# DAV Class 8 Maths Chapter 8 Worksheet 2 Solutions

The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 8 Worksheet 2 Solutions of Polynomials offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 8 WS 2 Solutions

Question 1.
Divide the following monomials by the given monomial:

(i) 6x3 by 3x2
Solution:
6x3 ÷ 3x2 = $$\frac{6 x^3}{3 x^2}$$
= $$\left(\frac{6}{3}\right)\left(\frac{x^3}{x^2}\right)$$
= 2 . x3 – 2
= 2x

(ii) – 35x4 by – 7x3
Solution:
– 35x4 ÷ (- 7x3) = $$\frac{-35 x^4}{-7 x^3}$$
= $$\left(\frac{-35}{-7}\right)\left(\frac{x^4}{x^3}\right)$$
= 5 . x4 – 3
= 5x

(iii) – 5z2 by √5z
Solution:
– 5z2 ÷ √5z = $$\frac{-5 z^2}{\sqrt{5} z}$$
= $$\left(\frac{-5}{\sqrt{5}}\right)\left(\frac{z^2}{z}\right)$$
= – √5z2 – 1
= √5z

(iv) 16p4 by – 6p2
Solution:
16p4 ÷ (- 6p2) = $$\frac{16 p^4}{-6 p^2}$$
= $$\left(\frac{16}{-6}\right)\left(\frac{p^4}{p^2}\right)$$
= $$\frac{-8}{3}$$ p4 – 2
= $$\frac{-8}{3}$$ p2

(v) 4√2 y3 by 3√2 y2
Solution:
4√2 y3 ÷ 3√2 y2 = $$\frac{4 \sqrt{2} y^3}{3 \sqrt{2} y^2}$$
= $$\left(\frac{4 \sqrt{2}}{3 \sqrt{2}}\right)\left(\frac{y^3}{y^2}\right)$$
= $$\frac{4}{3}$$ y3 – 2
= $$\frac{4}{3}$$ y

(vi) $$\frac{3}{4}$$ p2 by $$\frac{4}{3}$$2
Solution:
$$\frac{3}{4}$$ p2 ÷ $$\frac{4}{3}$$2 = $$\frac{\frac{3}{4} p^2}{\frac{4}{3} p^2}$$
= $$\left(\frac{3}{4} \times \frac{3}{4}\right)\left(\frac{p^2}{p^2}\right)$$
= $$\frac{9}{16}$$ p2 – 2
= $$\frac{9}{16}$$ p0
= $$\frac{9}{16}$$

Question 2.
Divide the following polynomials by a monomial:

(i) 6x4 – 24x3 + 15x2 + 9 by (- 3x2)
Solution:
6x4 – 24x3 + 15x2 + 9 ÷ (- 3x2) = $$\frac{6 x^4-24 x^3+15 x^2+9}{-3 x^2}$$
= $$\frac{6 x^4}{-3 x^2}-\frac{24 x^3}{-3 x^2}+\frac{15 x^2}{-3 x^2}+\frac{9}{-3 x^2}$$
= – 2x2 + 8x – 5 – $$\frac{3}{x^2}$$

(ii) – 12x + 22x2 – 16x3 + 4 by 2x
Solution:
– 12x + 22x2 – 16x3 + 4 ÷ 2x = $$\frac{-12 x+22 x^2-16 x^3+4}{2 x}$$
= $$\frac{-12 x}{2 x}+\frac{22 x^2}{2 x}-\frac{16 x^3}{2 x}+\frac{4}{2 x}$$
= – 6 + 11x – 8x2 + $$\frac{2}{x}$$

(iii) $$\frac{2}{3}$$ z4 – $$\frac{1}{3}$$ z2 – 1 by $$\frac{1}{3}$$ z
Solution:
$$\frac{2}{3}$$ z4 – $$\frac{1}{3}$$ z2 – 1 ÷ $$\frac{1}{3}$$ z = $$\frac{\frac{2}{3} z^4-\frac{1}{3} z^2-1}{\frac{1}{3} z}$$
= $$\frac{\frac{2}{3} z^4}{\frac{1}{3} z}-\frac{\frac{1}{3} z^2}{\frac{1}{3} z}-\frac{1}{\frac{1}{3} z}$$
= 2z3 – z – $$\frac{3}{z}$$

(iv) 2√2 q4 + 4√2 q3 + q2 by – 2√2 q2
Solution:
2√2 q4 + 4√2 q3 + q2 ÷ – 2√2 q2 = $$\frac{2 \sqrt{2} q^4+4 \sqrt{2} q^3+q^2}{-2 \sqrt{2} q^2}$$
= $$\frac{2 \sqrt{2} q^4}{-2 \sqrt{2} q^2}+\frac{4 \sqrt{2} q^3}{-2 \sqrt{2} q^2}+\frac{q^2}{-2 \sqrt{2} q^2}$$
= – q2 – 2q – $$\frac{1}{2 \sqrt{2}}$$

Question 3.
Divide the following polynomials by a monomial using long division method. Also, write the quotient and the remainder.

(i) – 8x3 + 6x4 – 4 + 12x by 2x2
Solution:
Here, dividend = – 8x3 + 6x4 – 4 + 12x
Dividend in standard form = 6x4 – 8x3 + 12x – 4
Divisor = 2x2

[∵ 6x4 ÷ 2x2 = 3x2,
– 8x3 ÷ 2x2 = – 4x]

Now, quotient = 3x2 – 4x
Remainder = 12x – 4

(ii) 5x10 – 9x8 – 9x5 + 7x by x5
Solution:
Here, dividend = 5x10 – 9x8 – 9x5 + 7x
Divisor = x5

[∵ 5x10 ÷ x5 = 5x5,
– 9x8 ÷ x5 = – 9x3
– 9x5 ÷ x5 = – 9]

So, quotient = 5x5 – 9x3 – 9
Remainder = 7x

(iii) 5z3 – 6z2 + 7z by 2z
Solution:
Here, dividend = 5z3 – 6z2 + 7z
Divisor = 2z

[∵ 5z3 ÷ 2z = $$\frac{5}{2}$$ z2,
– 6z2 ÷ 2z = – 3z,
7z ÷ 2z = $$\frac{7}{2}$$]

So, quotient = $$\frac{5}{2}$$ z2 – 3z + $$\frac{7}{2}$$
Remainder = 0