(i) p2 + 6p + 9= (p+3)2
(ii) 4z2 – 4z + 1=(2z+1)2
(iii) 4p4 + 9p4 + 12p2q2=(2p2+ 3q2)2
(iv) (a4 - 8a2b2 + 16b4) - 81=(a2-4b2)2
(v) 625 – x2 – 2xy – y2= 625 -(x2+ 2xy + y2) =(25-x-y)(25+x+y)
(i) 5(x+1)
(ii) 3(x+5)
(iii) 2(p+q)
(iv) 5(y2 + 5y + 3)
(v) (x+1)(x+3)
(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)
(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)
(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)
Mean = Sum of Observation/Number of Obervations
Therefore
$M= \frac {x^4 -625}{x+5}= \frac { (x^2 +25)(x-5)(x+5)}{ x+5}=(x^2 +25)(x-5)$
Area of Triangle= (1/2) x Height X Base
Therefore
$A= \frac {1}{2} (x^4 + y^4) \times 14xy=7xy(x^4 + y^4)$
$p^4 -(p - q)^4 =[p^2 - (p-q)^2] [p^2 + (p-q)^2]=[p -(p-q)][p+(p-q)][p^2 + p^2 + q^2 -2pq]=q(2p-q)(2p^2 + q^2 -2pq)$
$A= \pi r^2$ 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)
$63p^2q^2 r^2 s – 9pq^2 r^2 s^2 + 15p^2qr^2 s^2 – 60p^2 q^2 rs^2= 3pqrs(21pqr - 3qrs + 5prs -20pqs)$
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.
