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

i) x^{3} + x^{2} + x + 1 ii) x^{4} + x^{3} + x^{2} + x + 1 iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

iv) x^{3} - x^{2} - (2 + √2)x + √2

**Solution:**

When a polynomial p(x) is divided by (x - a) and if p(a) = 0 then (x - a) is a factor of p(x).

The root of x + 1 = 0 is -1.

**(i)** Let p(x) = x^{3} + x^{2} + x + 1

Therefore, p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

⇒ -1 + 1 - 1 + 1 = 0

Since the remainder of p(-1) is 0, we conclude that x + 1 is a factor of x^{3} + x^{2} + x + 1.

**(ii)** Let p(x) = x^{4} + x^{3} + x^{2} + x + 1

Therefore, p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1

= 1 - 1 + 1 - 1 + 1

= 1 ≠ 0

Since the remainder of p(-1) ≠ 0, we conclude that x + 1 is not a factor of x^{4} + x^{3} + x^{2} + x + 1.

**(iii)** Let p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

Therefore, p(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1

= 1 - 3 + 3 - 1 + 1

= 1 ≠ 0

Since the remainder of p(-1) ≠ 0, x + 1 is not a factor of x^{4} + 3x^{3} + 3x^{2} + x + 1.

**(iv)** Let p(x) = x^{3} - x^{2} - (2 + √2)x + √2

∴ p(-1) = (-1)^{3} - (-1)^{2} - (2 + √2)(-1) + √2

= -1 - 1 + 2 + √2 + √2

= 2√2 ≠ 0

Since the remainder of p(-1) ≠ 0, x + 1 is not a factor of x^{3} - x^{2} - (2 + √2)x + √2.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 2

**Video Solution:**

## Determine which of the following polynomials has (x + 1) a factor: i) x³ + x² + x + 1 ii) x⁴ + x³ + x² + x + 1 iii) x⁴ + 3x³ + 3x² + x +1 iv) x³ - x² - (2 + √2)x + √2

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.4 Question 1

**Summary:**

(x + 1) is a factor of x^{3} + x^{2} + x + 1 whereas, (x + 1) is not a factor of the polynomials x^{4 }+ x^{3} + x^{2 }+ x + 1, x^{4 }+ 3x^{3} + 3x^{2 }+ x + 1, and x^{3} - x^{2} - (2 + √2)x + √2.

**☛ Related Questions:**

- Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:(i) p(x) = 2x3 + x2 - 2x - 1, g(x) = x +1(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2(iii) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3
- Find the value of k, if x - 1 is a factor of p(x) in each of the following cases: (i) p(x) = x2 + x + k (ii) p(x) = 2x2 + kx + √2 (iii) p(x) = kx2 - √2x + 1 (iv) (x) = kx2 - 3x + k
- Factorise:(i) 12x2 - 7x + 1(ii) 2x2 + 7x + 3(iii) 6x2 + 5x - 6(iv) 3x2 - x - 4

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