You should be aware of the fundamental theorem of algebra (FTCoA). If a quadratic has roots at [tex]r_1[/tex] and [tex]r_2[/tex], then we can factorize it as
[tex](x - r_1) (x - r_2)[/tex]
Expanding, this is equivalent to
[tex]x^2 - (r_1 + r_2) x + r_1 r_2[/tex]
Now, if we can also write this
[tex]x^2 + bx + c[/tex]
then we must have
[tex]x^2 + bx + c = x^2 - (r_1 + r_2) x + r_1r_2 \\\\ ~~~~ \implies \begin{cases} b = -(r_1 + r_2) \\ c = r_1r_2 \end{cases}[/tex]
(By the way, these are known as Vieta's formulas.)
We're told that the quadratic,
[tex]p(x) = 3x^2 - 6x + 8,[/tex]
has roots [tex]x=\alpha[/tex] and [tex]x=\beta[/tex], so by the FTCoA, we can write
[tex]p(x) = 3 (x - \alpha) (x - \beta)[/tex]
Expanding this last form, we have the identity and exact values
[tex]3x^2 - 6x + 8 = 3 (x^2 - (\alpha + \beta) x + \alpha\beta) \\\\ ~~~~ \implies x^2 - 2x + \dfrac83 = x^2 - (\alpha + \beta) x + \alpha\beta \\\\ ~~~~ \implies \begin{cases} \boxed{\alpha + \beta = 2} \\\\ \boxed{\alpha\beta = \frac83} \end{cases}[/tex]
(i) We want to construct a new quadratic [tex]p_{\rm i}(x)[/tex] such that its roots are [tex]x=\alpha^2[/tex] and [tex]x=\beta^2[/tex]. By the FTCoA, we would have
[tex]p_{\rm i}(x) = (x - \alpha^2) (x - \beta^2)[/tex]
Expanding,
[tex]p_{\rm i}(x) = x^2 - (\alpha^2 + \beta^2) x + \alpha^2\beta^2[/tex]
Now, notice that
[tex]\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 2^2 - 2\cdot\dfrac83 = -\dfrac43[/tex]
[tex]\alpha^2\beta^2 = (\alpha\beta)^2 = \left(\dfrac83\right)^2 = \dfrac{64}9[/tex]
It follows that
[tex]p_{\rm i}(x) = x^2 + \dfrac43 x + \dfrac{64}9[/tex]
which we can rewrite with integer coefficients by scaling each term by a factor of 9 to get
[tex]p_{\rm i}(x) = \boxed{9x^2 + 12x + 64}[/tex]
(ii) Now we want a quadratic [tex]p_{\rm ii}(x)[/tex] with roots at [tex]2\alpha[/tex] and [tex]2\beta[/tex]. This means
[tex]p_{\rm ii}(x) = (x - 2\alpha) (x - 2\beta) \\\\ ~~~~~~~~ = x^2 - 2 (\alpha + \beta)x + 4\alpha\beta[/tex]
[tex]\implies p_{\rm ii}(x) = x^2 - 4x + \dfrac{32}3[/tex]
or more cleanly, scaling by 3,
[tex]p_{\rm ii}(x) = \boxed{3x^2 - 12x + 32}[/tex]
(iii) Now the roots are [tex]x=\frac\alpha2[/tex] and [tex]x=\frac\beta2[/tex], which gives
[tex]p_{\rm iii}(x) = \left(x - \dfrac\alpha2\right) \left(x - \dfrac\beta2\right) \\\\ ~~~~~~~~ = x^2 - \dfrac{\alpha + \beta}2 x + \dfrac{\alpha\beta}4[/tex]
[tex]\implies p_{\rm iii}(x) = x^2 - x + \dfrac8{12}[/tex]
[tex]\implies p_{\rm iii}(x) = \boxed{12x^2 - 12x + 8}[/tex]
(iv) If the roots are [tex]x=\frac\alpha{\alpha+1}[/tex] and [tex]x=\frac\beta{\beta+1}[/tex], we have
[tex]p_{\rm iv}(x) = \left(x - \dfrac\alpha{\alpha+1}\right) \left(x - \dfrac\beta{\beta+1}\right) \\\\ ~~~~~~~~~= x^2 - \left(\dfrac\alpha{\alpha+1} + \dfrac\beta{\beta+1}\right) x + \dfrac{\alpha\beta}{(\alpha+1)(\beta+1)}[/tex]
Note that
[tex]\dfrac\alpha{\alpha+1} + \dfrac\beta{\beta+1} = \dfrac{\alpha(\beta+1) + \beta(\alpha+1)}{(\alpha+1)(\beta+1)} \\\\ ~~~~~~~~~~~~~~~~~~~~ = \dfrac{2\alpha\beta + \alpha + \beta}{\alpha\beta + \alpha + \beta + 1} \\\\ ~~~~~~~~~~~~~~~~~~~~ = \dfrac{\frac{16}3 + 2}{\frac83 + 2 + 1} \\\\ ~~~~~~~~~~~~~~~~~~~~ = \dfrac{22}{17}[/tex]
[tex]\dfrac{\alpha\beta}{(\alpha+1)(\beta+1)} = \dfrac{\alpha\beta}{\alpha\beta + \alpha + \beta + 1} \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \dfrac{\frac83}{\frac83 + 2 + 1} \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \dfrac8{17}[/tex]
[tex]\implies p_{\rm iv}(x) = x^2 - \dfrac{22}{17}x + \dfrac8{17}[/tex]
[tex]\implies p_{\rm iv}(x) = \boxed{17x^2 - 22x + 8}[/tex]
