Question

In the following, circle the polynomial: (i) √?+ 2 (ii) t2 + 5t – 1 (iii) √3x2 – 2x (iv) 8 (v) 1 – √2x (vi) (vii) 0 (viii) y½ + 3y + 2​

Answer 1

Step-by-step explanation:

Given that:

(i) √x+ 2 (ii) t2 + 5t – 1 (iii) √3x2 – 2x (iv) 8 (v) 1 – √2x (vi) 0 (vii) y½ + 3y + 2

To prove:

Identify Polynomial

Explanation: The expression is said to be polynomial if degree /power of variable is +ve integer.

$(i) \sqrt{x} + 2 \\ \\ Ans:No,\:it\:is \: not \: a \: polynomial \: because \\ \: power \: of \: x \: is \: rational \: ( \frac{1}{2} )$

$(ii) {t}^{2} + 5t - 1 \\ \\ Ans:Yes, \: it \: is \: a \: polynomial \:$

$(iii)\sqrt{3} {x}^{2} - 3x \\ Ans: \: Yes ,\: it \: is \: a \: polynomial$

$(iv)8 \\ Ans: \: Yes, \: it \: is \: a \: polynomial \: of \: degree \: 0$

$(v)1 - \sqrt{2} x \\ Ans: \: Yes, \: it \: is \: a \: polynomial$

$(vi)0 \\ Ans: \: Yes, \: it \: is \: a \: polynomial \: of \: zero \: degree$

$(vii) {y}^{ \frac{1}{2} } + 3y + 2 \\ Ans: \: No, \: it \: is \: not \: a \: polynomial \\ because \: y \: has \: rational \: power$

Thus,(ii),(iii),(iv),(v) and (vi) are polynomial.

Hope it helps you.