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Types of Polynomials

Which of the ...

Question

Which of the following is NOT a constant polynomial?

p (x) = 15

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p (x) = 1

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p (x) = x

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p (x) = 20

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Solution

The correct option is B $p (x) = x$
Since $p (x) = x$ is a polynomial with variable $x$ and there is no constant term in it.
In the other given options, we can see that the polynomials have constant terms, none of them has any variable term.
Hence, $p (x) = x$ is not a constant polynomial.

Suggest Corrections

Similar questions

- \frac{1}{3}

p (x) = 5 x - π, x = \frac{4}{5}

p (x) = x^{2} - 1, x = 1, - 1

p (x) = x^{2}, x = 0

p (x) = i x + m, x = - \frac{m}{i}

p (x) = 3 x^{2} - 1, x = - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}

p (x) = 2 x + 1, x = \frac{1}{2}

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) (ii)

(iii) p(x) = x² − 1, x = 1, − 1 (iv) p(x) = (x + 1) (x − 2), x = − 1, 2

(v) p(x) = x², x = 0 (vi) p(x) = lx + m

(vii) (viii)

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1,x= $\frac{- 1}{3}$

(ii) p(x)=5x− $π$ ,x= $\frac{4}{5}$

(iii) p(x)= $x^{2}$ -1,x=1,−1

(iv) p(x)=(x+1)(x−2),x=−1,2

(v) p(x)= $x^{2}$ ,x=0

From the polynomial p(x) = x² + x + 1, do we get p(x) = 0 for any x? What about p(x) = 1? And p(x) = −1?

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