# RS Aggarwal Class 8 Math Sixth Chapter Operation on Algebric Expressions Exercise 6C Solution

## EXERCISE 6C

### (1) Divide:

**(i) 24x ^{2}y^{3} by 3xy**

= 8xy^{2}

**(ii) 36xyz ^{2} by – 9xz**

= – 4yz

**(iii) – 72x ^{2}y^{2}z by – 12xyz**

= 6xy

**(iv) – 56mnp ^{2} by 7mnp**

= – 8p

### (2) Divide:

**(i) 5m ^{3} – 30m^{2} + 45m by 5m**

**(ii) 8x ^{2}y^{2} – 6xy^{2} + 10x^{2}y^{2} by 2xy**

**(iii) 9x ^{2}y – 6xy + 12xy^{2} by – 3xy**

**(iv) 12x ^{4} + 8x^{3} – 6x^{2} by – 2x^{2}**

### Write the quotient and remainder when we divide:

**(3) (x ^{2} – 4x + 4) by (x – 2)**

**(4) (x ^{2} – 4) by (x + 2)**

(5) (x^{2} + 12x + 35) by (x + 7)

**(6) (15x ^{2} + x – 6) by (3x + 2)**

**(7) (14x ^{2} – 53x + 45) by (7x – 9)**

**(8) (6x ^{2} – 31x + 47) by (2x – 5)**

Therefore the quotient is (3x – 8) and the reminder is 7.

**(9) (2x ^{3} + x^{2} – 5x – 2) by (2x + 3)**

Therefore the quotient is (x^{2} – x – 1) and the remainder is 1.

**(10) (x ^{3} + 1) by (x + 1)**

Therefore the quotient is (x^{2} – x + 1) and the remainder is 0.

**(11) (x ^{4} – 2x^{3} + 2x^{2} + x + 4) by (x^{2} + x + 1)**

Therefore the quotient is (x^{2} – 3x + 4) and the remainder is 0.

**(12) (x ^{3} – 6x^{2} + 11x – 6) by (x^{2} – 5x + 6)**

Therefore the quotient is (x – 1) and the remainder is 0.

**(13) (5x ^{3} – 12x^{2} + 12x + 13) by (x^{2} – 3x + 4)**

Therefore the quotient is (5x + 3) and the remainder is (x + 1).

**(14) (2x ^{3} – 5x^{2} + 8x – 5) by (2x^{2} – 3x + 5)**

Therefore the quotient is (x – 1) and the remainder is 0.

**(15) (8x ^{4} + 10x^{3} – 5x^{2} – 4x + 1) by (2x^{2} + x – 1)**

Therefore the quotient is (4x^{2} + 3x – 2) and the remainder is (x – 1).