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- Class 8
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- division of algebraic expressions
Rd Sharma 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 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 Solutions. All Rd Sharma 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 + 6z2 − z by − 12z.
Answer:
Page No 8.11:
Question 3:
Divide 9x2y − 6xy + 12xy2 by −32xy.
Answer:
Page No 8.11:
Question 4:
Divide 3x3y2 + 2x2y + 15xy by 3xy.
Answer:
Page No 8.11:
Question 5:
Divide x2 + 7x + 12 by x + 4.
Answer:
Page No 8.11:
Question 6:
Divide 4y2 + 3y + 12 by 2y + 1.
Answer:
Page No 8.11:
Question 7:
Divide 3x3 + 4x2 + 5x + 18 by x + 2.
Answer:
Page No 8.11:
Question 8:
Divide 14x2 − 53x + 45 by 7x − 9.
Answer:
Page No 8.11:
Question 9:
Divide −21 + 71x − 31x2 − 24x3 by 3 − 8x.
Answer:
Page No 8.11:
Question 10:
Divide 3y4 − 3y3 − 4y2 − 4y by y2 − 2y.
Answer:
Page No 8.11:
Question 11:
Divide 2y5 + 10y4 + 6y3 + y2 + 5y + 3 by 2y3 + 1.
Answer:
Page No 8.11:
Question 12:
Divide x4 − 2x3 + 2x2 + x + 4 by x2 + x + 1.
Answer:
Page No 8.11:
Question 13:
Divide m3 − 14m2 + 37m − 26 by m2 − 12m +13.
Answer:
Page No 8.11:
Question 14:
Divide x4 + x2 + 1 by x2 + x + 1.
Answer:
Page No 8.11:
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 6x3 − x2 − 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 =10x
2-3x-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 andremainder = 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
= 14x2 + 21x - 8x - 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+10 3z+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 + 90x - 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
= 15y 4-6y3+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] + 4y2 + 25y + 4
= 6y5
+ 4y4 + 4y3 + 3y2 + 2y + 2 + 4y2 + 25y + 4
= 6y5
+ 4y4 + 4y3 + 7y2 + 27y + 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 x2 − x − 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] 2x2 − x + 3 is a factor of 6x5 − x4 + 4x3 − 5x2 − x − 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+4y 2+10y-35.
[v]
Remainder is zero; therefore, z2 + 3 is a factor of z 5 -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, [x - 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] x4 − x3 + 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] 10
x2-7x+85x-3=2x[5x-3]-15[5
x-3]+475[5x-3]=[5x-3][2x-1
5]+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+4 x+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] i s 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+1 4y+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:
ax2 − ay2 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:
x4 − y4 by x2 − y2
Answer:
x4-y4x2-y2=[x2]2-[y2]2[x2-y2]=[x2+y2][x2-y2][x2-y2]= x2+y 2
Page No 8.17:
Question 4:
Divide:
acx2 + [bc + ad]x + bd by [ax + b]
Answer:
acx2+[b c+ad]x+bd[ax+b]=acx2+b cx+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] 12
y7-12y6+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+12x2 y2-10y2+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] 3y3 − 5y + 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+9 x2+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+9y 3+y6]The degree of the polynomial is 6.[v] [a 3-38][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+1 Standard 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:
15m2n3 5m2n2=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]=-3 c
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:
-72a4b5c8-9a 2b2c3=-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 + 3x4 − x5 by 2x.
Answer:
x+2x2+3x4-x52x=x2x+2x22x +3x42x-x52x=12+x+32x3-12x4
Page No 8.6:
Question 2:
Divide y4-3 y3+12y2 by 3y.
Answer:
y4-3y3 +12y23y=y43y-3 y33y+12y23y=13 y[4-1]-y[3-1]+16y[2-1]=13y3-y2 +16y
Page No 8.6:
Question 3:
Divide −4a3 + 4a2 + a by 2a.
Answer:
-4 a3+4a2+a2a=-4a32a+4a22a+a2a=-2a [3-1]+2a[2-1]+1 2=-2a2+2a+12
Page No 8.6:
Question 4:
Divide -x6+2x4+4x3+2x2 by 2x2.
Answer:
-x6+2x4+4x3+2x22
x2=-x62x2+2x4
2x2+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]-3
z[2-1]+72=52z2
-3z+72
Page No 8.6:
Question 6:
Divide 3 a4+23 a3+3a2-6a by 3a.
Answer:
3a4+23a3+3a2-6a3
a=3a43a+23a33a
+3a23a-6a3a=13a[4-1]+23a[3-1
]+a[2-1]-2=13a
3+23a2+a-2
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