Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 6 Operations On Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Operations On Algebraic Expressions are extremely popular among Class 8 students for Math Operations On Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 84:

Question 1:

Add:
8ab, −5ab, 3ab, −ab

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

    8ab-5ab    3ab  -ab
________

   5ab

Page No 84:

Question 2:

Add:
7x, −3x, 5x, −x, −2x

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

    7x-3x    5x  -x-2x
_____
  6x

Page No 84:

Question 3:

Add:
3a − 4b + 4c, 2a + 3b − 8c, a − 6b + c

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

3a - 4b + 4c2a + 3b - 8c  a - 6b +  c
___________
6a -7b-3c

Page No 84:

Question 4:

Add:
5x − 8y + 2z, 3z − 4y − 2x, 6yzx and 3x − 2z − 3y

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

    5x-8y+2z-2x-4y+3z  -x+6y- z   3x-3y-2z   5x-9y+2z

Page No 84:

Question 5:

Add:
6ax − 2by + 3cz, 6by − 11axcz and 10cz − 2ax − 3by

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

       6ax-2by+ 3cz-11ax+ 6by-  cz  -2ax-3by+10cz   - 7ax+  by+12cz

Page No 84:

Question 6:

Add:
2x3 − 9x2 + 8, 3x2 − 6x − 5, 7x3 − 10x + 1 and 3 + 2x − 5x2 − 4x3

Answer:

On arranging the terms of the given expressions in the descending powers of x and adding column-wise:

     2x3- 9x2+  0x+8   0x3+ 3x2 - 6x-5   7x3+ 0x2-10x+1-4x3-5x2+ 2x+3     5x3-11x2-14x+7

Page No 84:

Question 7:

Add:
6p + 4q r + 3, 2r − 5p − 6, 11q − 7p + 2r − 1 and 2q − 3r + 4

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

    6p+  4q - r+3-5p+  0q+2r-6-7p+11q+2r-1   0p+  2q-3r+4-6p+17q+0r+0=-6p+17q

Page No 84:

Question 8:

Add:
4x2 − 7xy + 4y2 − 3, 5 + 6y2 − 8xy + x2 and 6 − 2xy + 2x2 − 5y2

Answer:

On arranging the terms of the given expressions in the descending powers of x and adding column-wise:

4x2+4y2-7xy-3 x2+ 6y2-8xy+02x2-5y2-2xy+67x2+5y2-17xy+3

Page No 84:

Question 9:

Subtract:
3a2b from −5a2b

Answer:

On arranging the terms of the given expressions in the descending powers of x and subtracting:

-5a2b   3a2b--8a2b

Page No 84:

Question 10:

Subtract:
−8pq from 6pq

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

                     6pq   -8pq   +                  14pq

Page No 84:

Question 11:

Subtract:
−2abc from −8abc

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

     -8abc -2abc  +     -6abc

Page No 84:

Question 12:

Subtract:
−16p from −11p

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

  -11p -16p +     5p

Page No 84:

Question 13:

Subtract:
2a − 5b + 2c − 9 from 3a − 4b c + 6

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

     3a-4b- c +6    2a-5b+2c-9 -    +     -    +      a + b-3c+15

Page No 84:

Question 14:

Subtract:
−6p + q + 3r + 8 from p − 2q − 5r − 8

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

      p-2q-5r-8-6p+  q+3r+8+    -     -   -   7p-3q-8r-16   

Page No 84:

Question 15:

Subtract:
x3 + 3x2 − 5x + 4 from 3x3x2 + 2x − 4

Answer:

On arranging the terms of the given expressions in the descending powers of x and subtracting column-wise:

   3x3-x2+2x-4  x3+3x2-5x+4-    -    +    -  2x3-4x2+7x-8

Page No 84:

Question 16:

Subtract:
5y4 − 3y3 + 2y2 + y − 1 from 4y4 − 2y3 − 6y2y + 5

Answer:

Arranging the terms of the given expressions in the descending powers of x and subtracting column-wise:

     4y4-2y3-6y2-y+5   5y4-3y3+2y2+y-1 -     +    -     -   +   -y4+ y3- 8y2-2y+6

Page No 84:

Question 17:

Subtract:
4p2 + 5q2 − 6r2 + 7 from 3p2 − 4q2 − 5r2 − 6

Answer:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

     3p2-4q2-5r2-6 4p2+5q2-6r2+7-    -      +      - -p2-9q2+r2-13

Page No 84:

Question 18:

What must be subtracted from 3a2 − 6ab − 3b2 − 1 to get 4a2 − 7ab − 4b2 + 1?

Answer:

Let the required number be x.
(3a2-6ab-3b2-1)-x=4a2-7ab-4b2+1
(3a2-6ab-3b2-1)-(4a2-7ab-4b2+1)=x

    3a2-6ab-3b2-1 4a2-7ab-4b2+1-     +    +     -  -a2+ ab+   b2-2

∴ Required number = -a2+ab+b2-2

Page No 84:

Question 19:

The two adjacent sides of a rectangle are 5x2 − 3y2 and x2 + 2xy. Find the perimeter.

Answer:

Sides of the rectangle are l and b.
l=5x2-3y2b=x2+2xy
Perimeter of the rectangle is (2l+2b).

Perimeter = 2(5x2-3y2) + 2(x2+2xy)                =10x2-6y2+2x2+4xy     10x2-6y2   2x2         + 4xy    12x2-6y2+4xyHence, the perimeter of the rectangle is 12x2-6y2+4xy.

Page No 84:

Question 20:

The perimeter of a triangle is 6p2 − 4p + 9 and two of its sides are p2 − 2p + 1 and 3p2 − 5p + 3. Find the third side of the triangle.

Answer:

Let a, b and c be the three sides of the triangle.

∴ Perimeter of the triangle =(a+b+c)

Given perimeter of the triangle = 6p2-4p+9
One side (a)  = p2-2p+1
Other side (b) = 3p2-5p+3
Perimeter = (a+b+c)
(6p2-4p+9)=(p2-2p+1)+(3p2-5p+3)+c6p2-4p+9-p2+2p-1-3p2+5p-3=c(6p2-p2-3p2)+(-4p+2p+5p)+(9-1-3)=c2p2+3p+5 = c

Thus, the third side is 2p2+3p+5.



Page No 87:

Question 1:

Find the product:
(5x + 7) × (3x + 4)

Answer:

By horizontal method:
(5x+7)×(3x+4)=5x(3x+4)+7(3x+4)=15x2+20x+21x+28=15x2+41x+28

Page No 87:

Question 2:

Find the product:
(4x + 9) × (x − 6)

Answer:

By horizontal method:

(4x+9)×(x-6)=4x(x-6)+9(x-6)=4x2-24x+9x-54=4x2-15x-54

Page No 87:

Question 3:

Find the product:
(2x + 5) × (4x − 3)

Answer:

By horizontal method:

(2x+5)×(4x-3)=2x(4x-3)+5(4x-3)=8x2-6x+20x-15=8x2+14x-15

Page No 87:

Question 4:

Find the product:
(3y − 8) × (5y − 1)

Answer:

By horizontal method:

(3y-8)×(5y-1)=3y(5y-1)-8(5y-1)=15y2-3y-40y+8=15y2-43y+8

Page No 87:

Question 5:

Find the product:
(7x + 2y) × (x + 4y)

Answer:

By horizontal method:

(7x+2y)×(x+4y)=7x(x+4y)+2y(x+4y)=7x2+28xy+2xy+8y2=7x2+30xy+8y2

Page No 87:

Question 6:

Find the product:
(9x + 5y) × (4x + 3y)

Answer:

By horizontal method:

(9x+5y)×(4x+3y)9x(4x+3y)+5y(4x+3y)=36x2+27xy+20xy+15y2=36x2+47xy+15y2

Page No 87:

Question 7:

Find the product:
(3m − 4n) × (2m − 3n)

Answer:

By horizontal method:

(3m-4n)×(2m-3n)=3m(2m-3n)-4n(2m-3n)=6m2-9mn-8mn+12n2=6m2-17mn+12n2

Page No 87:

Question 8:

Find the product:
(x2a2) × (xa)

Answer:

By horizontal method:

(x2-a2)×(x-a)=x2(x-a)-a2(x-a)=x3-ax2-a2x+a3
i.e (x3+a3)-ax(x-a)

Page No 87:

Question 9:

Find the product:
(x2y2) × (x + 2y)

Answer:

By horizontal method:

(x2-y2)×(x+2y)=x2(x+2y)-y2(x+2y)=x3+2x2y-xy2-2y3i.e(x3-2y3)+xy(2x-y)

Page No 87:

Question 10:

Find the product:
(3p2 + q2) × (2p2 − 3q2)

Answer:

By horizontal method:

(3p2+q2)×(2p2-3q2)=3p2(2p2-3q2)+q2(2p2-3q2)=6p4-9p2q2+2p2q2-3q4i.e6p4-7p2q2-3q4

Page No 87:

Question 11:

Find the product:
(2x2 − 5y2) × (x2 + 3y2)

Answer:

By horizontal method:

(2x2-5y2)×(x2+3y2)=2x2(x2+3y2)-5y2(x2+3y2)=2x4+6x2y2-5x2y2-15y4=2x4+x2y2-15y4

Page No 87:

Question 12:

Find the product:
(x3y3) × (x2 + y2)

Answer:

By horizontal method:

(x3-y3)×(x2+y2)=x3(x2+y2)-y3(x2+y2)=x5+x3y2-x2y3-y5=x5-y5+x2y2x-y

Page No 87:

Question 13:

Find the product:
(x4 + y4) × (x2y2)

Answer:

By horizontal method:
(x4+y4)×(x2-y2)=x4(x2-y2)+y4(x2-y2)=x6-x4y2+y4x2-y6=x6-y6-x2y2x2-y2

Page No 87:

Question 14:

Find the product:
x4+1x4×x+1x

Answer:

By horizontal method:

x4+1x4×x+1x=x4x+1x+1x4x+1x=x5+x3+1x3+1x5i.e x3(x2+1)+1x31+1x2

Page No 87:

Question 15:

Find the product:
(x2 − 3x + 7) × (2x + 3)

Answer:

By horizontal method:

(x2-3x+7)×(2x+3)=2x(x2-3x+7)+3(x2-3x+7)=2x3-6x2+14x+3x2-9x+21=2x3-3x2+5x+21

Page No 87:

Question 16:

Find the product:
(3x2 + 5x − 9) × (3x − 5)

Answer:

By horizontal method:
(3x2+5x-9)×(3x-5)=3x(3x2+5x-9)-5(3x2+5x-9)=9x3+15x2-27x-15x2-25x+45=9x3-52x+45

Page No 87:

Question 17:

Find the product:
(x2xy + y2) × (x + y)

Answer:

By horizontal method:
(x2-xy+y2)×(x+y)=x(x2-xy+y2)+y(x2-xy+y2)=x3-x2y+y2x+x2y-xy2+y3=x3+y3

Page No 87:

Question 18:

Find the product:
(x2 + xy + y2) × (xy)

Answer:

By horizontal method:

(x2+xy+y2)×(x-y)x(x2+xy+y2)-y(x2+xy+y2)=x3+x2y+xy2-x2y-xy2-y3=x3-y3

Page No 87:

Question 19:

Find the product:
(x3 − 2x2 + 5) × (4x − 1)

Answer:

By horizontal method:

(x3-2x2+5)×(4x-1)=4x(x3-2x2+5)-1(x3-2x2+5)=4x4-8x3+20x-x3+2x2-5=4x4-9x3+2x2+20x-5

Page No 87:

Question 20:

Find the product:
(9x2x + 15) × (x2 − 3)

Answer:

By horizontal method:

(9x2-x+15)×(x2-3)=x2(9x2-x+15)-3(9x2-x+15)=9x4-x3+15x2-27x2+3x-45=9x4-x3-12x2+3x-45

Page No 87:

Question 21:

Find the product:
(x2 − 5x + 8) × (x2 + 2)

Answer:

By horizontal method:

(x2-5x+8)×(x2+2)=x2(x2-5x+8)+2(x2-5x+8)=x4-5x3+8x2+2x2-10x+16=x4-5x3+10x2-10x+16

Page No 87:

Question 22:

Find the product:
(x3 − 5x2 + 3x + 1) × (x3 − 3)

Answer:

By horizontal method:

(x3-5x2+3x+1)×(x2-3)=x2(x3-5x2+3x+1)-3(x3-5x2+3x+1)=x5-5x4+3x3+x2-3x3+15x2-9x-3=x5-5x4+16x2-9x-3

Page No 87:

Question 23:

Find the product:
(3x + 2y − 4) × (xy + 2)

Answer:

By horizontal method:

(3x+2y-4)×(x-y+2)x(3x+2y-4)-y(3x+2y-4)+2(3x+2y-4)=3x2+2xy-4x-3xy-2y2+4y+6x+4y-8=3x2-2y2-xy+2x+8y-8

Page No 87:

Question 24:

Find the product:
(x2 − 5x + 8) × (x2 + 2x − 3)

Answer:

By horizontal method:

(x2-5x+8)×(x2+2x-3)=x2(x2-5x+8)+2x(x2-5x+8)-3(x2-5x+8)=x4-5x3+8x2+2x3-10x2+16x-3x2+15x-24=x4-3x3-5x2+31x-24

Page No 87:

Question 25:

Find the product:
(2x2 + 3x − 7) × (3x2 − 5x + 4)

Answer:

By horizontal method:

(2x2+3x-7)×(3x2-5x+4)=2x2(3x2-5x+4)+3x(3x2-5x+4)-7(3x2-5x+4)=6x4-10x3+8x2+9x3-15x2+12x-21x2+35x-28=6x4-x3-28x2+47x-28

Page No 87:

Question 26:

Find the product:
(9x2x + 15) × (x2x − 1)

Answer:

By horizontal method:

(9x2-x+15)×(x2-x-1)=x2(9x2-x+15)-x(9x2-x+15)-1(9x2-x+15)=9x4-x3+15x2-9x3+x2-15x-9x2+x-15=9x4-10x3+7x2-14x-15



Page No 90:

Question 1:

Divide:
(i) 24x2y3 by 3xy
(ii) 36xyz2 by −9xz
(iii) −72x2y2z by −12xyz
(iv) −56mnp2 by 7mnp

Answer:

(i) 24x2y3 by 3xy

24x2y3 3xy243x2-1y3-18xy2.

Therefore, the quotient is 8xy2.

(ii) 36xyz2 by −9xz

36xyz2 -9xz36-9x1-1y1-0z2-1-4yz


Therefore, the quotient is 4yz.

(iii)

-72x2y2z by -12xyz-72x2y2z -12xyz-72-12x2-1y2-1z1-16xy

Therefore, the quotient is 6xy.

(iv) −56mnp2 by 7mnp


-56mnp2 7mnp-567m1-1n1-1p2-1-8p

Therefore, the quotient is −8p.

Page No 90:

Question 2:

Divide:
(i) 5m3 − 30m2 + 45m by 5m
(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy
(iii) 9x2y − 6xy + 12xy2 by − 3xy
(iv) 12x4 + 8x3 − 6x2 by − 2x2

Answer:

(i) 5m3 − 30m2 + 45m by 5m

(5m3-30m2 +45m) ÷ 5m5m35m-30m25m+ 45m 5mm2 -6m + 9

Therefore, the quotient is m2 6m + 9.

(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy

(8x2y2 - 6xy2 + 10x2y3 )÷ 2xy8x2y22xy- 6xy22xy+ 10x2y3 2xy4xy - 3y + 5xy2

Therefore, the quotient is 4xy 3y + 5xy2.

(iii) 9x2y − 6xy + 12xy2 by − 3xy

(9x2y - 6xy + 12xy2 )÷ -3xy9x2y-3xy-6xy-3xy+12xy2 -3xy-3x + 2 -4y

Therefore, the quotient is −3x + 2 4y.

(iv) 12x4 + 8x3 − 6x2 by − 2x2

(12x4 + 8x3 - 6x2 )÷ -2x212x4 -2x2+8x3-2x2-6x2-2x2-6x2-4x+32 

Therefore the quotient is −6x2 4x + 3.

Page No 90:

Question 3:

Write the quotient and remainder when we divide:
(x2 − 4x + 4) by (x − 2)

Answer:



Therefore, the quotient is x-2 and the remainder is 0.

Page No 90:

Question 4:

Write the quotient and remainder when we divide:
(x2 − 4) by (x + 2)

Answer:



Therefore, the quotient is x−2 and the remainder is 0.

Page No 90:

Question 5:

Write the quotient and remainder when we divide:
(x2 + 12x + 35) by (x + 7)

Answer:

(x2 + 12x + 35) by (x + 7)



Therefore, the quotient is x+5 and the remainder is 0.

Page No 90:

Question 6:

Write the quotient and remainder when we divide:
(15x2 + x − 6) by (3x + 2)

Answer:



Therefore, the quotient is 5x-3 and the remainder is 0.

Page No 90:

Question 7:

Write the quotient and remainder when we divide:
(14x2 − 53x + 45) by (7x − 9)

Answer:



Therefore, the quotient is 2x - 5 and the remainder is 0.

Page No 90:

Question 8:

Write the quotient and remainder when we divide:
(6x2 − 31x + 47) by (2x − 5)

Answer:



Therefore, the quotient is 3x - 8 and the remainder is 7.

Page No 90:

Question 9:

Write the quotient and remainder when we divide:
(2x3 + x2 − 5x − 2) by (2x + 3)

Answer:



Therefore, the quotient is x2-x-1 and the remainder is 1.

Page No 90:

Question 10:

Write the quotient and remainder when we divide:
(x3 + 1) by (x + 1)

Answer:



Therefore, the quotient is x2-x+1 and the remainder is 0.

Page No 90:

Question 11:

Write the quotient and remainder when we divide:
(x4 − 2x3 + 2x2 + x + 4) by (x2 + x + 1)

Answer:



Therefore, the quotient is ( x2 - 3x + 4) and remainder is 0.

Page No 90:

Question 12:

Write the quotient and remainder when we divide:
(x3 − 6x2 + 11x − 6) by (x2 − 5x + 6)

Answer:



Therefore, the quotient is (x-1) and the remainder is 0.

Page No 90:

Question 13:

Write the quotient and remainder when we divide:
(5x3 − 12x2 + 12x + 13) by (x2 − 3x + 4)

Answer:



Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

Page No 90:

Question 14:

Write the quotient and remainder when we divide:
(2x3 − 5x2 + 8x − 5) by (2x2 − 3x + 5)

Answer:



Therefore, the quotient is (x-1) and the remainder is 0.

Page No 90:

Question 15:

Write the quotient and remainder when we divide:
(8x4 + 10x3 − 5x2 − 4x + 1) by (2x2 + x − 1)

Answer:



Therefore, the quotient is ( 4x2+ 3x -2) and the remainder is ( x-1).



Page No 93:

Question 1:

Find each of the following products:
(i) (x + 6)(x + 6)
(ii) (4x + 5y)(4x + 5y)
(iii) (7a + 9b)(7a + 9b)
(iv) 23x+45y23x+45y
(v) (x2 + 7)(x2 + 7)
(vi) 56a2+256a2+2

Answer:

(i) We have:

(x+6)(x+6)=(x+6)2=x2+62+2×x×6                [using (a+b)2=a2+b2+2ab]=x2+36+12x

(ii) We have:

(4x+5y)(4x+5y)=(4x+5y)2=4x2+5y2+2×4x×5y          [using (a+b)2=a2+b2+2ab]=16x2+25y2+40xy

(iii) We have:
(7a+9b)(7a+9b)=(7a+9b)2=7a2+9b2+2×7a×9b            [using (a+b)2=a2+b2+2ab]=49a2+81b2+126ab

(iv) We have:
23x+45y23x+45y=23x+45y2=23x2+45y2+2×23x×45y              [using (a+b)2=a2+b2+2ab]=49x2+1625y2+1615xy


(v) We have:
(x2+7)(x2+7)=(x2+7)2=x22+72+2×x2×7             [using (a+b)2=a2+b2+2ab]=x4+49+14x2

(vi) We have:
56a2+256a2+2=56a2+22=56a22+22+2×56a2×2               [using (a+b)2=a2+b2+2ab]=2536a4+4+103a2

Page No 93:

Question 2:

Find each of the following products:
(i) (x − 4)(x − 4)
(ii) (2x − 3y)(2x − 3y)
(iii) 34x-56y34x-56y
(iv) x-3xx-3x
(v) 13x2-913x2-9
(vi) 12y2-13y12y2-13y

Answer:

(i) We have:
(x-4)(x-4)=(x-4)2=x2-2×x×4+42                   [using (a-b)2=a2-2ab+b2]=x2-8x+16

(ii) We have:
(2x-3y)(2x-3y)=(2x-3y)2=2x2-2×2x×3y+3y2                [using (a-b)2=a2-2ab+b2]=4x2-12xy+9y2

(iii) We have:
34x-56y34x-56y=34x-56y2=34x2-2×34x×56y+56y2           [using (a-b)2=a2-2ab+b2]=916x2-1512xy+2536y2

(iv) We have:
x-3xx-3x=x-3x2=x2-2×x×3x+3x2              [using (a-b)2=a2-2ab+b2]=x2-6+9x2

(v) We have:
13x2-913x2-9=13x2-92=13x22-2×13x2×9+92             [using (a-b)2=a2-2ab+b2]=19x4-6x2+81

(vi) We have:
12y2-13y12y2-13y=12y2-13y2=12y22-2×12y2×13y+13y2            [using (a-b)2=a2-2ab+b2]=14y4-13y3+19y2

Page No 93:

Question 3:

Expand:
(i) (8a + 3b)2
(ii) (7x + 2y)2
(iii) (5x + 11)2
(iv) a2+2a2
(v) 3x4+2y92
(vi) (9x − 10)2
(vii) (x2yyz2)2
(viii) xy-yx2
(ix) 3m-45n2

Answer:

We shall use the identities (a+b)2 =a2 +b2 +2ab and (a-b)2 =a2 +b2 -2ab.

(i) We have:
(8a+3b)2=8a2+2×8a×3b+3b2=64a2+48ab+9b2

(ii)We have:
(7x+2y)2=7x2+2×7x×2y+2y2=49x2+28xy+4y2

(iii) We have :
(5x+11)2=5x2+2×5x×11+112=25x2+110x+121

(iv) We have:
a2+2a2=a22+2×a2×2a+2a2=a42+2+4a2

(v) We have:
3x4+2y92=3x42+2×3x4×2y9+2y92=9x162+13xy+4y281

(vi) We have:
(9x-10)29x2-2×9x×10+102=81x2-180x+100

(vii) We have:
(x2y-yz2)2x2y2-2×x2y×yz2+yz22=x4y2-2x2y2z2+y2z4

(viii) We have:
xy-yx2=xy2-2×xy×yx+yx2=x2y2-2+y2x2

(ix) We have:
3m-45n2=3m2-2×3m×45n+45n2=9m2-24mn5+1625n2



Page No 94:

Question 4:

Find each of the following products:
(i) (x + 3)(x − 3)
(ii) (2x + 5)(2x − 5)
(iii) (8 + x)(8 − x)
(iv) (7x + 11y)(7x − 11y)
(v) 5x2+34y25x2-34y2
(vi) 4x5-5y34x5+5y3
(vii) x+1xx-1x
(viii) 1x+1y1x-1y

Answer:

(i) We have:

(x+3)(x-3)=x2-9                                [using (a+b)(a-b)=a2-b2]

(ii) We have:

(2x+5)(2x-5)=4x2-25                              [using (a+b)(a-b)=a2-b2]

(iii) We have:

(8+x)(8-x)=64-x2                                 [using (a+b)(a-b)=a2-b2]

(iv) We have:

(7x+11y)(7x-11y)=49x2-121y2                        [using (a+b)(a-b)=a2-b2]

(v) We have:

5x2+34y25x2-34y2=25x4-916y4                       [using (a+b)(a-b)=a2-b2]

(vi) We have:

4x5-5y34x5+5y3=16x225-25y29                     [using (a+b)(a-b)=a2-b2)]

(vii) We have:
x+1xx-1x=x2-1x2                            [using (a+b)(a-b)=a2-b2]

(viii) We have:
1x+1y1x-1y=1x2-1y2                      [using (a+b)(a-b)=a2-b2]

(ix) We have:
2a+3b2a-3b=4a2-9b2                     [using (a+b)(a-b)=a2-b2]

Page No 94:

Question 5:

Using the formula for squaring a binomial, evaluate the following:
(i) (54)2
(ii) (82)2
(iii) (103)2
(iv) (704)2

Answer:

We shall use the identity (a+b)2 =a2 +b2 +2ab.

(i)
542=(50+4)2=502+2×50×4+42=2500+400+16=2916

(ii)
822=(80+2)2=802+2×80×2+22=6400+320+4=6724

(iii)
1032=(100+3)2=1002+2×100×3+32=10000+600+9=10609

(iv)
7042=(700+4)2=7002+2×700×4+42=490000+5600+16=495616

Page No 94:

Question 6:

Using the formula for squaring a binomial, evaluate the following:
(i) (69)2
(ii) (78)2
(iii) (197)2
(iv) (999)2

Answer:

We shall use the identity (a-b)2 = a2 +b2 -2ab.

(i)
692=(70-1)2=702-2×70×1+1=4900-140+1=4761

(ii)
782=(80-2)2=802-2×80×2+4=6400-320+4=6084

(iii)
1972=(200-3)2=2002-2×200×3+9=40000-1200+9=38809

(iv)
9992=(1000-1)2=10002-2×1000×1+1=1000000-2000+1=998001

Page No 94:

Question 7:

Find the value of:
(i) (82)2 − (18)2
(ii) (128)2 − (72)2
(iii) 197 × 203
(iv) 198×198-102×10296
(v) (14.7 × 15.3)
(vi) (8.63)2 − (1.37)2

Answer:

We shall use the identity (a-b) (a+b)=a2 - b2.

(i)
(82)2-(18)2=(82-18)(82+18)=(64)(100)=6400

(ii)
(128)2-(72)2=(128-72)(128+72)=(56)(200)=11200

(iii)
197×203=(200-3)(200+3)=2002-32=40000-9=39991

(iv)
198×198-102×10296=1982-102296=(198-102)(198+102)96=(96)(300)96=300

(v)
(14.7×15.3)=(15-0.3)×(15+0.3)=(15)2-(0.3)2=225-0.09=224.91

(vi)
(8.63)2-(1.37)2=(8.63-1.37)(8.63+1.37)=(7.26)(10)=72.6

Page No 94:

Question 8:

Find the value of the expression (9x2 + 24x + 16), when x = 12.

Answer:

9x2 + 24x + 16Given, x = 123x2 + 2 3x4 + 42  3x + 42312+4236 + 42402 = 1600

Therefore, the value of the expression (9x2 + 24x + 16), when x = 12, is 1600.

Page No 94:

Question 9:

Find the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y=43.

Answer:

64x2+81y2+144xyGiven: x=11  y =438x2 + 9y2 + 28x9y8x +9y 28(11) +9(43)288 +122100210000
                                                                                                                                               
Therefore, the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y = 43, is 10000.y=43

Page No 94:

Question 10:

Find the value of the expression (36x2 + 25y2 − 60xy), when x=23 and y=15.

Answer:

36x2+25y2-60xyx=23, y=15=6x2 + 5y2 - 26x5y=6x - 5y2=6(23) -5(15)2=4 - 12=329

Page No 94:

Question 11:

If x+1x=4, find the values of
(i) x2+1x2 and
(ii) x4+1x4.

Answer:

(i)  x+1x= 4Squaring both the sides:x+1x2= 42x2+1x2+2x1x= 16x2+1x2+2 =16x2+1x2=16-2x2+1x2= 14

Therefore, the value of  x2+1x2 is 14.

x2 + 1x2 = 14Squaring both the sides:x4 + 1x4+2x21x2= 142x4 + 1x4+2 = 196x4 + 1x4 = 196-2x4 + 1x4=194

Therefore, the value of x4 + 1x4 is 194.

Page No 94:

Question 12:

If x-1x=5, find the values of
(i) x2+1x2
(ii) x4+1x4.

Answer:

(i)  x-1x= 5Squaring both the sides:x-1x2= 52x2+1x2-2x1x= 25x2+1x2-2 =25x2+1x2=25+2x2+1x2= 27Therefore, the value of x2+1x2 is  27.

x2 + 1x2 = 27Squaring both the sides:x4 + 1x4-2x21x2= 272x4 + 1x4-2 = 729x4 + 1x4 = 729+2x4 + 1x4=731Therefore, the value of x4 + 1x4 is 731.

Page No 94:

Question 13:

Find the continued product:
(i) (x + 1)(x − 1)(x2 + 1)
(ii) (x − 3)(x + 3)(x2 + 9)
(iii) (3x − 2y)(3x + 2y)(9x2 + 4y2)
(iv) (2p + 3)(2p − 3)(4p2 + 9)

Answer:

i x+1x-1x2+1x2-x+x-1x2+1x2-1x2+1x22-122              according to the formula a2-b2 = a+ba-bx4-1.Therefore, the product of x+1x-1x2+1 is x4-1.

ii x-3x+3x2+9x2-32x2+9       according to the formula a2-b2 = a+ba-bx2-9x2+9x22-92                 according to the formula a2-b2 = a+ba-bx4-81Therefore, the product of x-3x+3x2+9 is x4-81.

iii 3x-2y3x+2y9x2+4y23x2-2y29x2+4y2        according to the formula a2-b2 = a+ba-b9x2-4y29x2+4y29x22-4y22                 according to the formula a2-b2 = a+ba-b81x4-16y4.Therefore, the product of 3x-2y3x+2y9x2+4y2 is 81x4-16y4.

iv 2p+32p-34p2+92p2-324p2+9       according to the formula a2-b2 = a+ba-b4p2-94p2+94p22-92                according to the formula a2-b2 = a+ba-b16p4-81.Therefore, the product of 2p+32p-34p2+9 is 16p4-81.

Page No 94:

Question 14:

If x + y = 12 and xy = 14, find the value of (x2 + y2).

Answer:

x+y = 12On squaring both the sides:x+y2 = 122x2+y2+2xy = 144x2+y2 = 144 - 2xyGiven:  xy = 14x2+y2 = 144 - 214x2+y2 = 144 - 28x2+y2 = 116Therefore, the value of x2+y2 is 116.

Page No 94:

Question 15:

If xy = 7 and xy = 9, find the value of (x2 + y2).

Answer:

x-y = 7On squaring both the sides:x-y2 = 72x2+y2-2xy = 49x2+y2 = 49 + 2xyGiven: xy = 9x2+y2 = 49 + 29x2+y2 = 49 + 18x2+y2 = 67.Therefore, the value of x2+y2 is 67.

Page No 94:

Question 1:

Tick (✓) the correct answer:
The sum of (6a + 4bc + 3), (2b − 3c + 4), (11b − 7a + 2c − 1) and (2c − 5a − 6) is
(a) (4a − 6b + 2)
(b) (−3a + 14b − 3c + 2)
(d) (−6a + 17b)
(d) (−6a + 6b + c −4)

Answer:

(c) (−6a + 17b)

   6a  +4b  -c   +3          +2b  -3c +4-7a +11b +2c -1-5a            +2c -6-6a +17b+ 0c +0¯



Page No 95:

Question 2:

Tick (✓) the correct answer:
(3q + 7p2 − 2r3 + 4) − (4p2 − 2q + 7r3 − 3) = ?
(a) (p2 + 2q + 5r3 + 1)
(b) (11p2 + q + 5r3 + 1)
(c) (−3p2 − 5q + 9r3 − 7)
(d) (3p2 + 5q − 9r3 +7)

Answer:

(d) (3p2 + 5q − 9r3 +7)

    7p2 +3q  -2r3 +4   4p2 -2q  +7r3 -3-       +       -      +   3p2+ 5q  -9r3 + 7    ¯

Page No 95:

Question 3:

Tick (✓) the correct answer:
(x + 5)(x − 3) = ?
(a) x2 + 5x − 15
(b) x2 − 3x − 15
(c) x2 + 2x + 15
(d) x2 + 2x − 15

Answer:

(d) x2 + 2x − 15

x+5x-3xx-3+5x-3x2-3x +5x -15x2+2x-15

Page No 95:

Question 4:

Tick (✓) the correct answer:
(2x + 3)(3x − 1) = ?
(a) (6x2 + 8x − 3)
(b) (6x2 + 7x − 3)
(c) 6x2 − 7x − 3
(d) (6x2 − 7x + 3)

Answer:

(b) (6x2 + 7x − 3)

2x+33x-12x3x-1+33x-16x2-2x +9x-36x2+7x-3

Page No 95:

Question 5:

Tick (✓) the correct answer:
(x + 4)(x + 4) = ?
(a) (x2 + 16)
(b) (x2 + 4x + 16)
(c) (x2 + 8x + 16)
(d) (x2 + 16x)

Answer:

(c) (x2 + 8x + 16)

x+4x+4x+42            (according to the formula a+b2 =a2+2ab+b2)x2+2x4+42x2+8x+16

Page No 95:

Question 6:

Tick (✓) the correct answer:
(x − 6)(x − 6) = ?
(a) (x2 − 36)
(b) (x2 + 36)
(c) (x2 − 6x + 36)
(d) (x2 − 12x + 36)

Answer:

(d) (x2 − 12x + 36)

x-6x-6x-62                             (according to the formula a-b2 =a2-2ab+b2)x2-2x6+62x2-12x+36

Page No 95:

Question 7:

Tick (✓) the correct answer:
(2x + 5)(2x − 5) = ?
(a) (4x2 + 25)
(b) (4x2 − 25)
(c) (4x2 − 10x + 25)
(d) (4x2 + 10x − 25)

Answer:

(b) (4x2 − 25)

2x+52x-52x2-52           according to the formula a+ba-b = a2-b24x2 - 25

Page No 95:

Question 8:

Tick (✓) the correct answer:
8a2b3 ÷ (−2ab) = ?
(a) 4ab2
(b) 4a2b
(c) −4ab2
(d) −4a2b

Answer:

(c) −4ab2

8a2b3 ÷ -2ab8-2a2-1b3-1-4ab2

Page No 95:

Question 9:

Tick (✓) the correct answer:
(2x2 + 3x + 1) ÷ (x + 1) = ?
(a) (x + 1)
(b) (2x + 1)
(c) (x + 3)
(d) (2x + 3)

Answer:

(b) (2x + 1)

Page No 95:

Question 10:

Tick (✓) the correct answer:
(x2 − 4x + 4) ÷ (x − 2) = ?
(a) (x − 2)
(b) (x + 2)
(c) (2 − x)
(d) (2 + x + x2)

Answer:

(a) (x − 2)

Page No 95:

Question 11:

Tick (✓) the correct answer:
(a + 1)(a − 1)(a2 + 1) = ?
(a) (a4 − 2a2 − 1)
(b) (a4a2 − 1)
(c) (a4 − 1)
(d) (a4 + 1)

Answer:

(c) (a4 − 1)

i a+1a-1a2+1a2 -12a2+1       according to the formula a2-b2 = a+ba-ba2-1a2+1a22-122                 according to the formula a2-b2 = a+ba-ba4-1

Page No 95:

Question 12:

Tick (✓) the correct answer:
1x+1y1x-1y=?
(a) 1x2-1y2
(b) 1x2+1y2
(c) 1x2+1y2-1xy
(d) 1x2-1y2+1xy

Answer:

a) 1x2-1y2

1x+1y1x-1yAccording to the formula a+ba-b=a2-b2:1x2-1y2(1x21y2)

Page No 95:

Question 13:

Tick (✓) the correct answer:
If x+1x=5, then x2+1x2=?
(a) 25
(b) 27
(c) 23
(d) 25125

Answer:

(c) 23

  x+1x= 5Squaring both the sides:x+1x2= 52x2+1x2+2x1x= 25x2+1x2+2 =25x2+1x2=25-2x2+1x2= 23

Page No 95:

Question 14:

Tick (✓) the correct answer:
If x-1x=6, then x2+1x2=?
(a) 36
(b) 38
(c) 32
(d) 36136

Answer:

(b) 38

x-1x= 6Squaring both the sides:x-1x2= 62x2+1x2-2x1x= 36x2+1x2-2 =36x2+1x2=36+2x2+1x2= 38

Page No 95:

Question 15:

Tick (✓) the correct answer:
(82)2 − (18)2 = ?
(a) 8218
(b) 6418
(c) 6400
(d) 7204

Answer:

(c) 6400

822-182=82 + 1882 - 18=10064=6400                           [using the identity (a-b)(a+b)=a2 -b2]

Page No 95:

Question 16:

Tick (✓) the correct answer:
(197 × 203) = ?
(a) 39991
(b) 39999
(c) 40009
(d) 40001

Answer:

(a) 39991

197×203200-3200+32002-3240000-939991                   [using the identity (a+b) (a-b) = a2  -b2]

Page No 95:

Question 17:

Tick (✓) the correct answer:
If (a + b) = 12 and ab = 14, then (a2 + b2) = ?
(a) 172
(b) 116
(c) 165
(d) 126

Answer:

(b) 116

a+b =12Squaring both the sides:a+b2 = 122a2+ b2 +2ab = 144a2+ b2 = 144-2aba2+ b2 = 144 -214a2+ b2 = 144 -28a2+ b2 = 116

Page No 95:

Question 18:

Tick (✓) the correct answer:
If (ab) = 7 and ab = 9, then (a2 + b2) = ?
(a) 67
(b) 31
(c) 40
(d) 58

Answer:

(a) 67

a-b =7Squaring both the sides:a-b2 = 72a2+ b2 -2ab = 49a2+ b2 = 49+2aba2+ b2 = 49 +29a2+ b2 =  49+18a2+ b2 = 67

Page No 95:

Question 19:

Tick (✓) the correct answer:
If x = 10, then the value of (4x2 + 20x + 25) = ?
(a) 256
(b) 425
(c) 625
(d) 575

Answer:

(c) 625

4x2+20x+252x2+22x5+522x + 52210+5220+52252625



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