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

Page No 8.11:

Question 1:

Divide 5x3 − 15x2 + 25x by 5x.

Answer:

Page No 8.11:

Question 2:

Divide 4z3 + 6z2z by − 12z.

Answer:

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Question 3:

Divide 9x2y − 6xy + 12xy2 by −32xy.

Answer:

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Question 4:

Divide 3x3y2 + 2x2y + 15xy by 3xy.

Answer:

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Question 5:

Divide x2 + 7x + 12 by x + 4.

Answer:

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Question 6:

Divide 4y2 + 3y + 12 by 2y + 1.

Answer:

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Question 7:

Divide 3x3 + 4x2 + 5x + 18 by x + 2.

Answer:

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Question 8:

Divide 14x2 − 53x + 45 by 7x − 9.

Answer:

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Question 9:

Divide −21 + 71x − 31x2 − 24x3 by 3 − 8x.

Answer:

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Question 10:

Divide 3y4 − 3y3 − 4y2 − 4y by y2 − 2y.

Answer:

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Question 11:

Divide 2y5 + 10y4 + 6y3 + y2 + 5y + 3 by 2y3 + 1.

Answer:

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Question 12:

Divide x4 − 2x3 + 2x2 + x + 4 by x2 + x + 1.

Answer:

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Question 13:

Divide m3 − 14m2 + 37m − 26 by m2 − 12m +13.

Answer:

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Question 14:

Divide x4 + x2 + 1 by x2 + x + 1.

Answer:

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Question 15:

Divide x5 + x4 + x3 + x2 + x + 1 by x3 + 1.

Answer:

Page No 8.11:

Question 16:

Divide 14x3 − 5x2 + 9x − 1 by 2x − 1 and find the quotient and remainder

Answer:


Quotient = 7x2 + x + 5Remainder = 4

Page No 8.11:

Question 17:

Divide 6x3x2 − 10x − 3 by 2x − 3 and find the quotient and remainder.

Answer:


Quotient = 3x2+ 4x + 1 Remainder = 0

Page No 8.11:

Question 18:

Divide 6x3 + 11x2 − 39x − 65 by 3x2 + 13x + 13 and find the quotient and remainder.

Answer:


Quotient = 2x-5Remainder =0



Page No 8.12:

Question 19:

Divide 30x4 + 11x3 − 82x2 − 12x + 48 by 3x2 + 2x − 4 and find the quotient and remainder.

Answer:

Quotient =10x2-3x-12Remainder= 0

Page No 8.12:

Question 20:

Divide 9x4 − 4x2 + 4 by 3x2 − 4x + 2 and find the quotient and remainder.

Answer:


 Quotient = 3x2 4x 2 and remainder = 0.

Page No 8.12:

Question 21:

Verify the division algorithm i.e. Dividend = Divisor × Quotient + Remainder, in each of the following. Also, write the quotient and remainder.

Dividend Divisor
(i) 14x2 + 13x − 15 7x − 4
(ii) 15z3 − 20z2 + 13z − 12 3z − 6
(iii) 6y5 − 28y3 + 3y2 + 30y − 9 2y2 − 6
(iv) 34x − 22x3 − 12x4 − 10x2 − 75 3x + 7
(v) 15y4 − 16y3 + 9y2 − 103y + 6 3y − 2
(vi) 4y3 + 8y + 8y2 + 7 2y2 − y + 1
(vii) 6y5 + 4y4 + 4y3 + 7y2 + 27y + 6 2y3 + 1

Answer:

(i)

Quotient = 2x + 3
Remainder = -3
Divisor = 7x - 4
Divisor × Quotient + Remainder = (7x - 4) (2x + 3) - ​3 
                                                = 14x+ 21- 8- 12 - ​3 
                                                = 14x2 + 13x - 15
                                                = Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.

(ii)

Quotient = 5z2+103z+11Remainder = 54Divisor = 3z-6Divisor × Quotient +Remainder = (3z-6) 5z2+103z+11+54                                                           = 15z3+10z2+33z-30z2-20z-66+54                                                           = 15z3-20z2+13z-12                                                           = DividendThus,Divisor × Quotient + Remainder = Dividend                                                            
Hence verified.

(iii)


Quotient = 3y3-5y+32
Remainder = 0
Divisor = 2y2 - 6
Divisor × Quotient + Remainder =
(2y2-6) 3y3-5y+32+0=6y5-10y3+3y2-18y3+30y-9=6y5-28 y3+3y2+30y-9
= Dividend
 
Thus, Divisor × Quotient + Remainder = Dividend
Hence verified.

(iv)

Quotient  = - 4x3 + 2x2 - 8x + 30
Remainder  = - 285 
Divisor  = 3x + 7
Divisor × Quotient + Remainder =  (3x + 7) (- 4x3 + 2x2 - 8x + 30) - 285 
                                                 = - 12x4 + 6x3 - 24x2 + 90- 28x3 + 14x2 - 56x + 210 - ​285
                                                 = - 12x 4 - 22x3 - 10x2 + 34x - 75
                                                 =  Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.

(v)


Quotient =  5y3-2y2+53y
Remainder =  6
Divisor = 3y - 2
Divisor × Quotient  + Remainder = (3y - 2) (5y3 - 2y2 53y) + 6
                                                = 15y4-6y3+5y2-10y3+4y2-103y+6
                                                = 15y4-16y3+9y2-103y+6
                                                =  Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.

(vi)

Quotient =  2y + 5
Remainder =  11y + 2
Divisor =  2y2 - y + 1
Divisor × Quotient + Remainder =  (2y2 - y + 1) (2y + 5)11y + 2
                                                =  4y3 +10y2 - 2y2 - 5y + 2y + 5 + 11y + 2
                                                =  4y3 + 8y2 + 8y + 7
                                                =  Dividend
Thus,
Divisor × Quotient + Remainder  = Dividend
Hence verified.

(vii)




Quotient = 3y2 + 2y + 2
Remainder = 4y2 + 25y + 4
Divisor = 2y3 + 1
Divisor × Quotient + Remainder = (2y3 + 1) (3y2 2y + 2)4y225y + 4
                                                = 6y54y44y33y22y + 4y225y + 4
                                                6y54y44y37y227y + 6
                                                = Dividend
Thus,
Divisor × Quotient + Remainder = Dividend
Hence verified.

Page No 8.12:

Question 22:

Divide 15y4 + 16y3 + 103y − 9y2 − 6 by 3y − 2. Write down the coefficients of the terms in the quotient.

Answer:


 Quotient = 
5y3 + (26/3)y2 + (25/9)y + (80/27)
Remainder = (- 2/27)
Coefficient of y3 = 5
Coefficient
 of y2 = (26/3)
Coefficient of y = (25/9)
Constant = (80/27)

Page No 8.12:

Question 23:

Using division of polynomials, state whether
(i) x + 6 is a factor of  x2x − 42
(ii) 4x − 1 is a factor of 4x2 − 13x − 12
(iii) 2y − 5 is a factor of 4y4 − 10y3 − 10y2 + 30y − 15
(iv) 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y − 35
(v) z2 + 3 is a factor of z5 − 9z
(vi) 2x2x + 3 is a factor of 6x5x4 + 4x3 − 5x2x − 15

Answer:

(i)

Remainder is zero. Hence (x+6) is a factor of x2 -x-42
(ii)

As the remainder is non zero . Hence ( 4x-1) is not a factor of 4x2 -13x-12



(iii)




 The remainder is non zero,
 2y - 5 is not a factor of 4y4-10y3-10y2+30y-15.

(iv)

Remainder is zero.  Therefore, 3y2 + 5 is a factor of 6y5+15y4+16y3+4y2+10y-35.


(v)

Remainder is zero; therefore, z2 + 3 is a factor of z5 -9z.

(vi)



Remainder is zero ; therefore, 2x2-x+3 is a factor of 6x5-x4 +4x3-5x2-x-15.

Page No 8.12:

Question 24:

Find the value of a, if x + 2 is a factor of 4x4 + 2x3 − 3x2 + 8x + 5a.

Answer:

We have to find the value of a if (x+2) is a factor of (4x4+2x3-3x2+8x+5a).Substituting x=-2 in 4x4+2x3-3x2+8x+5a, we get:4(-2)4+2(-2)3-3(-2)2+8(-2)+5a=0or, 64-16-12-16+5a=0or, 5a=-20or, a=-4 If (x+2) is a factor of (4x4+2x3-3x2+8x+5a), a=-4.

Page No 8.12:

Question 25:

What must be added to x4 + 2x3 − 2x2 + x − 1 , so that the resulting polynomial is exactly divisible by x2 + 2x − 3?

Answer:


Thus, (- 2) should be added to (x4+2x3-2x2+x-1) to make the resulting polynomial exactly divisible by (x2+2x-3).



Page No 8.15:

Question 1:

Divide the first polynomial by the second in each of the following. Also, write the quotient and remainder:
(i) 3x2 + 4x + 5, x − 2
(ii) 10x2 − 7x + 8, 5x − 3
(iii) 5y3 − 6y2 + 6y − 1, 5y − 1
(iv) x4x3 + 5x, x − 1
(v) y4 + y2, y2 − 2

Answer:

(i) 3x2+4x+5x-2=3x(x-2)+10(x-2)+25(x-2)=(x-2)(3x+10)+25(x-2)=(3x+10)+25(x-2)Therefore, quotient=3x+10 and remainder=25.(ii) 10x2-7x+85x-3=2x(5x-3)-15(5x-3)+475(5x-3)=(5x-3)(2x-15)+475(5x-3)=(2x-15)+4755x-3Therefore, quotient=2x-15 and remainder=475.(iii) 5y3-6y2+6y-15y-1=y2(5y-1)-y(5y-1)+1(5y-1)(5y-1)=(5y-1)(y2-y+1)(5y-1)=(y2-y+1)Therefore, Quotient = y2-y+1 and remainder = 0

(iv) x4-x3+5xx-1=x3(x-1)+5(x-1)+5x-1=(x-1)(x3+5)+5x-1=(x3+5)+5x-1Therefore, quotient = x3+5 and remainder = 5.
(v) y4+y2y2-2=y2(y2-2)+3(y2-2)+6y2-2=(y2-2)(y2+3)+6y2-2=(y2+3)+6y2-2Therefore, quotient = y2+3 and remainder = 6.

Page No 8.15:

Question 2:

Find whether the first polynomial is a factor of the second.
(i) x + 1, 2x2 + 5x + 4
(ii) y − 2, 3y3 + 5y2 + 5y + 2
(iii) 4x2 − 5, 4x4 + 7x2 + 15
(iv) 4 − z, 3z2 − 13z + 4
(v) 2a − 3, 10a2 − 9a − 5
(vi) 4y + 1, 8y2 − 2y + 1

Answer:

(i) 2x2+5x+4x+1=2x(x+1)+3(x+1)+1x+1=(x+1)(2x+3)+1(x+1)=(2x+3)+1x+1 Remainder=1Therefore, (x+1) is not a  factor of 2x2+5x+4

(ii) 3y3+5y2+5y+2y-2=3y2(y-2)+11y(y-2)+27(y-2)+56y-2=(y-2)(3y2+11y+27)+56y-2=(3y2+11y+27)+56y-2 Remainder = 56 (y-2) is not a factor of 3y3+5y2+5y+2.


(iii)  4x4+2+154x2-5= x2(4x2-5)+3(4x2-5)+304x2-5= (4x2-5)(x2+3)+304x2-5=(x2+3)+304x2-5 Remainder = 30Therefore, (4x2-5) is not a factor of 4x4+7x2+15

(iv) 3z2-13z+44-z=3z2-12z-z+44-z=3z(z-4)-1(z-4)4-z=(z-4)(3z-1)4-z=(4-z)(1-3z)4-z=1-3z Remainder = 0 (4-z) is a factor of 3z2-13z+4.

(V) 10a2-9a-52a-3=5a(2a-3)+3(2a-3)+42a-3=(2a-3)(5a+3)+42a-3=(5a+3)+42a-3 Remainder = 4 ( 2a-3) is not a factor of 10a2-9a-5.

(vi) 8y2-2y+14y+1=2y(4y+1)-1(4y+1)+24y+1=(4y+1)(2y-1)+24y+1=(2y-1)+24y+1 Remainder = 2 (4y+1) is not a factor of 8y2-2y+1.



Page No 8.17:

Question 1:

Divide:
x2 − 5x + 6 by x − 3

Answer:

x2-5x+6x-3=x2-3x-2x+6x-3=x(x-3)-2(x-3)(x-3)=(x-3)(x-2)(x-3)= x-2

Page No 8.17:

Question 2:

Divide:
ax2ay2 by ax + ay

Answer:

ax2-ay2ax+ay=a(x2-y2)a(x+y)=a(x+y)(x-y)a(x+y)= x-y

Page No 8.17:

Question 3:

Divide:
x4y4 by x2y2

Answer:

 x4-y4x2-y2=(x2)2-(y2)2(x2-y2)=(x2+y2)(x2-y2)(x2-y2)= x2+y2

Page No 8.17:

Question 4:

Divide:
acx2 + (bc + ad)x + bd by (ax + b)

Answer:

acx2+(bc+ad)x+bd(ax+b)=acx2+bcx+adx+bd(ax+b)=cx(ax+b)+d(ax+b)(ax+b)=(ax+b)(cx+d)(ax+b)= cx+d

Page No 8.17:

Question 5:

Divide:
(a2 + 2ab + b2) − (a2 + 2ac + c2) by 2a + b + c

Answer:

(a2+2ab+b2)-(a2+2ac+c2)(2a+b+c)=(a+b)2-(a+c)2(2a+b+c)=(a+b+a+c)(a+b-a-c)(2a+b+c)=(2a+b+c)(b-c)(2a+b+c)=b-c

Page No 8.17:

Question 6:

Divide:
14x2-12x-12 by 12x-4

Answer:

14x2-12x-1212x-4=12x(12x-4)+3(12x-4)12x-4=(12x-4)(12x+3)(12x-4)=12x+3



Page No 8.2:

Question 1:

Write the degree of each of the following polynomials.
(i) 2x2 + 5x2 − 7
(ii) 5x2 − 3x + 2
(iii) 2x + x2 − 8
(iv) 12y7-12y6+48y5-10
(v) 3x3 + 1
(vi) 5
(vii) 20x3 + 12x2y2 − 10y2 + 20

Answer:

(i)  Correction  : It is 2x3+5x2-7  instead of 2x2+5x2-7.     The degree of the polymonial  2x3+5x2-7 is 3.(ii) The degree of the polymonial 5x2-35x+2 is 2.(iii) The degree of the polymonial 2x+x2-8 is 2.(iv) The degree of the polymonial 12y7-12y6+48y5-10 is 7.(v) The degree of the polymonial 3x3+1 is 3.(vi) 5 is a constant polynomial and its degree is 0.(vii) The degree of the polymonial 20x3+12x2y2-10y2+20 is 4.

Page No 8.2:

Question 2:

Which of the following expressions are not polynomials?
(i) x2 + 2x−2
(ii) ax+x2-x3
(iii) 3y35y + 9
(iv) ax1/2 + ax + 9x2 + 4
(v) 3x−2 + 2x−1 + 4x +5

Answer:

(i) x2+2x-2 is not a polynomial because -2 is the power of variable x is not a non negative integer.(ii) ax+x2-x3 is not a polynomial because 12 is the power of variable x is not a non negative integer.(iii) 3y3-5y+9 is a polynomial because the powers of variable y are non negative integers.(iv) ax12+ax+9x2+4 is not a polynomial because 12 is the power of variable x is not a non negative integer.(v) 3x-2+2x-1+4x+5 is not a polynomial because -2 and -1 are the powers of variable x are not non negative integers.

Page No 8.2:

Question 3:

Write each of the following polynomials in the standard form. Also, write their degree.
(i) x2 + 3 + 6x + 5x4
(ii) a2 + 4 + 5a6
(iii) (x3 − 1)(x3 − 4)
(iv) (y3 − 2)(y3 + 11)
(v) a3-38a3+1617
(vi) a+34a+43

Answer:

(i) Standard form of the given polynomial can be expressed as:(5x4+x2+6x+3) or (3+6x+x2+5x4) The degree of the polynomial is 4.(ii) Standard form of the given polynomial can be expressed as:(5a6+a2+4) or (4+a2+5a6) The degree of the polynomial is 6.(iii) (x3-1)(x3-4)=x6-5x3+4Standard form of the given polynomial can be expressed as:(x6-5x3+4) or (4-5x3+x6)The degree of the polynomial is 6.(iv) (y3-2)(y3+11)=y6+9y3-22Standard form of the given polynomial can be expressed as:(y6+9y3-22) or (-22+9y3+y6)The degree of the polynomial is 6.(v) (a3-38)(a3+1617)=a6+77136a3-617Standard form of the given polynomial can be expressed as:(a6+77136a3-617) or (-617+77136a3+a6)The degree of the polynomial is 6.(vi) (a+34)(a+43)=a2+2512a+1Standard form of the given polynomial can be expressed as:(a2+2512a+1) or (1+2512a+a2)The degree of the polynomial is 2.



Page No 8.4:

Question 1:

Divide 6x3y2z2 by 3x2yz.

Answer:

6x3y2z23x2yz=6×x×x×x×y×y×z×z3×x×x×y×z = 2x(3-2)y(2-1)z(2-1)=2xyz

Page No 8.4:

Question 2:

Divide 15m2n3 by 5m2n2.

Answer:

15m2n35m2n2=15×m×m×n×n×n5×m×m×n×n=3m(2-2)n(3-2)=3m0n1=3n

Page No 8.4:

Question 3:

Divide 24a3b3 by −8ab.

Answer:

24a3b3-8ab= 24×a×a×a×b×b×b-8×a×b=-3a(3-1)b(3-1)=-3a2b2

Page No 8.4:

Question 4:

Divide −21abc2 by 7abc.

Answer:

-21abc27abc= -21×a×b×c×c7×a×b×c=-3a(1-1)b(1-1)c(2-1)=-3c

Page No 8.4:

Question 5:

Divide 72xyz2 by −9xz.

Answer:

72xyz2-9xz=72×x×y×z×z-9×x×z=-8x(1-1)yz(2-1)=-8yz

Page No 8.4:

Question 6:

Divide −72a4b5c8 by −9a2b2c3.

Answer:

-72a4b5c8-9a2b2c3=-72×a×a×a×a×b×b×b×b×b×c×c×c×c×c×c×c×c-9×a×a×b×b×c×c×c=8a(4-2)b(5-2)c(8-3)=8a2b3c5

Page No 8.4:

Question 7:

Simplify:
16m3y24m2y

Answer:

16m3y24m2y=16×m×m×m×y×y4×m×m×y=4m(3-2)y(2-1)=4my

Page No 8.4:

Question 8:

Simplify:
32m2n3p24mnp

Answer:

32m2n3p24mnp=32×m×m×n×n×n×p×p4×m×n×p=8m(2-1)n(3-1)p(2-1)=8mn2p



Page No 8.6:

Question 1:

Divide x + 2x2 + 3x4x5 by 2x.

Answer:

x+2x2+3x4-x52x=x2x+2x22x+3x42x-x52x=12+x+32x3-12x4                                                                                   

Page No 8.6:

Question 2:

Divide y4-3y3+12y2 by 3y.

Answer:

y4-3y3+12y23y=y43y-3y33y+12y23y=13y(4-1)-y(3-1)+16y(2-1)=13y3-y2+16y

Page No 8.6:

Question 3:

Divide −4a3 + 4a2 + a by 2a.

Answer:

-4a3+4a2+a2a=-4a32a+4a22a+a2a=-2a(3-1)+2a(2-1)+12=-2a2+2a+12

                                                                       

Page No 8.6:

Question 4:

Divide -x6+2x4+4x3+2x2 by 2x2.

Answer:

-x6+2x4+4x3+2x22x2=-x62x2+2x42x2+4x32x2+2x22x2=-12x(6-2)+2x(4-2)+22x(3-2)+2x(2-2)=-12x4+2x2+22x+2                                            
                                            

Page No 8.6:

Question 5:

Divide 5z3 − 6z2 + 7z by 2z.

Answer:

5z3-6z2+7z2z=5z32z-6z22z+7z2z=52z(3-1)-3z(2-1)+72=52z2-3z+72
                                                                

Page No 8.6:

Question 6:

Divide 3 a4+23 a3+3a2-6a by 3a.

Answer:

3a4+23a3+3a2-6a3a=3a43a+23a33a+3a23a-6a3a=13a(4-1)+23a(3-1)+a(2-1)-2=13a3+23a2+a-2                                           
                                          



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