RD Sharma Solutions for Class 8 Math Chapter 8 - Division of Algebraic Expressions (Part-3) | RD Sharma Solutions for Class 8 Mathematics PDF Download

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 1: 

(i)

(3x+10)
Therefore, quotient=3x+10 and remainder=25. (ii)

=(2x−1/5)+
Therefore,
quotient =2x−1/5 and remainder=47/5.
(iii)

=(y²−y+1)
Therefore, 

Quotient = y²−y+1 and remainder = 0
 

Therefore, quotient = x3+5 and remainder = 5. 
(v)


=(y²+3)+
Therefore, quotient = y2+3 and remainder = 6. 
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 2: 

(i)

=(2x+3)
∵ Remainder=1
Therefore, (x+1) is not a  factor of 2x2+5x+4
(ii)  

=(3y2+11y+27)+ 56/y-2
∵ Remainder = 56

∴ (y−2) is not a factor of 3y³+5y²+5y+2.
(iii)

=(x2+3
)
∵ Remainder = 30

Therefore, (4x²−5) is not a factor of 4x⁴+7x²+15
(iv)



=1−3z
∵ Remainder = 0
∴ (4−z) is a factor of 3z²−13z+4.
(v)

=(5a+3)+ 
∵ Remainder = 4

∴ ( 2a−3) is not a factor of 10a²−9a−5.
(vi) 

=(2y−1)
∵ Remainder = 2

∴ (4y+1) is not a factor of 8y²−2y+1.
Question 3: Divide:
x² − 5x + 6 by x − 3
Answer 3: 

= x−2 
Question 4:Divide:
ax² − ay² by ax + ay
Answer 4:
 

= x−y 
Question 5: Divide:
x⁴ − y⁴ by x² − y²
Answer 5:
 

= x2+y2 
Question 6: Divide:
acx² + (bc + ad)x + bd by (ax + b)
Answer 6: 

= cx+d 
Question 7: Divide:
(a² + 2ab + b²) − (a² + 2ac + c²) by 2a + b + c
Answer 7: 

=b−c 
Question 8: Divide: 

Answer 8: 



=1/2 x+3