Verify whether the indicated numbers are zero of the polynomial given
$f (x) = 3 x + 1, x = - \frac{1}{3}$

Open in App

Solution

In the given polynomial $f (x) = 3 x + 1$ , substitute $x = - \frac{1}{3}$ as shown below:

$f (- \frac{1}{3}) = (3 \times (- \frac{1}{3})) + 1 = - 1 + 1 = 0$

Hence the indicated numbers are the zeroes of the polynomial $f (x) = 3 x + 1$ .

Suggest Corrections

Similar questions

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

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3, x = 3, x = - 1

x = - \frac{1}{3}

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

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

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases :

(i) f(x) = 3x + 1, x = - $\frac{1}{3}$

(ii) f(x) = $x^{2}$ - 1, x = 1, -1

(iii) g(x) = $3 x^{2}$ - 2, x = $\frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}$

(vi) p(x) = $x^{3} - 6 x^{2}$ + 11x - 6, x = 1,2,3

(v) f(x) = 5x - $π, x = \frac{4}{5}$

(vi) f(x) = $x^{2}$ ,x = 0

(vii) f(x) = lx + m, = - $\frac{m}{l}$

(viii) f(x) = 2x + 1, x = $\frac{1}{2}$

Join BYJU'S Learning Program