DAV Class 8 Maths Chapter 8 Worksheet 1 Solutions

The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 8 Worksheet 1 Solutions of Polynomials offer comprehensive answers to textbook questions.

DAV Class 8 Maths Ch 8 WS 1 Solutions

Question 1.
Find out whether the given expression is a polynomial or not. If not, give reasons.

(i) 5x³ – 4x² + \(\frac{1}{2}\)
Solution:
Yes, it is polynomial of degree 3.

(ii) √3z² – 5√z + 6
Solution:
Not, it is not a polynomial as the power of z is \(\frac{1}{2}\).

(iii) 6x⁴ + \(\frac{2}{3}\) x³ – \(\frac{3}{4}\) x² – 1
Solution:
Yes, it is a polynomial of degree 4.

(iv) 7x^{\(\frac{2}{3}\)} – 8x^{\(\frac{3}{2}\)} + x²
Solution:
Not, it is not a polynomial as it contains variables with rational powers.

(v) 5x – \(\frac{1}{x}+\frac{1}{x^2}\) – 2
Solution:
Not, it is not a polynomial as power of x is an negative integer.

(vi) p⁴ – 3p³ – p + 1
Solution:
Yes, it is a polynomial of degree 4.

Question 2.
Write each of the following polynomials in standard form and also write down their degree:

(i) p⁶ – 8p⁹ + p⁷ + 5
Solution:
Standard form of the given polynomial is – 8p⁹ + p⁷ + p⁶ + 5
degree = 9

(ii) 4z³ – 3z⁵ + 2z⁴ + z + 1
Solution:
The standard form of the given polynomial is – 3z⁵ + 2z⁴ + 4z³ + z + 1
degree = 5

(iii) (x + \(\frac{2}{3}\)) (x + \(\frac{3}{4}\))
Solution:
(x + \(\frac{2}{3}\)) (x + \(\frac{3}{4}\)) = (x²)² + \(\left(\frac{2}{3}+\frac{3}{4}\right)\) x + \(\frac{2}{3} \times \frac{3}{4}\)
= x² + \(\frac{17}{12}\) x + \(\frac{1}{2}\), which is in standard form
degree = 2

(iv) (x² – \(\frac{2}{3}\)) (x² + \(\frac{4}{3}\))
Solution:
(x² – \(\frac{2}{3}\)) (x² + \(\frac{4}{3}\)) = (x²)² + \(\left(-\frac{2}{3}+\frac{4}{3}\right)\) x² + \(\left(-\frac{2}{3}\right)\left(\frac{4}{3}\right)\)
= x⁴ + \(\frac{2}{3}\) x² – \(\frac{8}{9}\) which is in standard form.
degree = 4

(v) (z² + 5) (z² – 6)
Solution:
(z² + 5) (z² – 6) = (z²)² + (5 – 6) z² + 5 (- 6)
= z⁴ – z² – 30 which is in standard form.
degree = 4

(vi) (y³ – 4) (y³ – 5)
Solution:
(y³ – 4) (y³ – 5) = (y³)² + (- 4 – 5) y³ + (- 4) (- 5)
= y⁶ – 9y³ + 20, which is in standard form
degree = 6

(vii) (p² + 2) (p² + 7)
Solution:
(p² + 2) (p² + 7) = (p²)² + (2 + 7) p² + 2 × 7
= p⁴ + 9p² + 14, which is in standard form
degree = 4

(viii) (\(\frac{5}{6}\)z – \(\frac{3}{4}\)z² – \(\frac{2}{3}\)z³ + 1)
Solution:
Standard form of (\(\frac{5}{6}\)z – \(\frac{3}{4}\)z² – \(\frac{2}{3}\) z³ + 1) is – \(\frac{2}{3}\) z³ – \(\frac{3}{4}\) z² + \(\frac{5}{6}\) z + 1
degree = 3

(ix) 4p + 15p⁶ – p⁵ + 4p² + 3
Solution:
Standard form of 4p + 15p⁶ – p⁵ + 4p² + 3 is 15p⁶ – p⁵ + 4p² + 4p + 3
degree = 6.

(x) q¹⁰ + q⁶ – q⁴ + q⁸
Solution:
Standard form of q¹⁰ + q⁶ – q⁴ + q⁸ is q¹⁰ + q⁸ + q⁶ – q⁴
degree = 10.

DAV Class 8 Maths Chapter 8 Worksheet 1 Notes

1. Monomials: The expression that contains only one term is called a monomial.
e.g. 4x², – 9, 32mp, etc.

2. Binomials: The expression that contains two terms is called a binomial.
e.g. a + b, a + 4, lm + ms, 5 – 2xy, z² – x², etc.

3. Trinomial: The expression that contains three terms is called a trinomial.
e.g. a + b + c, 2mn + 5m + n, 2a + 3x + 5z, etc.

4. Polynomial: The expression that contains many number of terms is called a polynomial.
e.g. 3x + y + 4z + 3t – 5w, 2n + 5y – 3, 4a – 5b – c – d, etc.

Example of polynomial in one variable:
P(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + ……………….. + a_nxⁿ
where x is a variable and a₀, a₁, a₂, a₃, a₄, …………., a_n are the coefficients.

Degree of a polynomial:

The highest power of the variable in a polynomial of one variable is called its degree.

• If degree is 1, then it is called as linear polynomial.
  e.g. P(x) = ax + b
• If degree is 2, then it is called as quadratic polynomial.
  e.g. P(x) = ax² + bx + e
• If degree is 3, then it is called as cubic polynomial.
  e.g. P(x) = ax³ + bx² + cx + d.
• If degree is 4, then it is called as biquadratic polynomial.
  e.g. P(x) = ax⁴ + bx³ + cx² + dx + e

Example 1.
Classify the following polynomials as monomials, binomials and trinomials:
(i) x + y
(ii) 100
(iii) x² + x³ + 1
(iv) 2y – 3y² + 4y³
(v) ab + bc + cd
(vi) 3x
Solution:
(i) binomial
(ii) monomial
(iii) trinomial
(iv) trinomial
(v) trinomial
(vi) monomial

Example 2.
Write the following polynomials in standard form:
(i) – √3x + 5 – 2√2x³ + 5x² – x⁴
(ii) 5 + 3x – 4x² + 7x³ + x⁶
(iii) – 6 + x⁹ – 3x⁸ – 2x³
(iv) 3x – 5x² + 7x³ + 8x⁵.
(v) x⁵ – x⁴ + 3x³ + x + 2.
Solution:
(i) – √3x + 5 – 2√2x³ + 5x² – x⁴
Standard form is: – x⁴ – 2√2x³ + 5x² – √3x + 5

(ii) 5 + 3x – 4x² + 7x³ + x⁶
Standard form is: x⁶ + 7x³ – 4x² + 3x + 5

(iii) – 6 + x⁹ – 3x⁸ – 2x³
Standard form is: x⁹ – 3x⁸ – 2x³ – 6

(iv) 3x – 5x² + 7x³ + 8x⁵
Standard form is: 8x⁵ + 7x³ – 5x² + 3x

(v) x⁵ – x⁴ + 3x³ + x + 2
Standard form is: x⁵ – x⁴ + 3x³ + x + 2

Example 3:
Divide 8x³ – 4x² – 2x by 4x.
Solution:
8x³ – 4x² – 2x ÷ 4x
= \(\frac{8 x^3-4 x^2-2 x}{4 x}\)
= \(\frac{8 x^3}{4 x}-\frac{4 x^2}{4 x}-\frac{2 x}{4 x}\)
= 2x² – x – \(\frac{1}{2}\)

Example 4.
Divide 8y³ + 10y – 6y² + 3 by 1 + 4y by long division method and verify the
answer.
Solution:
Let us write the dividend and the divisor in standard form:
8y³ – 6y² + 10y + 3 and 4y + 1

Verification:
Remainder = 0
∴ Dividend = Divisor × Quotient
8y³ – 6y² + 10y + 3 = (4y + 1) × (2y² – 2y + 3)
= 4y (2y² – 2y + 3) + 1 (2y² – 2y + 3)
= 8y³ – 8y² + 12y + 2y² – 2y + 3
= 8y³ – 6y² + 10y + 3