Polynomials — Question 1

(i) x³ + x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x³ + x² + x + 1

p(-1) = (-1)³ + (-1)² + (-1) + 1

= -1 + 1 -1 + 1

= 0

Remainder is zero (0), so (x + 1) is factor of this polynomial.

(ii) x⁴ + x³ + x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x⁴ + x³ + x² + x + 1

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

= 1 -1 + 1 -1 + 1

= 1

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

(iii) x⁴ + 3x³ + 3x² + x + 1

⇒ x + 1 = 0

⇒ x = -1

p(x) = x⁴ + 3x³ + 3x² + x + 1

p(-1) = (-1)⁴ + 3 x (-1)³ + 3 x (-1²) + (-1) + 1

= 1 -3 + 3 -1 + 1

= 1

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

(iv) x³ - x² - (2 + $\sqrt{2}$)x + $\sqrt{2}$

⇒ x + 1 = 0

⇒ x = -1

p(x) = x³ - x² - (2 + $\sqrt{2}$)x + $\sqrt{2}$

p(-1) = (-1)³ - (-1)² - (2 + $\sqrt{2}$)(-1) + $\sqrt{2}$

= -1 - 1 + 2 + $\sqrt{2}$ + $\sqrt{2}$

= -2 + 2 + 2 $\sqrt{2}$

= 2 $\sqrt{2}$

Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.