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 . x^{3 – 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 . x^{4 – 3}
= 5x

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

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

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

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