Question 1039621
{{{p=x^2}}}


{{{p^2-23p+49}}}


That method seems not useful.


Try Rational Roots Theorem and synthetic division to check roots -1, -7, 1, 7.


No remainders found (synthetic divisions done on paper).
The polynomial does not have rational roots.  Binomial factors are not found.
It does have four real but irrational roots.  Back to the substitution for {{{x^2}}}, you can find them.



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{{{p=x^2}}}

{{{p^2-23p+49}}}

{{{p=(23+- 3*sqrt(37))/2}}}


Substitute back the {{{x^2}}}.


{{{x=0+- sqrt((23+- 3*sqrt(37))/2)}}}-----which is FOUR values; the 0 included so that the rendering on the page will work.  I have not finished putting this into the final desired factored form.