The problem consists of three parts: (i) Rationalize the denominator of the fraction $\frac{3}{\sqrt{2}-1}$. (ii) Express the repeating decimal $2.\overline{143}$ as a fraction $\frac{a}{b}$ in its lowest terms, where $a$ and $b$ are integers. (iii) Show that $x+1$ is a factor of the polynomial $f(x) = x^3 - 5x^2 - 2x + 8$, and then completely factorize $f(x)$.

1. Problem Description

The problem consists of three parts:

(i) Rationalize the denominator of the fraction $\frac{3}{\sqrt{2}-1}$.

(ii) Express the repeating decimal $2.\overline{143}$ as a fraction $\frac{a}{b}$ in its lowest terms, where $a$ and $b$ are integers.

(iii) Show that $x+1$ is a factor of the polynomial $f(x) = x^3 - 5x^2 - 2x + 8$, and then completely factorize $f(x)$.

2. Solution Steps

(i) Rationalizing the denominator:

To rationalize the denominator of $\frac{3}{\sqrt{2}-1}$, we multiply both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}+1$.

$$\frac{3}{\sqrt{2}-1} = \frac{3(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$$

The denominator becomes $(\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$.

$$\frac{3(\sqrt{2}+1)}{1} = 3(\sqrt{2}+1) = 3\sqrt{2} + 3$$

(ii) Expressing the repeating decimal as a fraction:

Let $x = 2.\overline{143} = 2.143143143...$

Then $1000x = 2143.143143143...$

Subtracting $x$ from $1000x$, we get

$1000x - x = 2143.143143... - 2.143143...$

$999x = 2141$

$x = \frac{2141}{999}$

(iii) Showing that $x+1$ is a factor of $f(x)$ and complete the factorization:

$f(x) = x^3 - 5x^2 - 2x + 8$

If $x+1$ is a factor of $f(x)$, then $f(-1) = 0$.

$f(-1) = (-1)^3 - 5(-1)^2 - 2(-1) + 8 = -1 - 5 + 2 + 8 = 4 \neq 0$

There appears to be a typo in the original question. The polynomial should be $f(x) = x^3 - 5x^2 - 2x + 8$.

Let's find the correct constant term to make x+1 a factor.

$f(x) = x^3 - 5x^2 - 2x + c$.

$f(-1) = (-1)^3 - 5(-1)^2 - 2(-1) + c = 0$

$-1 - 5 + 2 + c = 0$

$-4 + c = 0$

$c = 4$

So, the correct polynomial should be $f(x) = x^3 - 5x^2 - 2x + 8$ (as originally stated). However, let's continue the problem assuming $f(x) = x^3 - 5x^2 - 2x + 8$ and demonstrate polynomial long division or synthetic division, given that $x+1$ is NOT a factor. Since the question instructs us to show $x+1$ is a factor and then factorize, and it is NOT a factor, there's an error.

Revised polynomial:

If the problem meant $x^3+x^2 -6x^2 -6x + 4x+4$

Assuming instead the problem meant show that x-1 is a factor of $x^3-5x^2-2x+8$.

$f(1) = 1-5-2+8 = 2$. so (x-1) is not a factor either

$f(x)=x^3-5x^2-2x+8$

Then trying $x-2$ as a possible factor, we have:

$f(2)= 8-20-4+8=-8$.

Let's correct the polynomial instead: $x^3-3x^2-6x+8$.

Try $x-1$. $1-3-6+8 = 0$. So $x-1$ is a factor.

Then divide $x^3 - 5x^2 - 2x + 8$ by $x+1$ using polynomial division or synthetic division:

x^2 - 6x + 4

x+1 | x^3 - 5x^2 - 2x + 8

-(x^3 + x^2)

----------------

-6x^2 - 2x

-(-6x^2 - 6x)

----------------

4x + 8

-(4x + 4)

----------

4

So $x^3 - 5x^2 - 2x + 8 = (x+1)(x^2 - 6x + 4) + 4$

$x^2 - 6x + 4$ can be solved with the quadratic formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$

3. Final Answer

(i) $3\sqrt{2} + 3$

(ii) $\frac{2141}{999}$

(iii) $x+1$ is not a factor of $x^3 - 5x^2 - 2x + 8$. If the polynomial was $x^3 - 5x^2 - 2x + 8$, then $x^3 - 5x^2 - 2x + 8 = (x+1)(x^2-6x+4) + 4$. The quadratic factor has roots $3 \pm \sqrt{5}$