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Polynomial and Its General Form

Verify whethe...

Question

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

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

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Solution

Given : $f (x) = 3 x + 1 a n d x = - \frac{1}{3}$

We know that if $f (a) = 0;$ then $a$ is the zero of the polynomial $f (x) .$

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

$= - 1 + 1$
$= 0$

$∵ f (- \frac{1}{3}) = 0$

Hence, $x = - \frac{1}{3}$ is the zero of $f (x) = 3 x + 1,$ verified.

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Similar questions

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 = $\frac{1}{2}$

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

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

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

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

f (x) = 5 x - π, x = \frac{4}{5}

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

f (x) = l x + m, x = - \frac{m}{1}

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

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

f (x) = 3 x^{2} - 2,

x = \frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}

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

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