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 + 6z² − z by − $\frac{1}{2}$z.

Answer 2:

Question 3:

Divide 9x²y − 6xy + 12xy² by −$\frac{3}{2}$xy.

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 + $\frac{1}{2}$ 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:

$$\begin{gathered} \mathrm{Quotient} = 7x^2 + x + 5 \\ \mathrm{Remainder} = 4 \end{gathered}$$

Question 17:

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

Answer 17:

$$\begin{gathered} \mathrm{Quotient} = 3x^2+ 4x + 1 \\ \mathrm{Remainder} = 0 \end{gathered}$$

Question 18:

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

Answer 18:

$$\begin{gathered} \mathrm{Quotient} = 2x-5 \\ \mathrm{Remainder} =0 \end{gathered}$$

Page-8.12

Question 19:

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

Answer 19:

$$\begin{gathered} \mathrm{Quotient} =10x^2-3x-12 \\ \mathrm{Remainder}= 0 \end{gathered}$$

Question 20:

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

Answer 20:

$\therefore$ 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 − $\frac{10}{3}$y + 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 $\times$ Quotient + Remainder = (7x $-$ 4) (2x + 3) $-$ ​3 
                                                = 14x² + 21x $-$ 8x $-$ 12 $-$ ​3 
                                                = 14x² + 13x $-$ 15
                                                = Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(ii)

$$\begin{gathered} \mathrm{Quotient} = 5z^2+\frac{10}{3}z+11 \\ \mathrm{Remainder} = 54 \\ \mathrm{Divisor} = 3z-6 \\ \mathrm{Divisor} \times \mathrm{Quotient} +\mathrm{Remainder} = (3z-6)\left( 5z^2+\frac{10}{3}z+11\right)+54 \\ = 15z^3+10z^2+33z-30z^2-20z-66+54 \\ = 15z^3-20z^2+13z-12 \\ = \mathrm{Dividend} \\ \text{Thus}, \\ \mathrm{Divisor} \times \mathrm{Quotient} + \mathrm{Remainder} = \mathrm{Dividend} \end{gathered}$$
Hence verified.

(iii)


Quotient = $3y^3-5y+\frac{3}{2}$
Remainder = 0
Divisor = 2y² $-$ 6
Divisor $\times$ Quotient + Remainder =
$$\begin{gathered} (2y^2-6) \left(3y^3-5y+\frac{3}{2}\right)+0 \\ =6y^5-10y^3+3y^2-18y^3+30y-9 \\ =6y^5-28 y^3+3y^2+30y-9 \end{gathered}$$
= Dividend
 
Thus, Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(iv)

Quotient  = $-$ 4x³ + 2x² $-$ 8x + 30
Remainder  = $-$ 285 
Divisor  = 3x + 7
Divisor $\times$ 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 $\times$ Quotient + Remainder = Dividend
Hence verified.

(v)


Quotient =  $5y^3-2y^2+\frac{5}{3}y$
Remainder =  6
Divisor = 3y $-$ 2
Divisor $\times$ Quotient  + Remainder = (3y $-$ 2) (5y³ $-$ 2y² + $\frac{5}{3}y$) + 6
                                                = $15y^4-6y^3+5y^2-10y^3+4y^2-\frac{10}{3}y+6$
                                                = $15y^4-16y^3+9y^2-\frac{10}{3}y+6$
                                                =  Dividend
Thus,
Divisor $\times$ Quotient + Remainder = Dividend
Hence verified.

(vi)

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

(vii)




Quotient = 3y² + 2y + 2
Remainder = 4y² + 25y + 4
Divisor = 2y³ + 1
Divisor $\times$ 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 $\times$ Quotient + Remainder = Dividend
Hence verified.

Question 22:

Divide 15y⁴ + 16y³ + $\frac{10}{3}$y − 9y² − 6 by 3y − 2. Write down the coefficients of the terms in the quotient.

Answer 22:

$\therefore$ 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  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 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)




$\because$ The remainder is non zero,
 2y $-$ 5 is not a factor of $4y^4-10y^3-10y^2+30y-15$.

(iv)

Remainder is zero.  Therefore, 3y² + 5 is a factor of $6y^5+15y^4+16y^3+4y^2+10y-35$.


(v)

Remainder is zero; therefore, z² + 3 is a factor of $z^{5 }-9z$.

(vi)



Remainder is zero ; therefore, $2x^2-x+3$ is a factor of $6x^5-x^{4 }+4x^3-5x^2-x-15$.

Question 24:

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

Answer 24:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{a} \mathrm{if} (\mathrm{x}+2) \mathrm{is} \mathrm{a} \mathrm{factor} \mathrm{of} (4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}). \\ S\mathrm{ubstituting} \mathrm{x}=-2 \mathrm{in} 4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}, \mathrm{we} \mathrm{get}: \\ 4(-2{)}^4+2(-2{)}^3-3(-2{)}^2+8(-2)+5\mathrm{a}=0 \\ \mathrm{or}, 64-16-12-16+5\mathrm{a}=0 \\ \mathrm{or}, 5\mathrm{a}=-20 \\ \mathrm{or}, \mathrm{a}=-4 \\ \therefore \mathrm{If} (\mathrm{x}+2) \mathrm{is} \mathrm{a} \mathrm{factor} \mathrm{of} (4{\mathrm{x}}^4+2{\mathrm{x}}^3-3{\mathrm{x}}^2+8\mathrm{x}+5\mathrm{a}), \mathrm{a}=-4. \end{gathered}$$

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, (x $-$ 2) should be added to ($x^4+2x^3-2x^2+x-1$) to make the resulting polynomial exactly divisible by ($x^2+2x-3$).