Polynomials — Question 4

(i) p(x) = x + 5

⇒ p(x) = 0

⇒ x + 5 = 0

⇒ x = -5

Hence, -5 is a zero of polynomial x + 5

(ii) p(x) = x - 5

⇒ p(x) = 0

⇒ x - 5 = 0

⇒ x = 5

Hence, 5 is a zero of polynomial x - 5

(iii) p(x) = 2x + 5

⇒ p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = $-\dfrac{5}{2}$

Hence, $-\dfrac{5}{2}$ is a zero of polynomial 2x + 5

(iv) p(x) = 3x - 2

⇒ p(x) = 0

⇒ 3x - 2 = 0

⇒ 3x = 2

⇒ x = $\dfrac{2}{3}$

Hence, $\dfrac{2}{3}$ is a zero of polynomial 3x - 2

(v) p(x) = 3x

⇒ p(x) = 0

⇒ 3x = 0

⇒ x = $\dfrac{0}{3}$

⇒ x = 0

Hence, 0 is a zero of polynomial 3x

(vi) p(x) = ax

⇒ p(x) = 0

⇒ ax = 0

⇒ x = $\dfrac{0}{a}$

⇒ x = 0

Hence, 0 is a zero of polynomial ax

(vii) p(x) = cx + d

⇒ p(x) = 0

⇒ cx + d = 0

⇒ cx = -d

x = $\dfrac{-d}{c}$

Hence, $-\dfrac{d}{c}$ is a zero of polynomial cx + d