Rd Sharma Solutions for Class 8 Math Chapter 7 Factorization are provided here with simple step-by-step explanations. These solutions for Factorization are extremely popular among Class 8 students for Math Factorization Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 7.12:

Question 1:

Factorize each of the following expressions:
qr − pr + qs − ps

Answer:

qr-pr+qs-ps =(qr-pr)+(qs-ps)  [Grouping the expressions]=r(q-p)+s(q-p)=(r+s)(q-p)              [Taking (q-p) as the common factor]

Page No 7.12:

Question 2:

Factorize each of the following expressions:
p2qpr2 − pq + r2

Answer:

p2q-pr2-pq+r2=(p2q-pq)+(r2-pr2)   [Grouping the expressions]=pq(p-1)+r2(1-p)=pq(p-1)-r2(p-1)      [(1-p)=-(p-1)]=(pq-r2)(p-1)              [Taking (p-1) as the common factor]

Page No 7.12:

Question 3:

Factorize each of the following expressions:
1 + x + xy + x2y

Answer:

1+x+xy+x2y=(1+x)+(xy+x2y)   [Grouping the expressions]=(1+x)+xy(1+x)=(1+xy)(1+x)           [Taking (1+x) as the common factor]

Page No 7.12:

Question 4:

Factorize each of the following expressions:
ax + ay − bx − by

Answer:

ax+ay-bx-by=(ax+ay)-(bx+by)   [Grouping the expressions]= a(x+y)-b(x+y)= (a-b)(x+y)              [Taking (x+y) as the common factor]

Page No 7.12:

Question 5:

Factorize each of the following expressions:
xa2 + xb2ya2yb2

Answer:

xa2+xb2-ya2-yb2=(xa2+xb2)-(ya2+yb2)   [Grouping the expressions]=x(a2+b2)-y(a2+b2)=(x-y)(a2+b2)                   [Taking (a2+b2) as the common factor]

Page No 7.12:

Question 6:

Factorize each of the following expressions:
x2 + xy + xz + yz

Answer:

x2+xy+xz+yz=(x2+xy)+(xz+yz)   [Grouping the expressions]=x(x+y)+z(x+y)=(x+z)(x+y)              [Taking (x+y) as the common factor]= (x+y)(x+z)

Page No 7.12:

Question 7:

Factorize each of the following expressions:
2ax + bx + 2ay + by

Answer:

2ax+bx+2ay+by=(2ax+bx)+(2ay+by)   [Grouping the expressions]=x(2a+b)+y(2a+b)=(x+y)(2a+b)                 [Taking (2a+b) as the common factor]

Page No 7.12:

Question 8:

Factorize each of the following expressions:
ab − by − ay + y2

Answer:

ab-by-ay+y2=(ab-ay)+(y2-by)              [Grouping the expressions]= a(b-y)+y(y-b)=a(b-y)-y(b-y)                  [(y-b)=-(b-y)]=(a-y)(b-y)                          [Taking (b-y) as the common factor]

Page No 7.12:

Question 9:

Factorize each of the following expressions:
axy + bcxy − az − bcz

Answer:

axy+bcxy-az-bcz=(axy+bcxy)-(az+bcz)       [Grouping the expressions]=xy(a+bc)-z(a+bc)=(xy-z)(a+bc)                       [Taking (a+bc) as the common factor]

Page No 7.12:

Question 10:

Factorize each of the following expressions:
lm2mn2lm + n2

Answer:

lm2-mn2-lm+n2=(lm2-lm)+(n2-mn2)   [Regrouping the expressions]                                  = lm(m-1)+n2(1-m)                                  =lm(m-1)-n2(m-1)      [(1-m)=-(m-1)]                                  =(lm-n2)(m-1)               [Taking (m-1) as the common factor]

Page No 7.12:

Question 11:

Factorize each of the following expressions:
x3y2 + xx2y2

Answer:

x3-y2+x-x2y2=(x3+x)-(x2y2+y2)         [Regrouping the expressions]=x(x2+1)-y2(x2+1)=(x-y2)(x2+1)                  [Taking (x2+1) as the common factor]

Page No 7.12:

Question 12:

Factorize each of the following expressions:
6xy + 6 − 9y − 4x

Answer:

6xy+6-9y-4x=(6xy-4x)+(6-9y)     [Regrouping the expressions]                             =2x(3y-2)+3(2-3y)                             =2x(3y-2)-3(3y-2)   [(2-3y)=-(3y-2)]                             =(2x-3)(3y-2)             [Taking (3y-2) as the common factor]

Page No 7.12:

Question 13:

Factorize each of the following expressions:
x2 − 2ax − 2ab + bx

Answer:

x2-2ax-2ab+bx=(x2-2ax)+(bx-2ab)    [Regrouping the expressions]=x(x-2a)+b(x-2a)=(x+b)(x-2a)                  [Taking (x-2a) as the common factor]=(x-2a)(x+b)

Page No 7.12:

Question 14:

Factorize each of the following expressions:
x3 − 2x2y + 3xy2 − 6y3

Answer:

x3− 2x2y + 3xy2− 6y3=(x3-2x2y)+(3xy2-6y3)           [Grouping the expressions]=x2(x-2y)+3y2(x-2y)=(x2+3y2)(x-2y)                         [Taking (x-2y) as the common factor]

Page No 7.12:

Question 15:

Factorize each of the following expression:
abx2 + (ay − b) x − y

Answer:

abx2+(ay-b)x-y=abx2+axy-bx-y                                  =(abx2-bx)+(axy-y)    [Regrouping the expressions]                                  =bx(ax-1)+y(ax-1)                                  =(bx+y)(ax-1)                [Taking (ax-1) as the common factor]

Page No 7.12:

Question 16:

Factorize each of the following expression:
(ax + by)2 + (bx − ay)2

Answer:

(ax+by)2+(bx-ay)2=a2x2+2abxy+b2y2+b2x2-2abxy+a2y2                                      =a2x2+b2y2+b2x2+a2y2                                      =(a2x2+a2y2)+(b2x2+b2y2)   [Regrouping the expressions]                                      =a2(x2+y2)+b2(x2+y2)                                      =(a2+b2)(x2+y2)                        [Taking (x2+y2) as the common factor]

Page No 7.12:

Question 17:

Factorize each of the following expression:
16(a − b)3 − 24 (a − b)2

Answer:

16(a-b)3-24(a-b)2=8(a-b)2[2(a-b)-3]         {Taking [8(a-b)2] as the common factor}=8(a-b)2(2a-2b-3)

Page No 7.12:

Question 18:

Factorize each of the following expression:
ab(x2 + 1) + x(a2 + b2)

Answer:

ab(x2+1)+x(a2+b2)=abx2+ab+a2x+b2x                                      =(abx2+a2x)+(b2x+ab)   [Regrouping the expressions]                                      =ax(bx+a)+b(bx+a)                                      =(ax+b)(bx+a)                   [Taking (bx+a) as the common factor]

Page No 7.12:

Question 19:

Factorize each of the following expression:
a2x2 + (ax2 + 1)x + a

Answer:

a2x2+(ax2+1)x+a=a2x2+ax3+x+a                                   =(ax3+a2x2)+(x+a)   [Regrouping the expressions]                                   =ax2(x+a)+(x+a)                                   =(ax2+1)(x+a)             [Taking (x+a) as the common factor]

Page No 7.12:

Question 20:

Factorize each of the following expression:
a(a − 2bc) + 2bc

Answer:

a(a-2b-c)+2bc=a2-2ab-ac+2bc                                =(a2-ac)+(2bc-2ab)             [Regrouping the terms]                                =a(a-c)+2b(c-a)                                =a(a-c)-2b(a-c)                   [(c-a)=-(a-c)]                                =(a-2b)(a-c)                            [Taking (a-c)  as the common factor]

Page No 7.12:

Question 21:

Factorize each of the following expression:
a(a + b − c) − bc

Answer:

a(a+b-c)-bc=a2+ab-ac-bc                            =(a2-ac)+(ab-bc)   [Regrouping the expressions]                            =a(a-c)+b(a-c)                            =(a+b)(a-c)               [Taking (a-c) as the common factor]

Page No 7.12:

Question 22:

Factorize each of the following expression:
x2 − 11xyx + 11y

Answer:

x2-11xy-x+11y=(x2-x)+(11y-11xy)   [Regrouping the expressions]                                 =x(x-1)+11y(1-x)                                 =x(x-1)-11y(x-1)       [(1-x)=-(x-1)]                                 =(x-11y)(x-1)               [Taking out the common factor (x-1)]

Page No 7.12:

Question 23:

Factorize each of the following expression:
ab − a − b + 1

Answer:

ab-a-b+1=(ab-b)+(1-a)   [Regrouping the expressions]                       =b(a-1)+(1-a)                         =b(a-1)-(a-1)   [(1-a)=-(a-1)]                       =(a-1)(b-1)         [Taking out the common factor (a-1)]

Page No 7.12:

Question 24:

Factorize each of the following expression:
x2 + y − xy − x

Answer:

x2+y-xy-x=(x2-xy)+(y-x)   [Regrouping the expressions]                        =x(x-y)+(y-x)                        =x(x-y)-(x-y)     [(y-x)=-(x-y)]                        =(x-1)(x-y)           [Taking (x-y) as the common expression]



Page No 7.17:

Question 1:

Factorize each of the following expression:
16x2 − 25y2

Answer:

16x2-25y2=(4x)2-(5y)2=(4x-5y)(4x+5y)

Page No 7.17:

Question 2:

Factorize each of the following expression:
27x2 − 12y2

Answer:

27x2-12y2=3(9x2-4y2)=3[(3x)2-(2y)2]=3(3x-2y)(3x+2y)

Page No 7.17:

Question 3:

Factorize each of the following expression:
144a2 − 289b2

Answer:

144a2-289b2=(12a)2-(17b)2=(12a-17b)(12a+17b)

Page No 7.17:

Question 4:

Factorize each of the following expression:
12m2 − 27

Answer:

12m2-27=3(4m2-9)=3[(2m)2-32]=3(2m-3)(2m+3)

Page No 7.17:

Question 5:

Factorize each of the following expression:
125x2 − 45y2

Answer:

125x2-45y2=5(25x2-9y2)=5[(5x)2-(3y)2]=5(5x-3y)(5x+3y)

Page No 7.17:

Question 6:

Factorize each of the following expression:
144a2 − 169b2

Answer:

144a2-169b2=(12a)2-(13b)2=(12a-13b)(12a+13b)

Page No 7.17:

Question 7:

Factorize each of the following expression:
(2a − b)2 − 16c2

Answer:

(2a-b)2-16c2=(2a-b)2-(4c)2=[(2a-b)-4c][(2a-b)+4c]=(2a-b-4c)(2a-b+4c)

Page No 7.17:

Question 8:

Factorize each of the following expression:
(x + 2y)2 − 4(2x − y)2

Answer:

(x+2y)2-4(2x-y)2=(x+2y)2-[2(2x-y)]2                                   =[(x+2y)-2(2x-y)][(x+2y)+2(2x-y)]                                   =(x+2y-4x+2y)(x+2y+4x-2y)                                   =5x(4y-3x)

Page No 7.17:

Question 9:

Factorize each of the following expression:
3a5 − 48a3

Answer:

3a5-48a3=3a3(a2-16)=3a3(a2-42)=3a3(a-4)(a+4)

Page No 7.17:

Question 10:

Factorize each of the following expression:
a4 − 16b4

Answer:

a4-16b4=a4-24b4=(a2)2-(22b2)2                                     =(a2-22b2)(a2+22b2)                                     =[a2-(2b)2](a2+4b2)                                     =(a-2b)(a+2b)(a2+4b2)

Page No 7.17:

Question 11:

Factorize each of the following expression:
x8 − 1

Answer:

x8-1=(x4)2-12=(x4-1)(x4+1)=[(x2)2-12](x4+1)=(x2-1)(x2+1)(x4+1)=(x2-12)(x2+1)(x4+1)=(x-1)(x+1)(x2+1)(x4+1)

Page No 7.17:

Question 12:

Factorize each of the following expression:
64 − (a + 1)2

Answer:

64-(a+1)2=(8)2-(a+1)2=[8-(a+1)][8+(a+1)]=(8-a-1)(8+a+1)=(7-a)(9+a)

Page No 7.17:

Question 13:

Factorize each of the following expression:
36l2 − (m + n)2

Answer:

36l2-(m+n)2=(6l)2-(m+n)2=[6l-(m+n)][6l+(m+n)]=(6l-m-n)(6l+m+n)

Page No 7.17:

Question 14:

Factorize each of the following expression:
25x4y4 − 1

Answer:

25x4y4-1=(5x2y2)2-1=(5x2y2-1)(5x2y2+1)

Page No 7.17:

Question 15:

Factorize each of the following expression:
a4-1b4

Answer:

   a1/b4
(a2)1/(b2)2
a2- 1/b2a2 1/b2
1/ba 1/ba2 1/b2

Page No 7.17:

Question 16:

Factorize each of the following expression:
x3 − 144x

Answer:

x3-144x=x(x2-144)=x(x2-122)=x(x-12)(x+12)

Page No 7.17:

Question 17:

Factorize each of the following expression:
(x - 4y)2 − 625

Answer:

(x-4y)2-625=(x-4y)2-252=[(x-4y)-25][(x-4y)+25]=(x-4y-25)(x-4y+25)

Page No 7.17:

Question 18:

Factorize each of the following expression:
9(a − b)2 − 100(x − y)2

Answer:

9(a-b)2-100(x-y)2=[3(a-b)]2-[10(x-y)]2=[3(a-b)-10(x-y)][3(a-b)+10(x-y)]=(3a-3b-10x+10y)(3a-3b+10x-10y)

Page No 7.17:

Question 19:

Factorize each of the following expression:
(3 + 2a)2 − 25a2

Answer:

(3+2a)2-25a2=(3+2a)2-(5a)2=[(3+2a)-5a][(3+2a)+5a]=(3+2a-5a)(3+2a+5a)=(3-3a)(3+7a)=3(1-a)(3+7a)

Page No 7.17:

Question 20:

Factorize each of the following expression:
(x + y)2 − (a − b)2

Answer:

(x+y)2-(a-b)2=[(x+y)-(a-b)][(x+y)+(a-b)]=(x+y-a+b)(x+y+a-b)

Page No 7.17:

Question 21:

Factorize each of the following expression:
116x2y2-449y2z2

Answer:

116x2y2-449y2z2=y2116x2-449z2=y214x2-27z2=y214x-27z14x+27z=y2x4-27zx4+27z

Page No 7.17:

Question 22:

Factorize each of the following expression:
75a3b2 - 108ab4

Answer:

75a3b2-108ab4=3ab2(25a2-36b2)=3ab2[(5a)2-(6b)2]=3ab2(5a-6b)(5a+6b)

Page No 7.17:

Question 23:

Factorize each of the following expression:
x5 − 16x3

Answer:

x5-16x3=x3(x2-16)=x3(x2-42)=x3(x-4)(x+4)

Page No 7.17:

Question 24:

Factorize each of the following expression:
50x2-2x281

Answer:

 50x2-2x281=225x2-x281=25x2-x92=25x-x95x+x9

Page No 7.17:

Question 25:

Factorize each of the following expression:
256x5 − 81x

Answer:

256x5-81x=x(256x4-81)=x[(16x2)2-92]=x(16x2+9)(16x2-9)=x(16x2+9)[(4x)2-32]=x(16x2+9)(4x+3)(4x-3)

Page No 7.17:

Question 26:

Factorize each of the following expression:
a4 − (2b + c)4

Answer:

a4-(2b+c)4=(a2)2-[(2b+c)2]2=[a2+(2b+c)2][a2-(2b+c)2]=[a2+(2b+c)2]{[a+(2b+c)][a-(2b+c)]}=[a2+(2b+c)2](a+2b+c)(a-2b-c)

Page No 7.17:

Question 27:

Factorize each of the following expression:
(3x + 4y)4x4

Answer:

(3x+4y)4-x4=[(3x+4y)2]2-(x2)2=[(3x+4y)2+x2][(3x+4y)2-x2]=[(3x+4y)2+x2][(3x+4y)+x][(3x+4y)-x]={(3x+4y)2+x2}(3x+4y+x)(3x+4y-x)=3x+4y2+x2(4x+4y)(2x+4y)=3x+4y2+x24(x+y)2(x+2y)=83x+4y2+x2(x+y)(x+2y)

Page No 7.17:

Question 28:

Factorize each of the following expression:
p2q2p4q4

Answer:

p2q2-p4q4=p2q2(1-p2q2)=p2q2[1-(pq)2]=p2q2(1-pq)(1+pq)

Page No 7.17:

Question 29:

Factorize each of the following expression:
3x3y − 243xy3

Answer:

 3x3y-243xy3=3xy(x2-81y2)=3xy[x2-(9y)2]=3xy(x-9y)(x+9y)

Page No 7.17:

Question 30:

Factorize each of the following expression:
a4b4 − 16c4

Answer:

a4b4-16c4=[(a2b2)2-(4c2)2]=(a2b2+4c2)(a2b2-4c2)=(a2b2+4c2)[(ab)2-(2c)2]=(a2b2+4c2)(ab+2c)(ab-2c)

Page No 7.17:

Question 31:

Factorize each of the following expression:
x4 − 625

Answer:

x4-625=(x2)2-252=(x2+25)(x2-25)=(x2+25)(x2-52)=(x2+25)(x+5)(x-5)

Page No 7.17:

Question 32:

Factorize each of the following expression:
x4 − 1

Answer:

 x4-1=(x2)2-1=(x2+1)(x2-1)=(x2+1)(x+1)(x-1)

Page No 7.17:

Question 33:

Factorize each of the following expression:
49(a − b)2 − 25(a + b)2

Answer:

49(a-b)2-25(a+b)2=[7(a-b)]2-[5(a+b)]2=[7(a-b)-5(a+b)][7(a-b)+5(a+b)]=(7a-7b-5a-5b)(7a-7b+5a+5b)=(2a-12b)(12a-2b)=2(a-6b)2(6a-b)=4(a-6b)(6a-b)

Page No 7.17:

Question 34:

Factorize each of the following expression:
x − yx2 + y2

Answer:

x-y-x2+y2=(x-y)+(y2-x2)               [Regrouping the terms]=(x-y)+(y+x)(y-x)=(x-y)-(y+x)(x-y)        [(y-x)=-(x-y)]=(x-y)[1-(y+x)]=(x-y)(1-x-y)

Page No 7.17:

Question 35:

Factorize each of the following expression:
16(2x − 1)2 − 25y2

Answer:

16(2x-1)2-25y2=[4(2x-1)]2-(5y)2=[4(2x-1)-5y][4(2x-1)+5y]=(8x-4-5y)(8x-4+5y)=(8x-5y-4)(8x+5y-4)

Page No 7.17:

Question 36:

Factorize each of the following expression:
4(xy + 1)2 − 9(x − 1)2

Answer:

4(xy+1)2-9(x-1)2=[2(xy+1)]2-[3(x-1)]2=[2(xy+1)-3(x-1)][2(xy+1)+3(x-1)]=(2xy+2-3x+3)(2xy+2+3x-3)=(2xy-3x+5)(2xy+3x-1)

Page No 7.17:

Question 37:

Factorize each of the following expression:
(2x + 1)2 − 9x4

Answer:

(2x+1)2-9x4=(2x+1)2-(3x2)2=[(2x+1)-3x2][(2x+1)+3x2]=(-3x2+2x+1)(3x2+2x+1)We can factorise the quadratic expressions in the curved brackets as:(-3x2+3x-x+1)(3x2+2x+1)=3x(-x+1)+1(-x+1)(3x2+2x+1)=(-x+1)(3x+1)(3x2+2x+1)=-(x-1)(3x+1)(3x2+2x+1)

Page No 7.17:

Question 38:

Factorize each of the following expression:
x4 − (2y − 3z)2

Answer:

 x4-(2y-3z)2=(x2)2-(2y-3z)2=[x2-(2y-3z)][x2+(2y-3z)]=(x2-2y+3z)(x2+2y-3z)

Page No 7.17:

Question 39:

Factorize each of the following expression:
a2b2 + a − b

Answer:

a2-b2+a-b=(a2-b2)+(a-b)            [Grouping the terms]                         =(a+b)(a-b)+(a-b)                         =(a-b)(a+b+1)             [Taking out the common factor (a-b)]

Page No 7.17:

Question 40:

Factorize each of the following expression:
16a4b4

Answer:

16a4-b4=(4a2)2-(b2)2=(4a2+b2)(4a2-b2)=(4a2+b2)[(2a)2-b2]=(4a2+b2)(2a+b)(2a-b)

Page No 7.17:

Question 41:

Factorize each of the following expression:
a4 − 16(b − c)4

Answer:

a4-16(b-c)4=(a2)2-[4(b-c)2]2=[a2+4(b-c)2][a2-4(b-c)2]=[a2+4(b-c)2]{a2-[2(b-c)]2}=[a2+4(b-c)2][a+2(b-c)][a-2(b-c)]=[a2+4(b-c)2](a+2b-2c)(a-2b+2c)

Page No 7.17:

Question 42:

Factorize each of the following expression:
2a5 − 32a

Answer:

2a5-32a=2a(a4-16)=2a[(a2)2-42]=2a(a2+4)(a2-4)=2a(a2+4)(a2-22)=2a(a2+4)(a+2)(a-2)=2a(a-2)(a+2)(a2+4)

Page No 7.17:

Question 43:

Factorize each of the following expression:
a4b4 − 81c4

Answer:

a4b4-81c4=(a2b2)2-(9c2)2=(a2b2+9c2)(a2b2-9c2)=(a2b2+9c2)[(ab)2-(3c)2]=(a2b2+9c2)(ab+3c)(ab-3c)

Page No 7.17:

Question 44:

Factorize each of the following expression:
xy9yx9

Answer:

xy9-yx9=xy(y8-x8)=xy[(y4)2-(x4)2]=xy(y4+x4)(y4-x4)=xy(y4+x4)[(y2)2-(x2)2]=xy(y4+x4)(y2+x2)(y2-x2)=xy(y4+x4)(y2+x2)(y+x)(y-x)

Page No 7.17:

Question 45:

Factorize each of the following expression:
x3x

Answer:

x3-x=x(x2-1)=x(x-1)(x+1)

Page No 7.17:

Question 46:

Factorize each of the following expression:
18a2x2 − 32

Answer:

 18a2x2-32=2(9a2x2-16)=2[(3ax)2-42]=2(3ax-4)(3ax+4)



Page No 7.22:

Question 1:

Factorize each of the following algebraic expression:
4x2 + 12xy +9y2

Answer:

    4x2+12xy+9y2=(2x)2+2×2x×3y+(3y)2=(2x+3y)2=(2x+3y)(2x+3y)

Page No 7.22:

Question 2:

Factorize each of the following algebraic expression:
9a2 − 24ab + 16b2

Answer:

    9a2-24ab+16b2=(3a)2-2×3a×4b+(4b)2=(3a-4b)2=(3a-4b)(3a-4b)

Page No 7.22:

Question 3:

Factorize each of the following algebraic expression:
p2q2 − 6pqr + 9r2

Answer:

    p2q2-6pqr+9r2=(pq)2-2×pq×3r+(3r)2=(pq-3r)2=(pq-3r)(pq-3r)

Page No 7.22:

Question 4:

Factorize each of the following algebraic expression:
36a2 + 36a + 9

Answer:

36a2+36a+9=9(4a2+4a+1)=9(2a)2+2×2a×1+12=9(2a+1)2=9(2a+1)(2a+1)



Page No 7.23:

Question 5:

Factorize each of the following algebraic expression:
a2 + 2ab + b2 − 16

Answer:

     a2+2ab+b2-16=a2+2×a×b+b2-16=(a+b)2-42=(a+b-4)(a+b+4)

Page No 7.23:

Question 6:

Factorize each of the following algebraic expression:
9z2x2 + 4xy − 4y2

Answer:

   9z2-x2+4xy-4y2=9z2-(x2-4xy+4y2)=9z2-[x2-2×x×2y+(2y)2]=(3z)2-(x-2y)2=[3z-(x-2y)][3z+(x-2y)]=(3z-x+2y)(3z+x-2y)=(x-2y+3z)(-x+2y+3z)

Page No 7.23:

Question 7:

Factorize each of the following algebraic expression:
9a4 − 24a2b2 + 16b4 − 256

Answer:

    9a4-24a2b2+16b4-256=(9a4-24a2b2+16b4)-256=[(3a2)2-2×3a2×4b2+(4b2)2]-162=(3a2-4b2)2-162=[(3a2-4b2)-16][(3a2-4b2)+16]=(3a2-4b2-16)(3a2-4b2+16)

Page No 7.23:

Question 8:

Factorize each of the following algebraic expression:
16 − a6 + 4a3b3 − 4b6

Answer:

    16-a6+4a3b3-4b6=16-(a6-4a3b3+4b6)=42-[(a3)2-2×a3×2b3+(2b3)2]=42-(a3-2b3)2=[4-(a3-2b3)][4+(a3-2b3)]=(4-a3+2b3)(4+a3-2b3)=(a3-2b3+4)(-a3+2b3+4)

Page No 7.23:

Question 9:

Factorize each of the following algebraic expression:
a2 − 2ab + b2c2

Answer:

    a2-2ab+b2-c2=(a2-2ab+b2)-c2=(a2-2×a×b+b2)-c2=(a-b)2-c2=[(a-b)-c][(a-b)+c]=(a-b-c)(a-b+c)

Page No 7.23:

Question 10:

Factorize each of the following algebraic expression:
x2 + 2x + 1 − 9y2

Answer:

    x2+2x+1-9y2=(x2+2x+1)-9y2=(x2+2×x×1+1)-9y2=(x+1)2-(3y)2=[(x+1)-3y][(x+1)+3y]=(x+1-3y)(x+1+3y)=(x+3y+1)(x-3y+1)

Page No 7.23:

Question 11:

Factorize each of the following algebraic expression:
a2 + 4ab + 3b2

Answer:

    a2+4ab+3b2=a2+4ab+4b2-b2=[a2+2×a×2b+(2b)2]-b2=(a+2b)2-b2=[(a+2b)-b][(a+2b)+b]=(a+2b-b)(a+2b+b)=(a+b)(a+3b)

Page No 7.23:

Question 12:

Factorize each of the following algebraic expression:
96 − 4xx2

Answer:

    96-4x-x2=100-4-4x-x2=100-(x2+4x+4)=100-(x2+2×x×2+22)=102-(x+2)2=[10-(x+2)][10+(x+2)]=(10-x-2)(10+x+2)=(8-x)(12+x)=(x+12)(-x+8)

Page No 7.23:

Question 13:

Factorize each of the following algebraic expression:
a4 + 3a2 +4

Answer:

    a4+3a2+4=a4+4a2-a2+4=(a4+4a2+4)-a2=[(a2)2+2×a2×2+22]-a2=(a2+2)2-a2=[(a2+2)-a][(a2+2)+a]=(a2-a+2)((a2+a+2)

Page No 7.23:

Question 14:

Factorize each of the following algebraic expression:
4x4 + 1

Answer:

    4x4+1=4x4+4x2+1-4x2=[(2x2)2+2×2x2×1+1]-4x2=(2x2+1)2-(2x)2=[(2x2+1)-2x][(2x2+1)+2x]=(2x2-2x+1)(2x2+2x+1)

Page No 7.23:

Question 15:

Factorize each of the following algebraic expression:
4x4 + y4

Answer:

    4x4+y4=4x4+4x2y2+y4-4x2y2=[(2x2)2+2×2x2×y+(y2)2]-(2xy)2=(2x2+y2)2-(2xy)2=[(2x2+y2)-2xy][(2x2+y2)+2xy]=(2x2-2xy+y2)(2x2+2xy+y2)

Page No 7.23:

Question 16:

Factorize each of the following algebraic expression:
(x + 2)2 − 6(x + 2) + 9

Answer:

   (x+2)2-6(x+2)+9=(x+2)2-2×(x+2)×3+32=[(x+2)-3]2=(x+2-3)2=(x-1)2=(x-1)(x-1)

Page No 7.23:

Question 17:

Factorize each of the following algebraic expression:
25 − p2q2 − 2pq

Answer:

    25-p2-q2-2pq=25-(p2+2pq+q2)=52-(p2+2×p×q+q2)=52-(p+q)2=[5-(p+q)][5+(p+q)]=(5-p-q)(5+p+q)=-(p+q-5)(p+q+5)

Page No 7.23:

Question 18:

Factorize each of the following algebraic expression:
x2 + 9y2 − 6xy − 25a2

Answer:

   x2+9y2-6xy-25a2=(x2-6xy+9y2)-25a2=[x2-2×x×3y+(3y)2]-25a2=(x-3y)2-(5a)2=[(x-3y)-5a][(x-3y)+5a]=(x-3y-5a)(x-3y+5a)

Page No 7.23:

Question 19:

Factorize each of the following algebraic expression:
49 − a2 + 8ab − 16b2

Answer:

    49-a2+8ab-16b2=49-(a2-8ab+16b2)=49-[a2-2×a×4b+(4b)2]=72-(a-4b)2=[7-(a-4b)][7+(a-4b)]=(7-a+4b)(7+a-4b)=-(a-4b-7)(a-4b+7)=-(a-4b+7)(a-4b-7)

Page No 7.23:

Question 20:

Factorize each of the following algebraic expression:
a2 − 8ab + 16b2 − 25c2

Answer:

 a2-8ab+16b2-25c2=(a2-8ab+16b2)-25c2=[a2-2×a×4b+(4b)2]-25c2=(a-4b)2-(5c)2=[(a-4b)-5c][(a-4b)+5c]=(a-4b-5c)(a-4b+5c)

Page No 7.23:

Question 21:

Factorize each of the following algebraic expression:
x2y2 + 6y − 9

Answer:

x2-y2+6y-9=x2-(y2-6y+9)=x2-(y2-2×y×3+32)=x2-(y-3)2=[x-(y-3)][x+(y-3)]=(x-y+3)(x+y-3)

Page No 7.23:

Question 22:

Factorize each of the following algebraic expression:
25x2 − 10x + 1 − 36y2

Answer:

25x2-10x+1-36y2=(25x2-10x+1)-36y2=[(5x)2-2×5x×1+1]-36y2=(5x-1)2-(6y)2=[(5x-1)-6y][(5x-1)+6y]=(5x-1-6y)(5x-1+6y)=(5x-6y-1)(5x+6y-1)

Page No 7.23:

Question 23:

Factorize each of the following algebraic expression:
a2 b2 + 2bc c2

Answer:

a2-b2+2bc-c2=a2-(b2-2bc+c2)=a2-(b2-2×b×c+c2)=a2-(b-c)2=[(a-(b-c)][(a+(b-c)]=(a-b+c)(a+b-c)

Page No 7.23:

Question 24:

Factorize each of the following algebraic expression:
a2 + 2ab + b2c2

Answer:

 a2+2ab+b2-c2=(a2+2ab+b2)-c2=(a2+2×a×b+b2)-c2=(a+b)2-c2=[(a+b)-c][(a+b)+c]=(a+b-c)(a+b+c)

Page No 7.23:

Question 25:

Factorize each of the following algebraic expression:
49 − x2y2 + 2xy

Answer:

49-x2-y2+2xy=49-(x2-2xy+y2)=49-(x2-2×x×y+y2)=72-(x-y)2=[7-(x-y)][7+(x-y)]=(7-x+y)(7+x-y)=(x-y+7)(y-x+7)

Page No 7.23:

Question 26:

Factorize each of the following algebraic expression:
a2 + 4b2 − 4ab − 4c2

Answer:

a2+4b2-4ab-4c2=(a2-4ab+4b2)-4c2=[a2-2×a×2b+(2b)2]-4c2=(a-2b)2-(2c)2=[(a-2b)-2c][(a-2b)+2c]=(a-2b-2c)(a-2b+2c)

Page No 7.23:

Question 27:

Factorize each of the following algebraic expression:
x2y2 − 4xz + 4z2

Answer:

x2-y2-4xz+4z2=(x2-4xz+4z2)-y2=[x2-2×x×2z+(2z)2]-y2=(x-2z)2-y2=[(x-2z)-y][(x-2z)+y]=(x-2z-y)(x-2z+y)=(x+y-2z)(x-y-2z)



Page No 7.27:

Question 1:

Factorize each of the following algebraic expression:
x2 + 12x − 45

Answer:

To factorise x2+12x-45, we will find two numbers p and q such that p+q=12 and pq=-45.Now, 15+(-3)=12 and 15×(-3)=-45Splitting the middle term 12x in the given quadratic as -3x+15x, we get: x2+12x-45=x2-3x+15x-45                       =(x2-3x)+(15x-45)                       =x(x-3)+15(x-3)                       =(x+15)(x-3)

Page No 7.27:

Question 2:

Factorize each of the following algebraic expression:
40 + 3xx2

Answer:

We have:40+3x-x2-(x2-3x-40) To factorise (x2-3x-40), we will find two numbers p and q such that p+q=-3 and pq=-40.Now,  5+(-8)=-3 and 5×(-8)=-40Splitting the middle term -3x in the given quadratic as 5x-8x, we get:40+3x-x2=-(x2-3x-40)                    =-(x2+5x-8x-40)                    =-[(x2+5x)-(8x+40)]                    =-[x(x+5)-8(x+5)]                    =-(x-8)(x+5)                    =(x+5)(-x+8)

Page No 7.27:

Question 3:

Factorize each of the following algebraic expression:
a2 + 3a − 88

Answer:

To factorise a2+3a-88, we will find two numbers p and q such that p+q=3 and pq=-88.Now,  11+(-8)=3 and 11×(-8)=-88Splitting the middle term 3a in the given quadratic as 11a-8a, we get:a2+3a-88=a2+11a-8a-88                    =(a2+11a)-(8a+88)                    =a(a+11)-8(a+11)                    =(a-8)(a+11)

Page No 7.27:

Question 4:

Factorize each of the following algebraic expression:
a2 − 14a − 51

Answer:

To factorise a2-14a-51, we will find two numbers p and q such that p+q=-14 and pq=-51.Now, 3+(-17)=-14 and 3×(-17)=-51Splitting the middle term -14a in the given quadratic as 3a-17a, we get:a2-14a-51=a2+3a-17a-51                       =(a2+3a)-(17a+51)                       =a(a+3)-17(a+3)                       =(a-17)(a+3)

Page No 7.27:

Question 5:

Factorize each of the following algebraic expression:
x2 + 14x + 45

Answer:

To factorise x2+14x+45, we will find two numbers p and q such that p+q=14 and pq=45.Now, 9+5=14 and 9×5=45Splitting the middle term 14x in the given quadratic as 9x+5x, we get:x2+14x+45=x2+9x+5x+45                      =(x2+9x)+(5x+45)                      =x(x+9)+5(x+9)                      =(x+5)(x+9)

Page No 7.27:

Question 6:

Factorize each of the following algebraic expression:
x2 − 22x + 120

Answer:

To factorise x2-22x+120, we will find two numbers p and q such that p+q=-22 and pq=120.Now, (-12)+(-10)=-22 and (-12)×(-10)=120Splitting the middle term -22x in the given quadratic as -12x-10x, we get:x2-22x+12=x2-12x-10x+120                      =(x2-12x)+(-10x+120)                      =x(x-12)-10(x-12)                      =(x-10)(x-12)

Page No 7.27:

Question 7:

Factorize each of the following algebraic expression:
x2 − 11x − 42

Answer:

To factorise x2-11x-42, we will find two numbers p and q such that p+q=-11 and pq=-42.Now,3+(-14)=-22 and 3×(-14)=42Splitting the middle term -11x in the given quadratic as-14x+3x, we get: x2-11x-42=x2-14x+3x-42                       =(x2-14x)+(3x-42)                       =x(x-14)+3(x-14)                       =(x+3)(x-14)

Page No 7.27:

Question 8:

Factorize each of the following algebraic expression:
a2 + 2a − 3

Answer:

To factorise a2+2a-3, we will find two numbers p and q such that p+q=2 and pq=-3.Now, 3+(-1)=2 and  3×(-1)=-3Splitting the middle term 2a in the given quadratic as-a+3a, we get:a2+2a-3=a2-a+3a-3                  =(a2-a)+(3a-3)                  =a(a-1)+3(a-1)                  =(a+3)(a-1)

Page No 7.27:

Question 9:

Factorize each of the following algebraic expression:
a2 + 14a + 48

Answer:

To factorise a2+14a+48, we will find two numbers p and q such that p+q=14 and pq=48.Now, 8+6=14 and 8×6=48Splitting the middle term 14a in the given quadratic as 8a+6a, we get:a2+14a+48=a2+8a+6a+48                       =(a2+8a)+(6a+48)                       =a(a+8)+6(a+8)                       =(a+6)(a+8)

Page No 7.27:

Question 10:

Factorize each of the following algebraic expression:
x2 − 4x − 21

Answer:

To factorise x2-4x-21, we will find two numbers p and q such that p+q=-4 and pq=-21.Now,3+(-7)=-4 and 3×(-7)=-21Splitting the middle term -4x in the given quadratic as -7x+3x, we get:x2-4x-21=x2-7x+3x-21                    =(x2-7x)+(3x-21)                    =x(x-7)+3(x-7)                    =(x+3)(x-7)

Page No 7.27:

Question 11:

Factorize each of the following algebraic expression:
y2 + 5y − 36

Answer:

To factorise y2+5y-36, we will find two numbers p and q such that p+q=5 and pq=-36.Now, 9+(-4)=5 and 9×(-4)=-36Splitting the middle term 5y in the given quadratic as -4y+9y, we get: y2+5y-36=y2-4y+9y-36                     =(y2-4y)+(9y-36)                     =y(y-4)+9(y-4)                     =(y+9)(y-4)

Page No 7.27:

Question 12:

Factorize each of the following algebraic expression:
(a2 − 5a)2 − 36

Answer:

(a2-5a)2-36=(a2-5a)2-62=[(a2-5a)-6][(a2-5a)+6]=(a2-5a-6)(a2-5a+6)In order to factorise a2-5a-6, we will find two numbers p and q such that p+q=-5 and pq=-6Now, (-6)+1=-5 and (-6)×1=-6Splitting the middle term -5 in the given quadratic as -6a+a, we get:a2-5a-6=a2-6a+a-6                   =(a2-6a)+(a-6)                   =a(a-6)+(a-6)                   =(a+1)(a-6)Now,In order to factorise a2-5a+6, we will find two numbers p and q such that p+q=-5 and pq=6Clearly,(-2)+(-3)=-5 and (-2)×(-3)=6Splitting the middle term -5 in the given quadratic as -2a-3a, we get:a2-5a+6=a2-2a-3a+6                   =(a2-2a)-(3a-6)                   =a(a-2)-3(a-2)                   =(a-3)(a-2) (a2-5a-6)(a2-5a+6)=(a-6)(a+1)(a-3)(a-2)                                                =(a+1)(a-2)(a-3)(a-6)

Page No 7.27:

Question 13:

Factorize each of the following algebraic expression:
(a + 7)(a − 10) + 16

Answer:

(a+7)(a-10)+16=a2-10a+7a-70+16=a2-3a-54To factorise a2-3a-54 , we will find two numbers p and q such that p+q=-3 and pq=-54.Now, 6+(-9)=-3 and 6×(-9)=-54Splitting the middle term -3a in the given quadratic as -9a+6a, we get: a2-3a-54=a2-9a+6a-54                      =(a2-9a)+(6a-54)                      =a(a-9)+6(a-9)                      =(a+6)(a-9)



Page No 7.30:

Question 1:

Resolve each of the following quadratic trinomial into factor:
2x2 + 5x + 3

Answer:

The given expression is 2x2+5x+3.                        (Coefficient of x2=2, coefficient of x=5 and constant term=3)We will split the coefficient of x into two parts such that their sum is 5 and their product equals the product of the coefficient of x2 and the constant term, i.e., 2×3=6.Now, 2+3=5 and 2×3=6Replacing the middle term 5x by 2x+3x, we have:2x2+5x+3=2x2+2x+3x+3                    =(2x2+2x)+(3x+3)                    =2x(x+1)+3(x+1)                    =(x+1)(2x+3)

Page No 7.30:

Question 2:

Resolve each of the following quadratic trinomial into factor:
2x2 − 3x − 2

Answer:

The given expression is 2x2-3x-2.                             (Coefficient of x2=2, coefficient of x=-3 and constant term=-2)We will split the coefficient of x into two parts such that their sum is -3 and their product equals the product of the coefficient of x2 and the constant term, i.e., 2×(-2)=-4.Now, (-4)+1=-3 and (-4)×1=-4Replacing the middle term 3x by -4x+x, we have:2x2-3x-2=2x2-4x+x-2                    =(2x2-4x)+(x-2)                    =2x(x-2)+(x-2)                    =(2x+1)(x-2)

Page No 7.30:

Question 3:

Resolve each of the following quadratic trinomial into factor:
3x2 + 10x + 3

Answer:

The given expression is 3x2+10x+3.          (Coefficient of x2=3, coefficient of x=10 and constant term=3)We will split the coefficient of x into two parts such that their sum is 10 and their product equals the product of the coefficient of x2 and the constant term,  i.e., 3×3=9.Now, 9+1=10 and 9×1=9Replacing the middle term 10x by 9x+x, we have:3x2+10x+3=3x2+9x+x+3                      =(3x2+9x)+(x+3)                     =3x(x+3)+(x+3)                     =(3x+1)(x+3)

Page No 7.30:

Question 4:

Resolve each of the following quadratic trinomial into factor:
7x − 6 − 2x2

Answer:

The given expression is 7x-6-2x2.                                        (Coefficient of x2=-2, coefficient of x=7 and constant term=-6)We will split the coefficient of x into two parts such that their sum is 7 and their product equals the product of the coefficient of x2 and the constant term,  i.e., (-2)×(-6)=12.Now, 4+3=7 and 4×3=12Replacing the middle term 7x by 4x+3x, we have:7x-6-2x2=-2x2+4x+3x-6                    =(-2x2+4x)+(3x-6)                    =2x(2-x)-3(2-x)                    =(2x-3)(2-x)

Page No 7.30:

Question 5:

Resolve each of the following quadratic trinomial into factor:
7x2 − 19x − 6

Answer:

The given expression is 7x2-19x-6.              (Coefficient of x2=7, coefficient of x=-19 and constant term=-6)We will split the coefficient of x into two parts such that their sum is -19 and their product equals the product of the coefficient of x2 and the constant term,  i.e., 7×(-6)=-42.Now, (-21)+2=-19 and (-21)×2=-42Replacing the middle term -19x by -21x+2x, we have:7x2-19x-6=7x2-21x+2x-6                      =(7x2-21x)+(2x-6)                      =7x(x-3)+2(x-3)                      =(7x+2)(x-3)

Page No 7.30:

Question 6:

Resolve each of the following quadratic trinomial into factor:
28 − 31x − 5x2

Answer:

The given expression is 28-31x-5x2.                (Coefficient of x2=-5, coefficient of x=-31 and constant term=28)We will split the coefficient of x into two parts such that their sum is -31 and their product equals the product of the coefficient of x2 and the constant term, i.e., (-5)×(28)=-140.Now,(-35)+4=-31 and (-35)×4=-140Replacing the middle term -31x by -35x+4x, we have:-5x2-31x+28=-5x2-35x+4x+28                           =(-5x2-35x)+(4x+28)                           =-5x(x+7)+4(x+7)                           =(4-5x)(x+7)

Page No 7.30:

Question 7:

Resolve each of the following quadratic trinomial into factor:
3 + 23y − 8y2

Answer:

The given expression is 3+23y-8y2.              (Coefficient of y2=-8, coefficient of y=23 and constant term=3)We will split the coefficient of y into two parts such that their sum is 23 and their product equals the product of the coefficient of y2 and the constant term, i.e., (-8)×3=-24.Now, (-1)+24=23 and (-1)×24=-24Replacing the middle term 23y by -y+24y, we have:3+23y-8y2=-8y2+23y+3=-8y2-y+24y+3=(-8y2-y)+(24y+3)=-y(8y+1)+3(8y+1)=(3-y)(8y+1)

Page No 7.30:

Question 8:

Resolve each of the following quadratic trinomial into factor:
11x2 − 54x + 63

Answer:

The given expression is 11x2-54x+63.                  (Coefficient of x2=11, coefficient of x=-54 and constant term=63)We will split the coefficient of x into two parts such that their sum is -54 and their product equals the product of the coefficient of x2 and the constant term, i.e., 11×63=693.Now, (-33)+(-21)=-54 and (-33)×(-21)=693Replacing the middle term-54x by -33x-21x, we have:11x2-54x+63= 11x2-33x-21x+63                          =(11x2-33x)+(-21x+63)                          =11x(x-3)-21(x-3)                          =(11x-21)(x-3)

Page No 7.30:

Question 9:

Resolve each of the following quadratic trinomial into factor:
7x − 6x2 + 20

Answer:

The given expression is 7x-6x2+20.                     (Coefficient of x2=-6, coefficient of x=7 and constant term=20)We will split the coefficient of x into two parts such that their sum is 7 and their product equals the product of the coefficient of x2 and the constant term, i.e., (-6)×20=-120.Now,15+(-8)=7 and 15×(-8)=-120Replacing the middle term 7x by 15x-8x, we get:7x-6x2+20=-6x2+7x+20=-6x2+15x-8x+20=(-6x2+15x)+(-8x+20)=3x(-2x+5)+4(-2x+5)=(3x+4)(-2x+5)

Page No 7.30:

Question 10:

Resolve each of the following quadratic trinomial into factor:
3x2 + 22x + 35

Answer:

The given expression is 3x2+22x+35.                                         (Coefficient of x2=3, coefficient of x=22 and constant term=35)We will split the coefficient of x into two parts such that their sum is 22 and their product equals the product of the coefficient of x2 and the constant term, i.e., 3×35=105.Now, 15+7=22 and 15×7=105Replacing the middle term 22x by 15x+7x, we get:3x2+22x+35=3x2+15x+7x+35                         =(3x2+15x)+(7x+35)                         =3x(x+5)+7(x+5)                         =(3x+7)(x+5)

Page No 7.30:

Question 11:

Resolve each of the following quadratic trinomial into factor:
12x2 − 17xy + 6y2

Answer:

The given expression is 12x2-17xy+6y2.                          (Coefficient of x2=12, coefficient of x=-17y and constant term=6y2)We willsplit the coefficient of x into two parts such that their sum is -17y and their product equals the product of the coefficient of x2 and the constant term i.e., 12×6y2=72y2.Now, (-9y)+(-8y)=-17y and (-9y)×(-8y)=72y2Replacing the middle term -17xy by -9xy-8xy, we get:12x2-17xy+6y2=12x2-9xy-8xy+6y2                              =(12x2-9xy)-(8xy-6y2)                              =3x(4x-3y)-2y(4x-3y)                              =(3x-2y)(4x-3y)

Page No 7.30:

Question 12:

Resolve each of the following quadratic trinomial into factor:
6x2 − 5xy − 6y2

Answer:

The given expression is 6x2-5xy-6y2.                           (Coefficient of x2=6, coefficient of x=-5y and constant term=-6y2)We will split the coefficient of x into two parts such that their sum is -5y and their product equals the product of the coefficient of x2 and the constant term, i.e., 6×(-6y2)=-36y2.Now, (-9y)+4y=-5y and(-9y)×4y=-36y2 Replacing the middle term -5xy by -9xy+4xy, we get:6x2-5xy-6y2= 6x2-9xy+4xy-6y2                          =(6x2-9xy)+(4xy-6y2)                          =3x(2x-3y)+2y(2x-3y)                          =(3x+2y)(2x-3y)

Page No 7.30:

Question 13:

Resolve each of the following quadratic trinomial into factor:
6x2 − 13xy + 2y2

Answer:

The given expression is 6x2-13xy+2y2.                        (Coefficient of x2=6, coefficient of x=-13y and constant term=2y2)We will split the coefficient of x into two parts such that their sum is -13y and their product equals the product of the coefficient of x2 and the constant term, i.e., 6×(2y2)=12y2.Now,(-12y)+(-y)=-13y and(-12y)×(-y)=12y2Replacing the middle term -13xy by -12xy-xy, we get: 6x2-13xy+2y2= 6x2-12xy-xy+2y2                            =(6x2-12xy)-(xy-2y2)                            =6x(x-2y)-y(x-2y)                            =(6x-y)(x-2y)

Page No 7.30:

Question 14:

Resolve each of the following quadratic trinomial into factor:
14x2 + 11xy − 15y2

Answer:

The given expression is 14x2+11xy-15y2.        (Coefficient of x2=14, coefficient of x=11y and constant term=-15y2)Now, we will split the coefficient of x into two parts such that their sum is 11y and their product equals the product of the coefficient of x2 and the constant term, i.e., 14×(-15y2)=-210y2.Now,21y+(-10y)=11y and21y×(-10y)=-210y2Replacing the middle term -11xy by -10xy+21xy, we get:14x2+11xy-15y2= 14x2-10xy+21xy-15y2                               =(14x2-10xy)+(21xy-15y2)                               =2x(7x-5y)+3y(7x-5y)                              =(2x+3y)(7x-5y)

Page No 7.30:

Question 15:

Resolve each of the following quadratic trinomial into factor:
6a2 + 17ab3b2

Answer:

The given expression is 6a2+17ab-3b2.             (Coefficient  of a2=6, coefficient of a=17b and constant term=-3b2)Now, we will split the coefficient of a into two parts such that their sum is 17b and their product equals the product of the coefficient of a2 and the constant term, i.e.,  6×(-3b2)=-18b2.Now,18b+(-b)=17b and18b×(-b)=-18b2Replacing the middle term 17ab by -ab+18ab, we get:16a2+17ab-3b2=6a2+-ab+18ab-3b2                               =(6a2-ab)+(18ab-3b2)                              =a(6a-b)+3b(6a-b)                             =(a+3b)(6a-b)

Page No 7.30:

Question 16:

Resolve each of the following quadratic trinomial into factor:
36a2 + 12abc − 15b2c2

Answer:

The given expression is 36a2+12abc-15b2c2.          (Coefficient of a2=36, coefficient of a=12bc and constant term=-15b2c2)Now, we will split the coefficient of a into two parts such that their sum is 12bc and their product equals the product of the coefficient of a2 and the constant term, i.e.,  36×(-15b2c2)=-540b2c2.Now,(-18bc)+30bc=12bc and(-18bc)×30bc=-540b2c2Replacing the middle term 12abc by -18abc+30abc, we get:36a2+12abc-15b2c2=36a2-18abc+30abc-15b2c2                                     =(36a2-18abc)+(30abc-15b2c2)                                     =18a(2a-bc)+15bc(2a-bc)                                     =(18a+15bc)(2a-bc)                                    =3(6a+5bc)(2a-bc)

Page No 7.30:

Question 17:

Resolve each of the following quadratic trinomial into factor:
15x2 − 16xyz − 15y2z2

Answer:

The given expression is 15x2-16xyz-15y2z2.(Coefficient of x2=15, coefficient of x=-16yz and constant term=-15y2z2)Now, we will split the coefficient of x into two parts such that their sum is -16yz and their product equals the product of the coefficient of x2 and the constant term, i.e.,  15×(-15y2z2)=-225y2z2.Now,(-25yz)+9yz=-16yx and (-25yz)×9yz=-225y2z2Replacing the middle term -16xyz by -25xyz+9xyz, we have:15x2-16xyz-15y2z2=15x2-25xyz+9xyz-15y2z2                                        =(15x2-25xyz)+(9xyz-15y2z2)                                        =5x(3x-5yz)+3yz(3x-5yz)                                        =(5x+3yz)(3x-5yz)

Page No 7.30:

Question 18:

Resolve each of the following quadratic trinomial into factor:
(x − 2y)2 − 5(x − 2y) + 6

Answer:

The given expression is a2-5a+6.Assuming a=x-2y, we have: (x-2y)2-5(x-2y)+6=a2-5a+6       (Coefficient of a2=1, coefficient of a=-5 and constant term=6)Now, we will split the coefficient of a into two parts such that their sum is -5 and their product equals the product of the coefficient of a2 and the constant term,  i.e.,  1×6=6.Clearly, (-2)+(-3)=-5 and(-2)×(-3)=6Replacing the middle term -5a by -2a-3a, we have:a2-5a+6=a2-2a-3a+6                  =(a2-2a)-(3a-6)                  =a(a-2)-3(a-2)                 =(a-3)(a-2)Replacing a by (x-2y), we get:(a-3)(a-2)=(x-2y-3)(x-2y-2)

Page No 7.30:

Question 19:

Resolve each of the following quadratic trinomial into factor:
(2a − b)2 + 2(2a − b) − 8

Answer:

Assuming x=2a-b, we have: (2a-b)2+2(2a-b)-8=x2+2x-8The given expression becomes x2+2x-8.    (Coefficient of x2=1 and that of x=2 ; constant term=-8)Now, we will split the coefficient of x into two parts such that their sum is 2 and their product equals the product of the coefficient of x2 and the constant term, i.e.,  1×(-8)=-8.Clearly,(-2)+4=2 and(-2)×4=-8Replacing the middle term 2x by -2x+4x, we get:x2+2x-8=x2-2x+4x-8                  =(x2-2x)+(4x-8)                  =x(x-2)+4(x-2)                  =(x+4)(x-2)Relacing x by 2a-b, we get:(x+4)(x-2)=(2a-b+4)(2a-b-2)



Page No 7.3:

Question 1:

Find the greatest common factor (GCF/HCF) of the following polynomial:
2x2 and 12x2

Answer:

The numerical coefficients of the given monomials are 2 and 12. So, the greatest common factor of 2 and 12 is 2.
The common literal appearing in the given monomials is x.
The smallest power of x in the two monomials is 2.
The monomial of the common literals with the smallest powers is x2.
Hence, the greatest common factor is 2x2.

Page No 7.3:

Question 2:

Find the greatest common factor (GCF/HCF) of the following polynomial:
6x3y and 18x2y3

Answer:

The numerical coefficients of the given monomials are 6 and 18. The greatest common factor of 6 and 18 is 6.
The common literals appearing in the two monomials are x and y. 
The smallest power of x in the two monomials is 2.
The smallest power of y in the two monomials is 1.
The monomial of the common literals with the smallest powers is x2y.
​Hence, the greatest common factor is 6x2y​.

Page No 7.3:

Question 3:

Find the greatest common factor (GCF/HCF) of the following polynomial:
7x, 21x2 and 14xy2

Answer:

The numerical coefficients of the given monomials are 7, 21 and 14. The greatest common factor of 7, 21 and 14 is 7.
The common literal appearing in the three monomials is x. 
The smallest power of x in the three monomials is 1.
The monomial of the common literals with the smallest powers is x.
​Hence, the greatest common factor is 7x.

Page No 7.3:

Question 4:

Find the greatest common factor (GCF/HCF) of the following polynomial:
42x2yz and 63x3y2z3

Answer:

The numerical coefficients of the given monomials are 42 and 63. The greatest common factor of 42 and 63 is 21.
The common literals appearing in the two monomials are x, y and z. 
The smallest power of x in the two monomials is 2.
The smallest power of y in the two monomials is 1.
The smallest power of z in the two monomials is 1
.
The monomial of the common literals with the smallest powers is x2yz.
​Hence, the greatest common factor is
21x2yz​.

Page No 7.3:

Question 5:

Find the greatest common factor (GCF/HCF) of the following polynomial:
12ax2, 6a2x3 and 2a3x5

Answer:

The numerical coefficients of the given monomials are 12, 6 and 2. The greatest common factor of 12, 6 and 2 is 2.
The common literals appearing in the three monomials are a and x. 
The smallest power of a in the three monomials is 1.
The smallest power of x in the three monomials is 2
.
The monomial of common literals with the smallest powers is ax2.
​Hence, the greatest common factor is
2ax2.

Page No 7.3:

Question 6:

Find the greatest common factor (GCF/HCF) of the following polynomial:
9x2, 15x2y3, 6xy2 and 21x2y2

Answer:

The numerical coefficients of the given monomials are 9, 15, 6 and 21. The greatest common factor of 9, 15, 6 and 21 is 3.
The common literal appearing in the three monomials is x. 
The smallest power of x in the four monomials is 1.
The monomial of common literals with the smallest powers is x.
​Hence, the greatest common factor is
3x.

Page No 7.3:

Question 7:

Find the greatest common factor (GCF/HCF) of the following polynomial:
4a2b3, −12a3b, 18a4b3

Answer:

The numerical coefficients of the given monomials are 4, -12 and 18. The greatest common factor of 4, -12 and 18 is 2.
The common literals appearing in the three monomials are a and b. 
The smallest power of a in the three monomials is 2.
The smallest power of b in the three monomials is 1.

The monomial of the common literals with the smallest powers is a2b.
​Hence, the greatest common factor is
2a2b​.

Page No 7.3:

Question 8:

Find the greatest common factor (GCF/HCF) of the following polynomial:
6x2y2, 9xy3, 3x3y2

Answer:

The numerical coefficients of the given monomials are 6, 9 and 3. The greatest common factor of 6, 9 and 3 is 3.
The common literals appearing in the three monomials are x and y. 
The smallest power of x in the three monomials is 1.
The smallest power of y in the three monomials is 2.

The monomial of common literals with the smallest powers is xy2.
​Hence, the greatest common factor is 3xy
2​.



Page No 7.32:

Question 1:

Factorize each of the following quadratic polynomials by using the method of  completing the square:
p2 + 6p + 8

Answer:

p2+6p+8=p2+6p+622-622+8    [Adding and subtracting 622, that is, 32]=p2+6p+32-32+8=p2+2×p×3+32-9+8=p2+2×p×3+32-1=(p+3)2-12                              [Completing the square]=[(p+3)-1][(p+3)+1]=(p+3-1)(p+3+1)=(p+2)(p+4)

Page No 7.32:

Question 2:

Factorize each of the following quadratic polynomials by using the method of completing the square:
q2 − 10q + 21

Answer:

q2-10q+21=q2-10q+1022-1022+21   [Adding and subtracting 1022, that is, 52]=q2-2×q×5+52-52+21=(q-5)2-4                                        [Completing the square]=(q-5)2-22 =[(q-5)-2][(q-5)+2]=(q-5-2)(q-5+2)=(q-7)(q-3)

Page No 7.32:

Question 3:

Factorize each of the following quadratic polynomials by using the method of completing the square:
4y2 + 12y + 5

Answer:

4y2+12y+5=4(y2+3y+54)                                    [Making the coefficient of y2=1]=4[y2+3y+322-322+54]        [Adding and subtracting 322]=4[(y+32)2-94+54]=4[(y+32)2-12]                                  [Completing the square]=4[(y+32)-1][(y+32)+1]=4(y+32-1)(y+32+1)=4(y+12)(y+52)=(2y+1)(2y+5)

Page No 7.32:

Question 4:

Factorize each of the following quadratic polynomials by using the method of completing the square:
p2 + 6p − 16

Answer:

p2+6p-16=p2+6p+622-622-16    [Adding and subtracting 622, that is, 32]=p2+6p+32-9-16=(p+3)2-25                                [Completing the square]=(p+3)2-52=[(p+3)-5][(p+3)+5]=(p+3-5)(p+3+5)=(p-2)(p+8)

Page No 7.32:

Question 5:

Factorize each of the following quadratic polynomials by using the method of completing the square:
x2 + 12x + 20

Answer:

x2+12x+20=x2+12x+1222-1222+20       [Adding and subtracting 1222, that is, 62]=x2+12x+62-62+20=(x+6)2-16                                          [Completing the square]=(x+6)2-42=[(x+6)-4][(x+6)+4]=(x+6-4)(x+6+4)=(x+2)(x+10)

Page No 7.32:

Question 6:

Factorize each of the following quadratic polynomials by using the method of completing the square:
a2 − 14a − 51

Answer:

a2-14a-51=a2-14a+1422-1422-51   [Adding and subtracting 1422, that is, 72]=a2-14a+72-72-51=(a-7)2-100                                    [Completing the square]=(a-7)2-102 =[(a-7)-10][(a-7)+10]=(a-7-10)(a-7+10)=(a-17)(a+3)



Page No 7.33:

Question 7:

Factorize each of the following quadratic polynomials by using the method of completing the square:
a2 + 2a − 3

Answer:

a2+2a-3=a2+2a+222-222-3   [Adding and subtracting 222, that is, 12]=a2+2a+12-12-3=(a+1)2-4                                    [Completing the square]=(a+1)2-22=[(a+1)-2][(a+1)+2]=(a+1-2)(a+1+2)=(a-1)(a+3)

Page No 7.33:

Question 8:

Factorize each of the following quadratic polynomials by using the method of completing the square:
4x2 − 12x + 5

Answer:

4x2-12x+5=4(x2-3x+54)                                     [Making the coefficient of x2=1]=4[x2-3x+322-322+54]      [Adding and subtracting 322]=4[(x-32)2-94+54]                                 [Completing the square]=4[(x-32)2-12]    =4[(x-32)-1][(x-32)+1]=4(x-32-1)(x-32+1)=4(x-52)(x-12)=(2x-5)(2x-1)

Page No 7.33:

Question 9:

Factorize each of the following quadratic polynomials by using the method of completing the square:
y2 − 7y + 12

Answer:

 y2-7y+12=y2-7y+722-722+12      [Adding and subtracting 722]=(y-72)2-494+484                   [Completing the square]=(y-72)2-14 =(y-72)2-122 =[(y-72)-12][(y-72)+12]=(y-72-12)(y-72+12)=(y-4)(y-3)

Page No 7.33:

Question 10:

Factorize each of the following quadratic polynomials by using the method of completing the square:
z2 − 4z − 12

Answer:

z2-4z-12=z2-4z+422-422-12       [Adding and subtracting 422, that is, 22]=z2-4z+22-22-12=(z-2)2-16                                [Completing the square]=(z-2)2-42=[(z-2)-4][(z-2)+4]=(z-6)(z+2)



Page No 7.4:

Question 9:

Find the greatest common factor (GCF/HCF) of the following polynomial:
a2b3, a3b2

Answer:

The common literals appearing in the three monomials are a and b. 
The smallest power of x in the two monomials is 2.
The smallest power of y in the two monomials is 2
.
The monomial of common literals with the smallest powers is a2b2.
​Hence, the greatest common factor is
a2b2.

Page No 7.4:

Question 10:

Find the greatest common factor (GCF/HCF) of the following polynomial:
36a2b2c4, 54a5c2, 90a4b2c2

Answer:

The numerical coefficients of the given monomials are 36, 54 and 90. The greatest common factor of 36, 54 and 90 is 18.
The common literals appearing in the three monomials are a and c. 

The smallest power of a in the three monomials is 2.
The smallest power of c in the three monomials is 2.

The monomial of common literals with the smallest powers is a2c2.
​Hence, the greatest common factor is
18a2c2.

Page No 7.4:

Question 11:

Find the greatest common factor (GCF/HCF) of the following polynomial:
x3, − yx2

Answer:

The common literal appearing in the two monomials is x. 
The smallest power of x in both the monomials is 2.
​Hence, the greatest common factor is x2.

Page No 7.4:

Question 12:

Find the greatest common factor (GCF/HCF) of the following polynomial:
15a3, − 45a2, − 150a

Answer:

The numerical coefficients of the given monomials are 15, -45 and -150. The greatest common factor of 15, -45 and -150 is 15.
The common literal appearing in the three monomials is a. 

The smallest power of a in the three monomials is 1.
​Hence, the greatest common factor is 15a.

Page No 7.4:

Question 13:

Find the greatest common factor (GCF/HCF) of the following polynomial:
2x3y2, 10x2y3, 14xy

Answer:

The numerical coefficients of the given monomials are 2, 10 and 14. The greatest common factor of 2, 10 and 14 is 2.
The common literals appearing in the three monomials are x and y. 

The smallest power of x in the three monomials is 1.
The smallest power of y in the three monomials is 1.

The monomial of common literals with the smallest powers is xy.
​Hence, the greatest common factor is 2xy.

Page No 7.4:

Question 14:

Find the greatest common factor (GCF/HCF) of the following polynomial:
14x3y5, 10x5y3, 2x2y2

Answer:

The numerical coefficients of the given monomials are 14, 10 and 2. The greatest common factor of 14, 10 and 2 is 2.
The common literals appearing in the three monomials are x and y. 

The smallest power of x in the three monomials is 2.
The smallest power of y in the three monomials is 2.

The monomial of common literals with the smallest powers is x2y2.
​Hence, the greatest common factor is 2x
2y2.

Page No 7.4:

Question 15:

Find the greatest common factor of the terms in each of the following expression:
5a4 + 10a3 − 15a2

Answer:

Terms are expressions separated by plus or minus signs. Here, the terms are 5a4, 10a3 and 15a2.
The numerical coefficients of the given monomials are 5, 10 and 15. The greatest common factor of 5, 10 and 15 is 5.
The common literal appearing in the three monomials is a. 

The smallest power of a in the three monomials is 2.
The monomial of common literals with the smallest powers is a2.
​Hence, the greatest common factor is 5a
2.

Page No 7.4:

Question 16:

Find the greatest common factor of the terms in each of the following expression:
2xyz + 3x2y + 4y2

Answer:

The expression has three monomials: 2xyz, 3x2y and 4y2
The numerical coefficients of the given monomials are 2, 3 and 4. The greatest common factor of 2, 3 and 4 is 1.
The common literal appearing in the three monomials is y. 

The smallest power of y in the three monomials is 1.
The monomial of common literals with the smallest powers is y.
​Hence, the greatest common factor is y.

Page No 7.4:

Question 17:

Find the greatest common factor of the terms in each of the following expression:
3a2b2 + 4b2c2 + 12a2b2c2

Answer:

The expression has three monomials: 3a2b2, 4b2c2 and 12a2b2c2
The numerical coefficients of the given monomials are 3, 4 and 12. The greatest common factor of 3, 4 and 12 is 1.
The common literal appearing in the three monomials is b. 

The smallest power of b in the three monomials is 2.
The monomial of common literals with the smallest powers is b2.
​Hence, the greatest common factor is b2.



Page No 7.5:

Question 1:

Factorize the following:
3x − 9

Answer:

The greatest common factor of the terms 3x and -9 of the expression 3x - 9 is 3. 
Now.
3x = 3x
and
-9 = 3.-3
Hence, the expression 3x - 9 can be factorised as 3(x - 3).

Page No 7.5:

Question 2:

Factorize the following:
5x − 15x2

Answer:

The greatest common factor of the terms 5x and 15x2 of the expression 5x - 15x2 is 5x. 
Now,
5x = 5x ×
and
-15x2 = 5x × -3x
Hence, the expression 5x - 15x2 can be factorised as 5x(1 - 3x)​.

Page No 7.5:

Question 3:

Factorize the following:
20a12b2 − 15a8b4

Answer:

The greatest common factor of the terms 20a12b2 and -15a8b4 of the expression 20a12b2 - 15a8b4 is 5a8b2.
20a12b2 = 5×4×a8×a4×b2 = 5a8×b2×4a4 and -15a8b4 = 5×-3×a8×b2×b2 = 5a8b2 × -3b2

Hence, the expression 20a12b2 - 15a8b4 can be factorised as 5a8b2(4a4-3b2)​​

Page No 7.5:

Question 4:

Factorize the following:
72x6y7 − 96x7y6

Answer:

The greatest common factor of the terms 72x6y7 and -96x7y6 of the expression 72x6y7 - 96x7y64 is 24x6y6.

Now,
72x6y7 = 24x6y6× 3y
and 
-96x7y6​ = 24x6y6 × -4x

Hence, the expression 72x6y7 - 96x7y6 can be factorised as 24x6y6(3y - 4x)​.

Page No 7.5:

Question 5:

Factorize the following:
20x3 − 40x2 + 80x

Answer:

The greatest common factor of the terms 20x3​, -40x2​ and 80x​ of the expression 20x3 - 40x2 + 80x​ is 20x.
Now,
20x3​ = 20x × x2
-40x2​ = 20x × -2x
and
80x ​= 20x × 4
Hence, the expression 20x3 - 40x2 + 80x​ ​can be factorised as 20x(x2 - 2x + 4)​.

Page No 7.5:

Question 6:

Factorize the following:
2x3y2 − 4x2y3 + 8xy4

Answer:

The greatest common factor of the terms 2x3y2, -4x2y3 and 8xy4 of the expression 2x3y2 - 4x2y3+ 8xy4y64 is 2xy2.

Now,
2x3y2 = 2xy2  × x2 
-4x2y3 = 2xy2 × -2xy
8xy4​ = 2xy2 × 4y2

Hence, the expression 2x3y2 - 4x2y3 + 8xy4 ​can be factorised as 2xy2(x2 - 2xy + 4y2)​.

Page No 7.5:

Question 7:

Factorize the following:
10m3n2 + 15m4n − 20m2n3

Answer:

The greatest common factor of the terms 10m3n2, 15m4n and -20m2n3 of the expression 10m3n2 + 15m4n - 20m2n3 is 5m2n​.

Now,
10m3n= 5m2n ​× 2mn
15m4n = 5m2n × 3m2
-20m2n​= 5m2n × -4n2

Hence, 10m3n2 + 15m2n - 20m2n3 ​can be factorised as 5m2n(2mn + 3m2 - 4n2).​

Page No 7.5:

Question 8:

Factorize the following:
2a4b4 − 3a3b5 + 4a2b5

Answer:

The greatest common factor of the terms 2a4b4, -3a3b5​ and 4a2b5 of the expression 2a4b4 - 3a3b5 + 4a2b5 is a2b4​.

Now,
2a4b= a2b4 × 2a2
-3a3b= a2​b4 × -3ab 
4a2b= a2​b4 × 4b

Hence, (2a4b4 - 3a3b5 + 4a2b5) ​can be factorised as [a2b4(2a2 - 3ab + 4b)]​.

Page No 7.5:

Question 9:

Factorize the following:
28a2 + 14a2b2 − 21a4

Answer:

The greatest common factor of the terms 28a2, 14a2b2 and 21a4 of the expression 28a2+14a2b2-21a4 is 7a2.Also, we can write 28a2=7a2×4, 14a2b2=7a2×2b2 and 21a4=7a2×3a2. 28a2+14a2b2-21a4=7a2×4+7a2×2b2-7a2×3a2                                           = 7a2(4+2b2-3a2)

Page No 7.5:

Question 10:

Factorize the following:
a4b − 3a2b2 − 6ab3

Answer:

The greatest common factor of the terms a4b, 3a2b2 and 6ab3 of the expression a4b-3a2b2-6ab3 is ab.Also, we can write a4b=ab×a3, 3a2b2=ab×3ab and 6ab3=ab×6b2. a4b-3a2b2-6ab3 = ab×a3-ab×3ab-ab×6b2                                        = ab(a3-3ab-6b2)

Page No 7.5:

Question 11:

Factorize the following:
2l2mn - 3lm2n + 4lmn2

Answer:

The greatest common factor of the terms 2l2mn, 3lm2n and 4lmn2 of the expression 2l2mn-3lm2n+4lmn2 is lmn.Also, we can write 2l2mn=lmn×2l, 3lm2n=lmn×3m and 4lmn2=lmn×4n. 2l2nm-3lm2n+4lmn2=lmn×2l-lmn×3m+lmn×4n                                             =lmn(2l-3m+4n)

Page No 7.5:

Question 12:

Factorize the following:
x4y2x2y4x4y4

Answer:

The greatest common factor of the terms x4y2, x2y4 and x4y4 of the expression x4y2-x2y4-x4y4 is x2y2.Also, we can write x4y2=x2y2×x2, x2y4=x2y2×y2 and x4y4=x2y2×x2y2. x4y2-x2y4-x4y4 = x2y2×x2-x2y2×y2-x2y2×x2y2                                     =x2y2(x2-y2-x2y2)

Page No 7.5:

Question 13:

Factorize the following:
9x2y + 3axy

Answer:

The greatest common factor of the terms 9x2y and 3axy of the expression 9x2y+3axy is 3xy.Also, we can write 9x2y=3xy×3x and 3axy=3xy×a. 9x2y+3axy =3xy×3x+3xy×a                           =3xy(3x+a)

Page No 7.5:

Question 14:

Factorize the following:
16m − 4m2

Answer:

The greatest common factor of the terms 16m and 4m2 of the expression 16m-4m2 is 4m.Also, we can write 16m=4m×4 and 4m2=4m×m. 16m-4m2=4m×4-4m×m                        =4m(4-m)

Page No 7.5:

Question 15:

Factorize the following:
−4a2 + 4ab − 4ca

Answer:

The greatest common factor of the terms -4a2, 4ab and- 4ca of the expression-4a2+4ab-4ca is -4a.Also, we can write -4a2=-4a×a, 4ab=-4a×(-b) and 4ca=-4a×c. -4a2+4ab-4ca=-4a×a+(-4a)×(-b)-4a×c                                     =-4a(a-b+c)

Page No 7.5:

Question 16:

Factorize the following:
x2yz + xy2z + xyz2

Answer:

The greatest common factor of the terms x2yz, xy2z and xyz2 of the expression x2yz+xy2z+xyz2 is xyz.Also, we can write x2yz=xyz×x, xy2z=xyz×y and xyz2=xyz×z. x2yz+xy2z+xyz2=xyz×x+xyz×y+xyz×z                                     =xyz(x+y+z)                                

Page No 7.5:

Question 17:

Factorize the following:
ax2y + bxy2 + cxyz

Answer:

The greatest common factor of the terms ax2y, bxy2 and cxyz of the expression ax2y+bxy2+cxyz is xy.Also, we can write ax2y=xy×ax, bxy2=xy×by and cxyz=xy×cz. ax2y+bxy2+cxyz = xy×ax+xy×by+xy×cz                                      =xy(ax+by+cz)



Page No 7.7:

Question 1:

Factorize each of the following algebraic expressions:
6x(2xy) + 7y(2xy)

Answer:

6x(2x-y)+7y(2x-y)= (6x+7y)(2x-y)         [Taking (2x-y) as the common factor]

Page No 7.7:

Question 2:

Factorize each of the following algebraic expressions:
2r(yx) + s(xy)

Answer:

2r(y-x)+s(x-y)=2r(y-x)-s(y-x)      [(x-y)=-(y-x)]=(2r-s)(y-x)              [Taking (y-x) as the common factor]

Page No 7.7:

Question 3:

Factorize each of the following algebraic expressions:
7a(2x − 3) + 3b(2x − 3)

Answer:

7a(2x-3)+3b(2x-3)=(7a+3b)(2x-3)            [Taking (2x-3) as the common factor]

Page No 7.7:

Question 4:

Factorize each of the following algebraic expressions:
9a(6a − 5b) −12a2(6a − 5b)

Answer:

9a(6a-5b)-12a2(6a-5b)=(9a-12a2)(6a-5b)     [Taking (6a-5b) as the common factor]=3a(3-4a)(6a-5b)       [Taking 3a as the common factor of the quadratic (9a-12a2)]

Page No 7.7:

Question 5:

Factorize each of the following algebraic expressions:
5(x − 2y)2 + 3(x − 2y)

Answer:

5(x-2y)2+3(x-2y)=[5(x-2y)+3](x-2y)     [Taking (x-2y) as the common factor]=(5x-10y+3)(x-2y)

Page No 7.7:

Question 6:

Factorize each of the following algebraic expressions:
16(2l − 3m)2 −12(3m − 2l)

Answer:

16(2l-3m)2-12(3m-2l)=16(2l-3m)2+12(2l-3m)            [(3m-2l)=-(2l-3m)]= [16(2l-3m)+12](2l-3m)           [Taking (2l-3m) as the common factor]=4[4(2l-3m)+3](2l-3m)              {Taking 4 as the common factor of [16(2l-3m)+12]}=4(8l-12m+3)(2l-3m)

Page No 7.7:

Question 7:

Factorize each of the following algebraic expressions:
3a(x − 2y) −b(x − 2y)

Answer:

3a(x-2y)-b(x-2y)=(3a-b)(x-2y)           [Taking (x-2y) as the common factor]

Page No 7.7:

Question 8:

Factorize each of the following algebraic expressions:
a2(x + y) +b2(x + y) +c2(x + y)

Answer:

a2(x+y)+b2(x+y)+c2(x+y)=(a2+b2+c2)(x+y)                      [Taking (x+y) as the common factor]

Page No 7.7:

Question 9:

Factorize each of the following algebraic expressions:
(xy)2 + (xy)

Answer:

(x-y)2+(x-y)=(x-y)(x-y)+(x-y)   [Taking (x-y) as the common factor]= (x-y+1)(x-y)

Page No 7.7:

Question 10:

Factorize each of the following algebraic expressions:
6(a + 2b) −4(a + 2b)2

Answer:

6(a+2b)-4(a+2b)2=[6-4(a+2b)](a+2b)      [Taking (a+2b) as the common factor]= 2[3-2(a+2b)](a+2b)   {Taking 2 as the common factor of [6-4(a+2b)]}=2(3-2a-4b)(a+2b)

Page No 7.7:

Question 11:

Factorize each of the following algebraic expressions:
a(xy) + 2b(yx) + c(xy)2

Answer:

a(x-y)+2b(y-x)+c(x-y)2=a(x-y)-2b(x-y)+c(x-y)2   [(y-x)=-(x-y)]= [a-2b+c(x-y)](x-y)= (a-2b+cx-cy)(x-y)

Page No 7.7:

Question 12:

Factorize each of the following algebraic expressions:
−4(x − 2y)2 + 8(x −2y)

Answer:

-4(x-2y)2+8(x-2y)= [-4(x-2y)+8](x-2y)   [Taking (x-2y) as the common factor]= 4[-(x-2y)+2](x-2y)   {Taking 4 as the common factor of [-4(x-2y)+8]}= 4(2y-x+2)(x-2y)

Page No 7.7:

Question 13:

Factorize each of the following algebraic expressions:
x3(a − 2b) + x2(a − 2b)

Answer:

x3(a-2b)+x2(a-2b)=(x3+x2)(a-2b)       [Taking (a-2b) as the common factor]= x2(x+1)(a-2b)      [Taking x2 as the common factor of (x3+x2)]

Page No 7.7:

Question 14:

Factorize each of the following algebraic expressions:
(2x − 3y)(a + b) + (3x − 2y)(a + b)

Answer:

(2x-3y)(a+b)+(3x-2y)(a+b)=(2x-3y+3x-2y)(a+b)   [Taking (a+b) as the common factor]=(5x-5y)(a+b)=5(x-y)(a+b)                      [Taking 5 as the common factor of (5x-5y)]

Page No 7.7:

Question 15:

Factorize each of the following algebraic expressions:
4(x + y) (3a − b) +6(x + y) (2b 3a)

Answer:

4(x+y)(3a-b)+6(x+y)(2b-3a) =2(x+y)[2(3a-b)+3(2b-3a)]                {Taking [2 (x+y)] as the common factor}=2(x+y)(6a-2b+6b-9a)=2(x+y)(4b-3a)



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