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

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).