Solve this

1. f(x) = 3x + 1, x = −1/3

we know that ,

f(x) = 3x + 1

substitute x = −1/3 in f(x)

f( −1/3) = 3(−1/3) + 1

= -1 + 1

= 0

Since, the result is $0 x=-1 / 3$ is the root of $3 x+1$

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

we know that,

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

Given that x = (1, -1)

substitute x = 1 in f(x)

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

= 1 - 1

= 0

Now, substitute x = (-1) in f(x)

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

= 1 - 1

= 0

Since, the results when $x=(1,-1)$ are 0 they are the roots of the polynomial $f(x)=x^{2}-1$

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

We know that

$g(x)=3 x^{2}-2$

Given that,

$x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$

Substitute x = 2/√3 in g(x)

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

= 3(4/3) - 2

= 4 - 2

= 2 ≠ 0

Now, Substitute x = - 2/√3 in g(x)

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

= 3(4/3) - 2

= 4 - 2

= 2 ≠ 0

Since, the results when

$x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$ are not 0, they are roots of $3 x^{2}-2$

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

We know that,

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

given that the values of x are 1, 2 , 3

substitute x = 1 in p(x)

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

= 1 - (6 * 1) + 11 - 6

= 1 - 6 + 11 - 6

= 0

Now, substitute x = 2 in p(x)

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

= (2 * 3) - (6 * 4) + (11 * 2) - 6

= 8 - 24 - 22 - 6

= 0

Now, substitute x = 3 in p(x)

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

= (3 * 3) - (6 * 9) + (11 * 3) - 6

= 27 - 54 + 33 - 6

= 0

Since, the result is 0 for $x=1,2,3$ these are the roots of $x^{3}-6 x^{2}+11 x-6$

(5) f(x) = 5x − π, x = 4/5

we know that,

f(x) = 5x − π

Given that, x = 4/5

Substitute the value of x in f(x)

f(4/5) = 5(4/5) - π

= 4 - π

≠ 0

Since, the result is not equal to zero, x = 4/5 is not the root of the polynomial 5x - π

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

we know that, $f(x)=x^{2}$

Given that value of x is '0'

Substitute the value of x in f(x)

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

= 0

Since, the result is zero, $x=0$ is the root of $x^{2}$

7. f(x) = lx + m, x = - m/l

We know that,

f(x) = lx + m

Given, that x = - m/l

Substitute the value of x in f(x)

$\mathrm{f}\left(-\frac{\mathrm{m}}{\mathrm{l}}\right)=\mathrm{I}\left(-\frac{\mathrm{m}}{\mathrm{l}}\right)+\mathrm{m}$

= - m + m

= 0

Since, the result is 0, x = - m/l is the root of lx + m

(8) $f(x)=2 x+1, x=1 / 2$

We know that,

f(x) = 2x + 1

Given that x = 1/2

Substitute the value of x and f(x)

f(1/2) = 2(1/2) + 1

= 1 + 1

= 2 ≠ 0

Since, the result is not equal to zero

x = 1/2 is the root of 2x + 1