Question

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

Answer 1

(i) x- 1 is a factor of p(x)

x + 1 = x – a

a = -1

For the value of p(a),

value of r(x) = 0.

∴ p(x) = x3 + x2 + x + 1

p(-1)= (-1)3 + (-1)2 + (-1) + 1

= -1 + 1 -1 + 1

p(-1)= 0

Here, p(a) = r(x) = 0

∴ x + 1 is a factor.

(ii) If x + 1 = x – a, then a = -1

p(x) = x4 + x3 + x2 + x + 1

p(-1)= (-1)4 + (-1)3 +(-1)2 + (-1)+ 1

= 1 – 1 + 1 – 1 + 1

= 3 – 2 p(-1)= 1

Here, r(x) = p(a)= 1

Reminder is not zero.

∴ x+1 is not a factor.

(iii) If x + 1 = x – a then a = -1

p(x) = x4 + 3x3 + 3x2 + x + 1

p(-1) = (-1)4 + 3(-1 )3 +3(-1 )2 + (-1) + 1

= 1 + 3(-1) + 3(1) + 1(-1) + 1

= 1- 3 + 3 – 1 + 1

= 5 – 4 P(-1)= 1

Here, r(x) = p(a) = 1

Remainder is not zero

∴ x+1 is not a factor.

(iv) If x + 1 = x – a then,

a = -1

p(x) = x3 – x2 – (2 + √2)x+ p(-1)

= (-1)3 – (-1)2 -(2 + √2)(-1) + √2

= -1 -(+1) – (2 – √2) + √2

= -1 – 1 + 2 + √2 +√2

= -2 + 2 + √2

= + 2√2

p(-1) = 2√2

Here, r(x) = p(a) = 2√2 Value of remainder r(x) is not zero.

∴ x + 1 is not a factor.