Question

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

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

(v) f(x) = 5x -$\pi ,x=\frac{4}{5}$

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

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

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

Answer 1

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

$f(-\frac{1}{3})=3(-\frac{1}{3})+4$

= - 1 + 1 = 0

$\therefore x=-\frac{1}{3}$ is the zero of f(x)

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

f(1) =$(1{)}^2$ - 1 = 1 -1 =0

x=1 is zero of f(x)

f(-1)=$(-1{)}^2$ - 1 = 1 = 0

x = - 1 is zero of f(x)

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

$g\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-2=3\times \frac{4}{3}-2$

=4-2=2

$\therefore x=\frac{2}{\sqrt{3}}$ is not its zero

$g\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-2$

=$3\times \frac{4}{3}-2=4-2=2$

$\therefore x=\frac{-2}{3}$ is not its zero

(iv) p(x) = $x^3-6x^3$ + 11x - 6, x = 1, 2,3

P(1) $=(1{)}^3-6(1{)}^2$ + 11(1) - 6

= 1 - 6 $\times 1+11\times$ 1 - 6 = 1 - 6 + 11 - 6

=12 - 12 = 0

$\therefore$ x = 1 is its zero

p(2) $=(2{)}^2-6(2{)}^2+11\times$ 2 - 6

= 8 - 6$\times$ 4 + 22 - 6 = 8 - 24 + 22 - 6

=30 - 30 = 0

$\therefore$ x = 2 is its zero

P(3) $=(3{)}^3-6(3{)}^2+11\times$ 3 - 6

= 27 - 6$\times$ 9 + 33 - 6

= 27 - 54 + 33 - 6 = 60 - 60 = 0

$\therefore$ x = 3 is its zero

Hence x = 1,2,3 are its zeros

(v) f(x) = 5x - $\pi ,x=\frac{4}{5}$

=$f\left(\frac{4}{5}\right)5\times \frac{4}{5}-\pi =4-\pi$

$\therefore x=\frac{4}{5}$ is not its zero

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

$\therefore$ f(0)$=(0{)}^2$ = 0

$\therefore$ x = 0 is its zero

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

$\therefore f(-\frac{m}{l})=1\times (-\frac{m}{l})+m$

= - m + m = 0

$\therefore =-\frac{m}{l}$ is its zero

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

$\therefore f\left(\frac{1}{2}\right)=2\times {\left(\frac{1}{2}\right)}^2+1=2\times \frac{1}{4}+1$

=$\frac{1}{2}+1=\frac{3}{2}$

$\therefore x=\frac{1}{2}$ is not its zero