RD Sharma Solutions for Class 8 Math Chapter 8 - Division of Algebraic Expressions (Part-2) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Divide x + 2x² + 3x⁴ − x⁵ by 2x.
Answer 1: 

Question 2: Divide y4−3y3+1/2 y2 by 3yy4-3y3+12y2 by 3y.
Answer 2:
 

Question 3: Divide −4a³ + 4a² + a by 2a.
Answer 3: 

=−2a2+2a+1/2
Question 4: Divide −x6+2x4+4x3+2x2 by
Answer 4:
 


 

Question 5: Divide 5z³ − 6z² + 7z by 2z.
Answer 5: 

Question 6: Divide +3a2−6a by 3a3 a4+23 a3+3a2-6a by 3a.
Answer 6:

Question 7: Divide 5x³ − 15x² + 25x by 5x.
Answer 7: 

Question 8: Divide 4z³ + 6z² − z by −1/2 12z.
Answer 8: 

Question 9: Divide 9x²y − 6xy + 12xy² by −3/2 32xy.
Answer 9: 

Question 10: Divide 3x³y² + 2x²y + 15xy by 3xy.

Answer 10:

Question 11: Divide x² + 7x + 12 by x + 4.
Answer 11:

Question 12: Divide 4y² + 3y + 1/212 by 2y + 1.
Answer 12:

 
Question 13: Divide 3x³ + 4x² + 5x + 18 by x + 2.
Answer 13: 

Question 14: Divide 14x² − 53x + 45 by 7x − 9.
Answer 14: 

Question 15: Divide −21 + 71x − 31x² − 24x³ by 3 − 8x.
Answer 15: 

Question 16: Divide 3y⁴ − 3y³ − 4y² − 4y by y² − 2y.
Answer 16: 

Question 17: Divide 2y⁵ + 10y⁴ + 6y³ + y² + 5y + 3 by 2y³ + 1.
Answer 17: 

Question 18: Divide x⁴ − 2x³ + 2x² + x + 4 by x² + x + 1.
Answer 18: 

Question 19: Divide m³ − 14m² + 37m − 26 by m² − 12m +13.
Answer 19:

Question 20: Divide x⁴ + x² + 1 by x² + x + 1.
Answer 20: 

Question 21: Divide x⁵ + x⁴ + x³ + x² + x + 1 by x³ + 1.
Answer 21: 

Question 22: Divide 14x³ − 5x² + 9x − 1 by 2x − 1 and find the quotient and remainder
Answer 22: 

Quotient = 7x2 + x + 5Remainder = 4

Question 23: Divide 6x³ − x² − 10x − 3 by 2x − 3 and find the quotient and remainder.
Answer 23: 

Quotient = 3x2+ 4x + 1 Remainder = 0Question 24: Divide 6x³ + 11x² − 39x − 65 by 3x² + 13x + 13 and find the quotient and remainder.
Answer 24: 

Quotient = 2x−5Remainder =0 

Question 25: Divide 30x⁴ + 11x³ − 82x² − 12x + 48 by 3x² + 2x − 4 and find the quotient and remainder.
Answer 25: Quotient =10x2−3x−12 
Remainder= 0 

Question 26: Divide 9x⁴ − 4x² + 4 by 3x² − 4x + 2 and find the quotient and remainder.
Answer 26: 

∴∴ Quotient = 3x² + 4x + 2 and remainder = 0.
Question 27: Verify the division algorithm i.e. Dividend = Divisor × Quotient + Remainder, in each of the following. Also, write the quotient and remainder.

Answer 27: (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+10/3 z+11Remainder = 54Divisor = 3z−6Divisor × Quotient +Remainder = (3z−6)( 5z2+10/3 z+11)+54= 15z3+10z2+33z−30z2−20z−66+54 

= 15z3−20z2+13z−12 

= Dividend 
Thus,Divisor × Quotient + Remainder = DividendHence verified. 
(iii) 

  
Quotient = 3y3−5y+323y3-5y+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+5/3 y5y3-2y2+53y
Remainder =  6
Divisor = 3y −- 2
Divisor ×× Quotient  + Remainder = (3y −- 2) (5y³ −- 2y² + 5/3 y53y) + 6
= 15y4−6y3+5y2−10y3+4y2−10/3 y+615y4-6y3+5y2-10y3+4y2-103y+6
= 15y4−16y3+9y2−10/3 y+615y4-16y3+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. 

Question 28: Divide 15y⁴ + 16y³ + 10/3 103y − 9y² − 6 by 3y − 2. Write down the coefficients of the terms in the quotient.
Answer 28: 

∴∴ 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) 

Question 29: 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 29: (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−154y4-10y3-10y2+30y-15. 
(iv) 

Remainder is zero.  Therefore, 3y² + 5 is a factor of 6y5+15y4+16y3+4y2+10y−356y5+15y4+16y3+4y2+10y-35. 
(v) 

Remainder is zero; therefore, z² + 3 is a factor of z5 −9zz5 -9z. 
(vi) 

Remainder is zero ; therefore, 2x2−x+32x2-x+3 is a factor of 6x5−x4 +4x3−5x2−x−156x5-x4 +4x3-5x2-x-15. 
Question 30: Find the value of a, if x + 2 is a factor of 4x⁴ + 2x³ − 3x² + 8x + 5a.

Answer 30:  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.Question 31: What must be added to x⁴ + 2x³ − 2x² + x − 1 , so that the resulting polynomial is exactly divisible by x² + 2x − 3?
Answer 31: 

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