11. Given that $$ \mathrm{f}(x)=3 x^{3}-4 x^{2}-5 x+2 $$ (i) Show that $$ (3 x-1) $$ is a factor of $$ \mathrm{f}(x) $$. (ii) Factorise $$ \mathrm{f}(x) $$ completely. (iii) Hence, state the number of real solutions for $$ 3 y^{6}-4 y^{4}-5 y^{2}+2=0 $$.

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**(i)**  
To show that $$3x-1$$ is a factor of $$\mathrm{f}(x)=3x^3-4x^2-5x+2$$, we use the Factor Theorem. First, set  
$$
3x-1=0 \quad \Longrightarrow \quad x=\frac{1}{3}.
$$
Now compute $$\mathrm{f}\left(\frac{1}{3}\right)$$:
$$
\mathrm{f}\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^3-4\left(\frac{1}{3}\right)^2-5\left(\frac{1}{3}\right)+2.
$$
Calculate each term:
$$
3\left(\frac{1}{27}\right)=\frac{1}{9}, \quad 4\left(\frac{1}{9}\right)=\frac{4}{9}, \quad 5\left(\frac{1}{3}\right)=\frac{5}{3}.
$$
Thus,
$$
\mathrm{f}\left(\frac{1}{3}\right)=\frac{1}{9}-\frac{4}{9}-\frac{5}{3}+2.
$$
Combine the first two terms:
$$
\frac{1}{9}-\frac{4}{9}=-\frac{3}{9}=-\frac{1}{3},
$$
so we have
$$
\mathrm{f}\left(\frac{1}{3}\right)=-\frac{1}{3}-\frac{5}{3}+2=-\frac{6}{3}+2=-2+2=0.
$$
Since $$\mathrm{f}\left(\frac{1}{3}\right)=0$$, the Factor Theorem confirms that $$3x-1$$ is a factor of $$\mathrm{f}(x)$$.

---

**(ii)**  
We know that $$\mathrm{f}(x)=3x^3-4x^2-5x+2$$ and that $$3x-1$$ is a factor. Thus, we can write:
$$
\mathrm{f}(x)=(3x-1)(Ax^2+Bx+C).
$$
Expanding the right-hand side:
$$
(3x-1)(Ax^2+Bx+C)=3Ax^3+3Bx^2+3Cx-Ax^2-Bx-C.
$$
Combine like terms:
$$
=3Ax^3+(3B-A)x^2+(3C-B)x-C.
$$
Now, equate the coefficients with $$\mathrm{f}(x)=3x^3-4x^2-5x+2$$:

1. Coefficient of $$x^3$$:  
   $$
   3A=3 \quad \Longrightarrow \quad A=1.
   $$

2. Coefficient of $$x^2$$:  
   $$
   3B-A=-4 \quad \Longrightarrow \quad 3B-1=-4 \quad \Longrightarrow \quad 3B=-3 \quad \Longrightarrow \quad B=-1.
   $$

3. Coefficient of $$x$$:  
   $$
   3C-B=-5 \quad \Longrightarrow \quad 3C-(-1)=-5 \quad \Longrightarrow \quad 3C+1=-5 \quad \Longrightarrow \quad 3C=-6 \quad \Longrightarrow \quad C=-2.
   $$

4. Constant term:  
   $$
   -C=2 \quad \Longrightarrow \quad C=-2 \quad (\text{which confirms our result}).
   $$

Thus, the factorisation is:
$$
\mathrm{f}(x)=(3x-1)(x^2-x-2).
$$
Now factorise the quadratic $$x^2-x-2$$. We look for two numbers which multiply to $$-2$$ and add to $$-1$$. These numbers are $$-2$$ and $$1$$. Therefore:
$$
x^2-x-2=(x-2)(x+1).
$$
The complete factorisation is:
$$
\mathrm{f}(x)=(3x-1)(x-2)(x+1).
$$

---

**(iii)**  
We now consider the equation:
$$
3y^6-4y^4-5y^2+2=0.
$$
Let $$x=y^2$$. Then, the equation becomes:
$$
3x^3-4x^2-5x+2=0.
$$
From part (ii), we have already factorised the cubic:
$$
3x^3-4x^2-5x+2=(3x-1)(x-2)(x+1)=0.
$$
Thus, the solutions for $$x$$ are:
$$
x=\frac{1}{3}, \quad x=2, \quad x=-1.
$$
Recall that $$x=y^2$$. So we set up the equations:

1. $$y^2=\frac{1}{3}$$  
   $$
   \Longrightarrow y=\pm \frac{1}{\sqrt{3}}.
   $$

2. $$y^2=2$$  
   $$
   \Longrightarrow y=\pm \sqrt{2}.
   $$

3. $$y^2=-1$$  
   This has no real solutions since the square of a real number cannot be negative.

Therefore, the total number of real solutions for $$y$$ is:
$$
2+2+0=4.
$$
**(i)** $$3x-1$$ is a factor of $$f(x)$$ because $$f\left(\frac{1}{3}\right) = 0$$. **(ii)** $$f(x) = (3x - 1)(x - 2)(x + 1)$$. **(iii)** There are 4 real solutions for $$y$$: $$y = \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \sqrt{2}, -\sqrt{2}$$.

Given that
(i) Show that is a factor of .
(ii) Factorise completely.
(iii) Hence, state the number of real solutions for
.

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Given that (i) Show that is a factor of . (ii) Factorise completely. (iii) Hence, state the number of real solutions for .