find zero of the polynomial :(a)(x)=x-5,(b)p(t)=2t-3,(c)p(x)=3x+1
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Answer:
p(x) = x + 5 Let p(x) =0, then, p(x) = x + 5 = 0 x = 0 – 5 ∴ x = -5 -5 is zero of p(x). (ii) p(x) = x – 5 If p(x) = 0, then p(x) = x – 5 = 0 x = 0 + 5 ∴ x = 5 5 is the zero of p(x). (iii) p(x) = 2x + 5 If p(x)= 0, then p(x) = 2x + 5 = 0 2x = – 5 ∴ x = −52−52 −52−52 is the zero of p(x). (iv) p(x) = 3x – 2 If p(x)= 0, then p(x) = 3x – 2 = 0 3x = 2 ∴ x = 2323 2323 is the zero of p(x). (v) p(x) = 3x If p(x) = 0, then p(x) = 3x = 0 ∴ x = 0303 0303 is the zero of p(x) (vi) p(x) = ax, a ≠ 0 If p(x)= 0, then p(x) = ax = 0 ∴ x = 0a0a ∴ x = ∞(infinity) ∞ is the zero of p(x). (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers If p(x)= 0, then p(x) = cx + d = 0 cx = 0 – d cx = -d ∴ x = - dcdc is the zero of p(x).Read more on Sarthaks.com - https://www.sarthaks.com/670167/find-the-zero-of-the-polynomial-in-each-of-the-following-cases-i-p-x-x-5