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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.

Please help me

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  • $\begingroup$ Try $x=3$ or $x=5$. $\endgroup$ – Math Lover Dec 1 '17 at 4:29
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    $\begingroup$ Is there a simpler method $\endgroup$ – user508848 Dec 1 '17 at 4:31
  • $\begingroup$ hint: $45 = 3\times 3\times 5$ $\endgroup$ – John Joy Dec 3 '17 at 2:14
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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$$P(x)=x^4-7x^3+4x^2+39x-45=(x-5)(x-3)(x^2+x-3)$$

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  • $\begingroup$ 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. $\endgroup$ – Xander Henderson Dec 1 '17 at 4:56
  • $\begingroup$ @Xander Yes that is true. In this case, I saw that the unit solutions wouldn’t work because of the coefficients of the quartic. So I tried the next one which was three. Lo’ and behold, it worked! Really it boils down to you being able to guesstimate the magnitude of the numbers and if they’ll cancel out $\endgroup$ – Crescendo Dec 1 '17 at 5:31
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I graphed it (Is that cheating?) and found four real roots. Two are integers and can easily be verified by substitution. Where do you think the other two come from? (That negative one makes me think of logarithms.)

enter image description here

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    $\begingroup$ Logarithms? In an integer polynomial? Why? $\endgroup$ – G Tony Jacobs Dec 1 '17 at 5:00
  • $\begingroup$ The other two are from the remaning quadratic factor. I don't know how you could possibly think of logs... $\endgroup$ – Dylan Dec 1 '17 at 21:03
  • $\begingroup$ -ln 10 = -2.303. This demonstrates the limitation of solving by graphing. Just because the other solutions look like integers, doesn't mean they are. You have yo test it by back substitution. Meanwhile, the algebraic method absolutely proves that there are two integer solutions and two real non-integer solutions. Could one be a log? It appears to be so from the graph, but it's unlikely. Still, it's easy to test. $\endgroup$ – Adam Hrankowski Dec 1 '17 at 22:03

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