Say for example I have the polynomial $x^2-8x-153$. It is not easy to immediately see the factors of this are $9$ and $17$. Is there a method to find these factors easily when the terms in the polynomials are this big? The reason I ask this is because I don't want to spend a lot of time on a test guessing for the factors. Thanks in advance!
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IF you have some practice with factoring, $153$ is not that bad, since $3$ can be seen to divide into $15$ and into $3$, so we get $153=3(51)$. You can then see $51=17*3$, (hint for this $51=30+21$, so factoring a three gives $3(10+7)$). That is if you have practice factoring integers. It well known however that factoring integers is hard, so hard in fact that it is the basis some encryption algorithms, and factoring polynomials is harder. – Baby Dragon Oct 25 '13 at 04:57
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SEE http://math.stackexchange.com/questions/522903/factoring-quadratics-asterisk-method/522957#522957 – Will Jagy Oct 25 '13 at 05:00
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2How dare you call 153 ugly! It is the smallest number which is the sum of the cubes of its digits! – Andrew Dudzik Oct 25 '13 at 05:08
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Sorry, ugly isn't an accurate name to call it. Let's say unique because I have rarely encountered this number before :P. – Kot Oct 25 '13 at 05:10
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1@user33433 That reminds me of the anecdotal story of someone saying "1729 is boring" in front of Ramanujan. – Dennis Meng Oct 25 '13 at 05:23
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@DennisMeng That's what I was going for :) – Andrew Dudzik Oct 25 '13 at 06:51
1 Answers
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Every quadratic equation can be solved by completing the square. Noting that $8/2 = 4$, we have
$$x^2 - 8x - 153 = x^2 - 8x + (4^2) - 4^2 - 153$$
or rearranging,
$$(x - 4)^2 = 153 + 4^2 = 169 = 13^2$$
Thus, $x - 4 = \pm 13$, leading to the two solutions $x = -9$ and $x = 17$. Then the polynomial factors as
$$(x + 9)(x - 17)$$
So this technique only requires knowing about square roots.
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3I completely forgot about completing the square! I also think you mean $(x+9)$ not $(x-9)$. Thanks for refreshing my memory! – Kot Oct 25 '13 at 04:57