Class 9 Maths

# Polynomials

## Exercise 2.4 Part 1

Question 1: Determine which of the following polynomials has (x + 1) a factor:

(i) x^3 + x^2 + x + 1

Answer: Let, x + 1 = 0
Or, x = - 1
i.e. zero of x + 1 = - 1
Let p(x) = x^3 + x^2 + x + 1
Hence, p( - 1) = ( - 1)^3 + (- 1)^2 + ( - 1) + 1
= - 1 + 1 – 1 + 1
Or, p( - 1) = 0
Therefore, according to factor theorem (x + 1) is the factor of x^3 + x^2 + x + 1.

(ii) x^4 + x^3 + x^2 + x + 1

Answer: Let, x + 1 = 0
Or, x = - 1
i.e. zero of x + 1 = - 1
Let p(x) = x^4 + x^3 + x^2 + x + 1
Hence, p( - 1) = ( -1)^4 + ( - 1)^3 + (- 1)^2 + ( - 1) + 1
= 1 -1 + 1 – 1 + 1
Or, p( - 1) = 1
Here, p( - 1) ≠ 0
Therefore, according to factor theorem (x + 1) is not the factor of x^4 + x^3 + x^2 + x + 1.

(iii) x^4 + 3x^3 + 3x^2 + x + 1

Answer: Let, x + 1 = 0
Or, x = - 1
i.e. zero of x + 1 = - 1
Let p(x) = x^4 + 3x^3 + 3x^2 + x + 1
Hence, p( - 1) = ( -1)^4 + 3( - 1)^3 + 3(- 1)^2 + ( - 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
Or, p( - 1) = 1
Here, p( - 1) ≠ 0
Therefore, according to factor theorem (x + 1) is not the factor of x^4 + 3x^3 + 3x^2 + x + 1.

(iv) x^3 - x^2 - (2 + sqrt2)x + sqrt2

Answer: Let, x + 1 = 0
Hence, x = - 1, i.e. zero of x + 1 = - 1
Let p(x) = x^3 - x^2 - (2 + sqrt2)x + sqrt2
Or, p( - 1) = ( - 1)^3 - ( - 1)^2 - (2 + sqrt2) xx 1 + sqrt2
= - 1 – 1 - (2 - sqrt2) + sqrt2
= - 1 – 1 + 2 + sqrt2 + sqrt2
= 2 + 2sqrt2
Here, p( - 1) ≠ 0
Therefore, (x + 1) is not the factor of x^3 - x^2 - (2 + sqrt2)x + sqrt2.