In this page we have Practice questions for Factorization Class 8 Chapter 14 . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Question 1
Factor the given expressions using identity:
(x+y)
2 = x
2 +y
2 +2xy
(x-y)
2 = x
2 +y
2 -2xy
x
2 –y
2 = (x-y) (x+y)
(i) p
2 + 6p + 9
(ii) 4z
2 – 4z + 1
(iii) 4p
4 + 9p
4 + 12p
2 q
2
(iv) (a
4 - 8a
2 b
2 + 16b
4 ) - 81
(v) 625 – x
2 – 2xy – y
2
Solution
(i) p2 + 6p + 9= (p+3)2
(ii) 4z2 – 4z + 1=(2z+1)2
(iii) 4p4 + 9p4 + 12p2 q2 =(2p2 + 3q2 )2
(iv) (a4 - 8a2 b2 + 16b4 ) - 81=(a2 -4b2 )2
(v) 625 – x2 – 2xy – y2 = 625 -(x2 + 2xy + y2 ) =(25-x-y)(25+x+y)
Question 2
Divide as directed.
(i) 5(x + 1) (2x + 5) ÷ (2x + 5)
(ii) 39x(x + 5) (y - 2) ÷ 13x(y - 2)
(iii) 12pq (p + q) (q + r) (r + p) ÷ 6pq(q + r) (r + p)
(iv) 60(y - 4) (y
2 + 5y + 3) ÷ 5(y - 4)
(v) x(x + 1) (x - 2) (x + 3) ÷ x(x - 2)
Solution
(i) 5(x+1)
(ii) 3(x+5)
(iii) 2(p+q)
(iv) 5(y2 + 5y + 3)
(v) (x+1)(x+3)
Question 3
Factorize the expressions.
(i) px² + qx
(ii) 7x² + 21y²
(iii) 2x³ + 2xy² + 2xz²
(iv) ap² + bq² + bp² + aq²
(v) nm + n
2 + m + n
(vi) y (y - z) + 16 (y - z)
(vii) 5y² – 20y – 8z + 2yz
(viii) 10xy + 4x+ 5y + 2
(ix) 6xy - 4y + 6 - 9x
Solution
(i) x(px+q)
(ii) 7(x²+3y²)
(iii) 2x(x²+y²+z²)
(iv) (a+b)(p²+q²)
(v) (n+1)(m+n)
(vi) (y-z)(y+16)
(vii) 5y² – 20y – 8z + 2yz=5y(y-4) +2z(y-4)=(y-4)(5y+2z)
(viii)10xy + 4x+ 5y + 2=2x(5y+2) + 1(5y+2)=(2x+1)(5y+2)
(ix) 6xy - 4y + 6 - 9x= 6xy-9x- 4y+6= 3x(2y-3)-2(2y-3)= (3x-2)(2y-3)
Question 4
Factor completely using the formula:
(i) 576 - p
2 + 8pq - 16q
2
(ii) 25x
2 - (y + z)
2
(iii) 4p
2 - 4p - 3
(iv) 81 - x
2 - y
2 - 2xy
(v) 121 - x
2 - y
2 - 2xy
Solution
(i) 576 - p2 + 8pq - 16q2 = 242 -(p2 - 8pq + 16q2 )=242 -(p-4q)2 =(24-p+4q)(24+p-4q)
(ii) 25x2 - (y + z)2 =(5x)2 -(y + z)2 =(5x-y-z)(5x+y+z)
(iii) 4p2 - 4p - 3=4p2 - 6p+2p - 3=2p(2p-3) +1(2p-3)=(2p+1)(2p-3)
(iv) 81 - x2 - y2 - 2xy= 92 -(x2 + y2 + 2xy)=92 -(x+y)2 =(9-x-y)(9+x+y)
(v) 121 - x2 - y2 - 2xy= 112 -(x2 + y2 + 2xy)=112 -(x+y)2 =(11-x-y)(11+x+y)
Question 5
Factorize the expressions:
(i) 4a
2 - 12ab + 9b
2
(ii) 36p
2 - 84pq + 49q
2
(iii) 4a
8 - b
8
(iv) (z - 1)
2 + 2 (z-1) (2z+3) + (2z +3)
2
(v) p
2 -9q
2
Solution
(i) 4a2 - 12ab + 9b2 =(2a-3b)2
(ii) 36p2 - 84pq + 49q2 =(6p-7q)2
(iii) 4a8 - b8 = (2a4 )2 - (b4 )2 =(a4 - b4 )(a4 + b4 )
=(a2 - b2 )(a2 + b2 )(a4 + b4 )
(iv) (z - 1)2 + 2 (z-1) (2z+3) + (2z +3)2 =[(z-1)+(2z+3)]2 =(3z+2)2
(v)p2 -9q2 =(p-3q)(p+3q)
Question 6
The sum of (x + 5) observations is x
4 - 625. Find the mean of the observations.
Solution
Mean = Sum of Observation/Number of Obervations
Therefore
$M= \frac {x^4 -625}{x+5}= \frac { (x^2 +25)(x-5)(x+5)}{ x+5}=(x^2 +25)(x-5)$
Question 7
The height of a triangle is x
4 + y
4 and its base is 14xy. Find the area of the triangle.
Solution
Area of Triangle= (1/2) x Height X Base
Therefore
$A= \frac {1}{2} (x^4 + y^4) \times 14xy=7xy(x^4 + y^4)$
Question 8
Factorise
p
4 -(p - q)
4
Solution
$p^4 -(p - q)^4 =[p^2 - (p-q)^2] [p^2 + (p-q)^2]=[p -(p-q)][p+(p-q)][p^2 + p^2 + q^2 -2pq]=q(2p-q)(2p^2 + q^2 -2pq)$
Question 9
The area of a circle is given by the expression πx
2 + 6πx + 9π. Find the radius of the circle.
Solution
$A= \pi r^2$ or
$r^2 =\frac {A}{\pi} = \frac {\pi x^2 + 6\pi x + 9 \pi}{\pi}=x^2 + 6x + 9=(x+3)^2$
$r=(x+3)
Question 10
Factorize the expressions.
63p
2 q
2 r
2 s - 9pq
2 r
2 s
2 + 15p
2 qr
2 s
2 -60p
2 q
2 rs
2
Solution
$63p^2q^2 r^2 s – 9pq^2 r^2 s^2 + 15p^2qr^2 s^2 – 60p^2 q^2 rs^2= 3pqrs(21pqr - 3qrs + 5prs -20pqs)$
Summary
This Practice questions for Factorization Class 8 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.
Download this Practice questions for Factorization as pdf
link to this page by copying the following text
Also Read
Notes
Worksheets
Ncert Solutions
Please enable JavaScript to view the comments powered by Disqus.
Class 8 Maths
Class 8 Science