Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p (x) = 3 x + 1, x = - \frac{1}{3}$
(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) $p (x) = l x + m, x = - \frac{m}{l}$
(vii) $p (x) = 3 x^{2} - 1, x = - \frac{1}{3}, \frac{2}{3}$
(viii) $p (x) = 2 x + 1, x = \frac{1}{2}$

Question

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

p (x) = 3 x + 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) = l x + m, x = - \frac{m}{l}

(vii)

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

(viii)

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

Filo · Accepted Answer

In order to verify the values are zeros of polynomial

p (x)

, we must replace the variable

x

with the given values.

If

p (x) = 0

, then that given value is zero of polynomial

p (x)

.

(i)

p (x) = 3 x + 1

:

Put

x = - \frac{1}{3}

, we get,

p (x) = p (\frac{- 1}{3}) = 3 (- \frac{1}{3}) + 1 = - 1 + 1 = 0

.

So,

x = - \frac{1}{3}

is the zero of the given polynomial

p (x)

.

(ii)

p (x) = 5 x - π

:

Put

x = \frac{4}{5}

, we get,

p (x) = p (\frac{4}{5}) = 5 (\frac{4}{5}) - π = 4 - π \neq = 0

.

So,

x = \frac{4}{5}

is not the zero of the given polynomial

p (x)

.

(iii)

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

:

Put

x = 1

, we get,

p (x) = p (1) = (1)^{2} - 1 = 1 - 1 = 0

.

So,

x = 1

is the zero of the given polynomial

p (x)

.

Now put

x = - 1

, we get,

p (x) p (- 1) = (- 1)^{2} - 1 = 1 - 1 = 0

.

So,

x = - 1

is the zero of the given polynomial

p (x)

.

(iv)

p (x) = (x + 1) (x - 2)

:

Put

x = - 1

, we get,

p (x) = (- 1 + 1) (- 1 - 2) = 0 (- 3) = 0

.

So,

x = - 1

is the zero of the given polynomial

p (x)

.

Now put

x = 2

, we get,

p (x) = (2 + 1) (2 - 2) = 3 (0) = 0

.

So,

x = 2

is the zero of the given polynomial

p (x)

.

(v)

p (x) = x^{2}

:

Put

x = 0

, we get,

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

.

So,

x = 0

is the zero of the given polynomial

p (x)

.

(vi)

p (x) = l x + m

:

Put

x = - \frac{m}{l}

, we get,

p (x) = p (- \frac{m}{l}) = l (- \frac{m}{l}) + m = - m + m = 0

.

So,

x = - \frac{m}{l}

is the zero of the given polynomial

p (x)

.

(vii)

p (x) = 3 x^{2} - 1

:

Put

x = - \frac{1}{3}

, we get,

p (x) = p (- \frac{1}{3}) = 3 (- \frac{1}{3})^{2} - 1 = (3 \times \frac{1}{3}) - 1 = 1 - 1 = 0

.

So,

x = - \frac{1}{3}

is the zero of the given polynomial

p (x)

.

Now put

x = \frac{2}{3}

, we get,

p (x) = p (\frac{2}{3}) = 3 (\frac{2}{3})^{2} - 1 = (3 \times \frac{4}{3}) - 1 = 4 - 1 = 3 \neq = 0

.

So,

x = \frac{2}{3}

is not the zero of the given polynomial

p (x)

.

(viii)

p (x) = 2 x + 1

:

Put

x = \frac{1}{2}

, we get,

p (x) = p (\frac{1}{2}) = 2 (\frac{1}{2}) + 1 = 1 + 1 = 2 \neq = 0

.

So,

x = \frac{1}{2}

is not the zero of the given polynomial

p (x)

.

Question Text	Verify whether the following are zeroes of the polynomial, indicated against them. (i) $p (x) = 3 x + 1, x = - \frac{1}{3}$ (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) $p (x) = l x + m, x = - \frac{m}{l}$ (vii) $p (x) = 3 x^{2} - 1, x = - \frac{1}{3}, \frac{2}{3}$ (viii) $p (x) = 2 x + 1, x = \frac{1}{2}$
Answer Type	Text solution:1
Upvotes	150

Text solutionVerified

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Text solutionVerified

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