# Solve $x^4 -7x^3 + 4x^2 +39x -45=0$

Solve $x^4 -7x^3 + 4x^2 +39x -45=0$

I tried this question by using the products of roots $= -45$. But factorization didn't go well. Trial and error method is not working.

• Try $x=3$ or $x=5$. – Math Lover Dec 1 '17 at 4:29
• Is there a simpler method – user508848 Dec 1 '17 at 4:31
• hint: $45 = 3\times 3\times 5$ – John Joy Dec 3 '17 at 2:14

Rational root theorem gives $x=3$ as a root. Factoring that out, we can verify that $x=5$ is another root. Factoring that out gives an irreducible quadratic that can be easily solved with the quadratic formula.
Hence$$P(x)=x^4-7x^3+4x^2+39x-45=(x-5)(x-3)(x^2+x-3)$$
• Minor pedantic nitpick: the rational root theorem tells us that if this polynomial has any rational roots, they must be in the set $\{\pm 1, \pm 3,\pm 5,\pm 15,\pm 45\}$. One can verify that both 3 and 5 get the job done, but there is no obvious reason to try those first. – Xander Henderson Dec 1 '17 at 4:56