RD Sharma solution class 8 chapter 8 Division of Algebraic Expression Exercise 8.4

Exercise 8.4

Page-8.11

Question 1:

Divide 5x3 − 15x2 + 25x by 5x.

Answer 1:

Question 2:

Divide 4z3 + 6z2z by − 12z.

Answer 2:

Question 3:

Divide 9x2y − 6xy + 12xy2 by −32xy.

Answer 3:

Question 4:

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

Answer 4:

Question 5:

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

Answer 5:

Question 6:

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

Answer 6:

Question 7:

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

Answer 7:

Question 8:

Divide 14x2 − 53x + 45 by 7x − 9.

Answer 8:

Question 9:

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

Answer 9:

Question 10:

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

Answer 10:

Question 11:

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

Answer 11:

Question 12:

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

Answer 12:

Question 13:

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

Answer 13:

Question 14:

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

Answer 14:

Question 15:

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

Answer 15:

Question 16:

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

Answer 16:


Quotient = 7x2 + x + 5Remainder = 4

Question 17:

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

Answer 17:


Quotient = 3x2+ 4x + 1 Remainder = 0

Question 18:

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

Answer 18:


Quotient = 2x-5Remainder =0

Page-8.12

Question 19:

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

Answer 19:

Quotient =10x2-3x-12Remainder= 0

Question 20:

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

Answer 20:


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

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 21:

(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.

Question 22:

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

Answer 22:


 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 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 23:

(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.

Question 24:

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

Answer 24:

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 25:

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

Answer 25:


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

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