# 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`.