Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial p(x) = x − 3.
(iii) $- \frac{1}{2}$ is a zero of the polynomial p(y) = 2y + 1.
(iv) $\frac{2}{5}$ is a zero of the polynomial p(x) = 2 − 5x.
(v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2).
(vi) 0 and 3 are the zeros of the polynomial p(x) = x² − 3x.
(vii) 2 and −3 are the zeros of the polynomial p(x) = x² + x − 6.

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Solution

$(i) p (x) = x - 4 \Rightarrow p (4) = 4 - 4$
= 0
Hence, 4 is the zero of the given polynomial.

$(ii) Disclaimer : In the question, instead of - 3, 3 should be the zero of the given polynomial . p (x) = x - 3 \Rightarrow p (3) = 3 - 3$
= 0
Hence, 3 is the zero of the given polynomial.

$(iii) p (y) = 2 y + 1 \Rightarrow p (- \frac{1}{2}) = 2 \times (- \frac{1}{2}) + 1 = - 1 + 1 = 0$

Hence, $- \frac{1}{2}$ is the zero of the given polynomial.

$(iv) p (x) = 2 - 5 x \Rightarrow p (\frac{2}{5}) = 2 - 5 \times (\frac{2}{5})$
$= 2 - 2 = 0$

Hence, $\frac{2}{5}$ is the zero of the given polynomial.

$(v) p (x) = (x - 1) (x - 2) \Rightarrow p (1) = (1 - 1) \times (1 - 2)$
$= 0 \times (- 1) = 0$
Also,
$p (2) = (2 - 1) (2 - 2)$
$= (- 1) \times 0 = 0$

Hence, 1 and 2 are the zeroes of the given polynomial.

$(vi) p (x) = x^{2} - 3 x \Rightarrow p (0) = 0^{2} - 3 \times 0$

Also,
$p (3) = 3^{2} - 3 \times 3 = 9 - 9 = 0$

Hence, 0 and 3 are the zeroes of the given polynomial.

$(vii) p (x) = x^{2} + x - 6 \Rightarrow p (2) = 2^{2} + 2 - 6$
$= 4 - 4 = 0$
Also,
$p (- 3) = {(- 3)}^{2} + (- 3) - 6 = 9 - 9 = 0$

Hence, 2 and $- 3$ are the zeroes of the given polynomial.

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Similar questions

Verify that
(i) 1 and 2 are the zeros of the polynomial, $p (x) = x^{2} - 3 x + 2$ .
(ii) 2 and -3 are the zeros of the polynomial, $q (x) = x^{2} + x - 6.$
(iii) 0 and 3 are the zeros of the polynomial, $r (x) = x^{2} - 3 x$ .

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

Q. If

x - 2

and

x - 1

are the only zeroes of the polynomial

p (x)

, which of the following is a zero of the polynomial

p (x) - 6

Q. Question 3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =

- \frac{1}{3}

(ii)

p (x) = 5 x - 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

(vi)

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

(vii)

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

(viii)

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

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