I've been trying to simplify $x^3 - 4x^2 + 10x - 125$ for a while now, and I don't seem to progress.
I know that the factors of $125$ are $1$, $5$, $25$ and $125$, but none of these seem to help here. So far I have managed to get $x(x^2 - 4x + 10) - 125$.
Can I go any further than this? Thank you! By the way... it's my first post here, so if I didn't provide something necessary - excuse me.
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By simplify, do you mean factorise? – Michael Albanese Aug 24 '14 at 19:41
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Oh god, yes I do. Sorry, I'm not learning maths in english so the terms are kind of confusing :) – Benskey Aug 24 '14 at 19:42
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1If you want to factorize it, it's irreducible. However, you can get things like $x(x(x-4)+10)-125$, which technically aren't factorizations. – Akiva Weinberger Aug 24 '14 at 19:45
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$x(x^2-4x+10)-125$ is not a factorization, however. A factorization is merely a product, no subtraction allowed afterwards. – Thomas Andrews Aug 24 '14 at 19:46
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Also try $-1,-5,-25,-125$ – Will Jagy Aug 24 '14 at 19:46
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1Okay, thanks. This means that for the past 2 hours I've been pretty much searching for nothing :) – Benskey Aug 24 '14 at 19:47
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1Have you typed in the polynomial correctly? – Will Jagy Aug 24 '14 at 19:48
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@columbus8myhw since the point was factorizing, you should post your comment as an answer...and we could upvote it! – MattAllegro Aug 24 '14 at 19:52
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1If you replace the $10$ you typed with a $20$ the result is much nicer. Different problem, of course, which is why I asked you to check whether you have typed yours correctly. – Will Jagy Aug 24 '14 at 19:55
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Hmm, didn't know this. So which comment should I post ? ^^ – Benskey Aug 24 '14 at 19:55
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If you want to factorize it, it's irreducible over the rationals. However, you can get things like $x(x(x−4)+10)−125$, which technically aren't factorizations.
If you don't want to restrict yourself to rationals, you can write it as $(x-5.89793)(x^2+1.89793 x+21.1938)$. (The exact form is extremely complicated.)
Akiva Weinberger
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2Yeah. (Of course, no cubics are irreducible over the reals, as my second paragraph demonstrates.) – Akiva Weinberger Aug 24 '14 at 20:00
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If you have access to any computer algebra system, you can solve for the roots of a cubic equation. (If you don't have access to a computer algebra system then writing everything out this fully is a waste of time!)
$$x^3 - 4x^2 + 10x - 125=(x-x_0)(x-x_1)(x-x_2)=\left(x-\frac{1}{3} 7^{2/3} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x-\frac{1}{3} 7^{2/3} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x+\frac{1}{3} \left(2\ 7^{2/3} \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}-\sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-4\right)\right)$$
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1Which is $(x-5.89793)(x^2+1.89793 x+21.1938)$ if you want a (real) decimal approximation. – Akiva Weinberger Aug 24 '14 at 19:56
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What do you get when you multiply together the first two factors? – Akiva Weinberger Aug 24 '14 at 19:57
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If I understood correctly, -4d^5 ? Or are you asking for the multiplication of a real decimal approximation? If so - its x^3 :) – Benskey Aug 24 '14 at 20:00