Solve the equation $$x^4-8x^3+23x^2-30x+15=0$$
As $x=0$ is obviously not a solution, we can consider $x\ne0$, so I have tried to divide both sides by $x^2$ to get $$x^2-8x+23-\dfrac{30}{x}+\dfrac{15}{x^2}=0$$ Clearly this does not help. I have also tried to find the rational roots using Horner's method. It seems that the polynomial in the left-hand side does not have rational roots.
Solve the equation $x^4-8x^3+23x^2-30x+15=0$
algebra-precalculuspolynomialsquarticsroots
Related Solutions
Well if $\frac pq; \gcd (p,q) =1;p,q$ integers is a root then
$30\frac {p^n}{q^n} = 91$. As $91$ is an integer, and $p$ and $q$ are relatively prime, $q^n|30$.
If $q = 1$ then $30p^n = 91$ which has no integer solutions.
If $q \ne 1$ then $q^n|30=2*3*5$. But $30 = 2*3*5$ has no prime factors of powers greater than $1$. So $n= 1$. and $\frac pq = \frac {91}{30}$. That is is the only rational solution and it only holds for $n=1$.
Using Maple's Groebner Basis package, we find that $a$ satisfies the equation $$ 256a^8-1024a^7-2080a^6+3520a^5+3105a^4-4060a^3+1158a^2-92a+1=0 $$ Here's the Maple code . . .
Best Answer
Note that, if $p(x)$ is your polynomial, then $p(x+2)=x^4-x^2-2x-1=x^4-(x+1)^2$. Can you take it from here?