# Class 8 RD Sharma Solutions – Chapter 8 Division Of Algebraic Expressions – Exercise 8.5

• Last Updated : 11 Feb, 2021

### (i) 3x2 + 4x + 5, x – 2

Solution:

3x2 + 4x + 5, x – 2

By using factorization method,  (Taking common (x-2) factor) ∴ the Quotient is 3x + 10 and the Remainder is 25.

### (ii) 10x2 – 7x + 8, 5x – 3

Solution:

10x2 – 7x + 8, 5x – 3

By using factorization method,  (Taking common (5x-3) factor) ∴ the Quotient is (2x – 1/5) and the Remainder is 37/5.

### (iii) 5y3 – 6y2 + 6y – 1, 5y – 1

Solution:

5y3 – 6y2 + 6y – 1, 5y – 1

By using factorization method,  (Taking common (5y-1) factor) ∴ the Quotient is (y2 – y + 1) and the Remainder is 0.

### (iv) x4 – x3 + 5x, x – 1

Solution:

x4 – x3 + 5x, x – 1

By using factorization method,  (Taking common (x-1) factor) ∴ the Quotient is x3 + 5 and the Remainder is 5.

### (v) y4 + y2, y2 – 2

Solution:

y4 + y2, y2 – 2

By using factorization method,  (Taking common (y2-2) factor) ∴ the Quotient is y2 + 3 and the Remainder is 6.

### (i) x + 1, 2x2 + 5x + 4

Solution:

x + 1, 2x2 + 5x + 4

Let us perform factorization method,  (Taking common (x+1) factor) Since remainder is 1, therefore the first polynomial is not a factor of the second polynomial.

### (ii) y – 2, 3y3 + 5y2 + 5y + 2

Solution:

y – 2, 3y3 + 5y2 + 5y + 2

Let us perform factorization method,  (Taking common (y-2) factor) Since remainder is 56 therefore the first polynomial is not a factor of the second polynomial.

### (iii) 4x2 – 5, 4x4 + 7x2 + 15

Solution:

4x2 – 5, 4x4 + 7x2 + 15

Let us perform factorization method,  (Taking common (4x2-5) factor) Since remainder is 30 therefore the first polynomial is not a factor of the second polynomial.

### (iv) 4 – z, 3z2 – 13z + 4

Solution:

4 – z, 3z2 – 13z + 4

Let us perform factorization method,  (Taking common (z-4) factor)   Since remainder is 0 therefore the first polynomial is a factor of the second polynomial.

### (v) 2a – 3, 10a2 – 9a – 5

Solution:

2a – 3, 10a2 – 9a – 5

Let us perform factorization method,  (Taking common (2a-3) common) Since remainder is 4 therefore the first polynomial is not a factor of the second polynomial.

### (vi) 4y + 1, 8y2 – 2y + 1

Solution:

4y + 1, 8y2 – 2y + 1

Let us perform factorization method,  (Taking common (4y+1) factor) Since remainder is 2 therefore the first polynomial is not a factor of the second polynomial.

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