Answer:
Explanation:At x = 3, we have y = 0
This means x = 3 is a root.
Let the cubic function be:
[tex]y=(x-3)(x-a)(x-b)_{}[/tex]When x = 1, y = -2, so
[tex]\begin{gathered} -2=(1-3)(1-a)(1-b) \\ 1=(1-a)(1-b) \\ 1=1-b-a+ab \\ a+b-ab=0\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\text{.}(1) \end{gathered}[/tex]SImilarly, when x = 2, y = -1
[tex]\begin{gathered} -1=(-1-3)(-1-a)(-1-b) \\ -1=4(1+a)(1+b)_{} \\ (1+a)(1+b)=-\frac{1}{4} \\ \\ 1+b+a+ab=-\frac{1}{4} \\ \\ a+b+ab=-\frac{5}{4}\ldots\ldots\ldots.....\ldots\ldots..\ldots.\ldots\ldots\text{.}(2) \end{gathered}[/tex]Adding (1) and (2)
[tex]\begin{gathered} 2a+2b=-\frac{5}{4} \\ \\ a+b=-\frac{5}{8}\ldots\ldots\ldots\ldots\ldots\ldots...........\ldots\ldots\ldots\ldots(3) \end{gathered}[/tex]Subtracting (1) from (2)
[tex]\begin{gathered} 2ab=-\frac{5}{4} \\ \\ ab=-\frac{5}{8}\ldots\ldots..\ldots.\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}(4) \\ \\ \Rightarrow b=-\frac{5}{8a}\ldots\ldots\ldots\ldots\ldots\ldots\ldots.\ldots\ldots.\text{.}(5) \end{gathered}[/tex]Using (5) in (3)
[tex]\begin{gathered} a-\frac{5}{8a}=-\frac{5}{8} \\ \\ 8a^2-5=-5a \\ 8a^2+5a-5=0 \\ a=\frac{-5}{16}-\sqrt[]{\frac{185}{16}} \\ \\ OR \\ \frac{-5}{16}+\sqrt[]{\frac{185}{16}} \end{gathered}[/tex]Therefore, for b, we have:
[tex]\begin{gathered} b=\frac{1}{16}(\sqrt{185}-5) \\ OR \\ b=\frac{1}{16}(-5-\sqrt[]{185}) \end{gathered}[/tex]Replacing a and b by these values obtained, we have the required cubic function
