For the polynomial x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 , find the coefficient of x 3 .

Let p x = x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = 1 5 x 3 + 2 5 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 As we know, a coefficient is a number or an alphabet that is multiplied by a variable of a single term or the terms of a polynomial. Therefore, the coefficient of x 3 in p ( x ) = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 is 1 5 .

For the polynomial x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 , find the coefficient of x 6 .

Let p x = x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = 1 5 x 3 + 2 5 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 As we know, a coefficient is a number or an alphabet that is multiplied by a variable of a single term or the terms of a polynomial. Therefore, the coefficient of x 6 in p ( x ) = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 is - 1 .

For the polynomial x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 , find the constant term.

Let p x = x 3 + 2 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = 1 5 x 3 + 2 5 x + 1 5 - 7 2 x 2 - x 6 ⇒ p x = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 As we know, a constant term is a term in an algebraic expression that does not contain any variables Therefore, the constant in p ( x ) = - x 6 + 1 5 x 3 - 7 2 x 2 + 2 5 x + 1 5 is 1 5 .

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All in One Mathematics>Chapter 2 - Polynomials>Exercise 2.1

Brijesh Dwevedi and Jitendra Gupta Solutions for Chapter: Polynomials, Exercise 1: Exercise 2.1

Author:Brijesh Dwevedi & Jitendra Gupta

Brijesh Dwevedi Mathematics Solutions for Exercise - Brijesh Dwevedi and Jitendra Gupta Solutions for Chapter: Polynomials, Exercise 1: Exercise 2.1

Attempt the practice questions on Chapter 2: Polynomials, Exercise 1: Exercise 2.1 with hints and solutions to strengthen your understanding. All in One Mathematics solutions are prepared by Experienced Embibe Experts.