Hey, I recognize the left-hand side of the first equation. A student once showed me a foreign text containing different root-finding methods for certain higher-order polynomials. They were not general-solution methods; they work only when the polynomial or its factorization has an appropriate form.
I've assumed that the OP has two different exercises in the same thread (maybe because the solution methods are similar).
[imath](x-1)(x-2)(x-4)(x-8) = 7x^2[/imath]
I think the approach is to recognize that we can get (on the left-hand side) two quadratic factors of the form:
[imath]x^2 + 8 + Bx[/imath]
Rearrange and group the given linear factors.
[imath][(x-1)(x-8)]*[(x-2)(x-4)] = 7x^2[/imath]
[imath](x^2 + 8 - 9x)(x^2 + 8 - 6x) = 7x^2[/imath]
On the left-hand side, divide each quadratic factor by [imath]x[/imath]. Divide the right-hand side by [imath]x^2[/imath].
[imath](x + \frac{8}{x} - 9)(x + \frac{8}{x} - 6) = 7[/imath]
Substitute [imath]t = x + \frac{8}{x}[/imath] and solve for [imath]t[/imath] using the Quadratic Formula.
[imath](t - 9)(t - 6) = 7[/imath]
[imath]t^2 - 15t + 47 = 0[/imath]
[imath]t = \frac{15 ± \sqrt{37}}{2}[/imath]
Using the first t-solution [imath]\frac{15 + \sqrt{37}}{2}[/imath], we have
[imath]x + \frac{8}{x} = \frac{15 + \sqrt{37}}{2}[/imath]
Clear the ratios.
[imath]2x^2 - (15 + \sqrt{37})x + 16 = 0[/imath]
Using the Quadratic Formula again, yields two Real solutions.
Repeat the last three steps, using the second t-solution. That yields two Complex conjugates.
Maybe there's another special method (rearranging, then using a substitution) for working the second equation, also.
?