Find the zeroes of the polynomial in the following :
$p (x) = x - 4$ .

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $4$
Zero of $p (x) = x - 4$ will be when, $p (x) = 0$ .
$=> x - 4 = 0$
$=> x = 4$ .
Thus, $4$ is the zero of the polynomial.

Hence, option $A$ is correct.

Suggest Corrections

Similar questions

Q. 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.

Q. Find the zeros of the following polynomial

p (x) = x^{2} + 3 x - 4

Q. 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}

Find the zero of the polynomial in each of the following cases:
(i) $p (x) = x + 5$
(ii) $p (x) = x - 5$
(iii) $p (x) = 2 x + 5$
(iv) $p (x) = 3 x - 2$
(v) $p (x) = 3 x$
(vi) $p (x) = a x, a \neq 0$
(vii) $p (x) = c x + d, c \neq 0$ and $c, d$ are real numbers.

Give zeroes of the polynomial p(x)=(x-1)(x-2)(x-3)(x-4)

Join BYJU'S Learning Program

Find the zeroes of the polynomial in the following :p(x)=x−4.

The correct option is D 4Zero of p(x)=x−4 will be when, p(x)=0.=>x−4=0=>x=4.Thus, 4 is the zero of the polynomial.Hence, option A is correct.

Find the zeroes of the polynomial in the following :
$p (x) = x - 4$ .

The correct option is D $4$
Zero of $p (x) = x - 4$ will be when, $p (x) = 0$ .
$=> x - 4 = 0$
$=> x = 4$ .
Thus, $4$ is the zero of the polynomial.

Hence, option $A$ is correct.