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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Try $x=3$ or $x=5$. – Math Lover Dec 01 '17 at 04:29
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1Is there a simpler method – Dec 01 '17 at 04:31
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hint: $45 = 3\times 3\times 5$ – John Joy Dec 03 '17 at 02:14
2 Answers
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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$$P(x)=x^4-7x^3+4x^2+39x-45=(x-5)(x-3)(x^2+x-3)$$
Crescendo
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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 01 '17 at 04:56
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@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 – Crescendo Dec 01 '17 at 05:31
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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.)
MathAdam
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The other two are from the remaning quadratic factor. I don't know how you could possibly think of logs... – Dylan Dec 01 '17 at 21:03
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-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. – MathAdam Dec 01 '17 at 22:03