 11
What should be added to 4c(-a + b + c) to obtain 3a (a + b + c) - 2b (a - b + c)?
Solution
Let y is to be Added, then
$y + 4c(-a + b + c)=3a (a + b + c) - 2b (a - b + c)$
$y=3a (a + b + c) - 2b (a - b + c) - 4c(-a + b + c)=3a^2 + 3ab + 3ac -2ab +2b^2 -2bc +4ac -4bc-4c^2=3a^2 + ab+ 7ac + 2b^2 - 6bc -4c^2$
Summary
This Important questions for Factorization Class 8 Maths Chapter 14 with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.
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