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Suppose $a$ is a root of $x^2 + 3x - 1.$ Find $\frac{2a^5 - 5a^4 + 2a^3 - 8a^2}{a^2 + 1}.$


I was thinking of factoring the fraction a bit first, than letting $a^2 = 1 - 3a.$ However, that leads nowhere.

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    $\begingroup$ Well I'd substratc $2a^5 + 6a^4 - 2a^3 = 0$ from the top and then keep subtracting multiples of $a^2 +3a -1 = 0$ and we what sort of thing I get in the end. $\endgroup$
    – fleablood
    Sep 16, 2020 at 3:09
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    $\begingroup$ $(2x^3-11x^2+37x-130)(x^2+3x-1)+427x-130$ is the factorization. What next? @fleablood $\endgroup$ Sep 16, 2020 at 3:22
  • $\begingroup$ I suggest something else. $\endgroup$
    – Spectre
    Sep 16, 2020 at 3:23
  • $\begingroup$ Take $a^4$ out of $2a^5 - 5a^4$. $\endgroup$
    – Spectre
    Sep 16, 2020 at 3:24
  • $\begingroup$ You have a linear factor there. Find that. $\endgroup$
    – Spectre
    Sep 16, 2020 at 3:24

2 Answers 2

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A complete solution. Let $a$ be a root of $x^2+3x-1$. Then

$a^2=1-3a$

$a^3=a-3a^2=a-3(1-3a)=10a-3$

$a^4=10a^2-3a=10(1-3a)-3a=10-33a$

$a^5=10a-33a^2=10a-33(1-3a)=-33+109a$

Therefore the numerator:$$2a^5 - 5a^4 + 2a^3 - 8a^2= -66+218a-5(10-33a)+2(10a-3)-8(1-3a)=-130+427a$$

and the denominator: $a^2+1=2-3a$.

Let $a'$ be another root of $x^2+3x-1=0$. Then $aa'=-1. a+a'=-3$. Multiply the numerator and the denominator by $2-3a'$. The denominator becomes $$(2-3a)(2-3a')=4+9-6(-3)=13-18=-5.$$ The numerator becomes $$(427a-130)(2-3a')= 854a-260+1281+390a'=1021-1170+464a=464a-149.$$ So the fraction is equal to $$\frac{464a-149}{-5}$$ where $$a=\frac{-3\pm \sqrt{13}}{2}.$$

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  • $\begingroup$ I guess I was a fool.... $\endgroup$
    – Spectre
    Sep 16, 2020 at 3:33
  • $\begingroup$ Fool for taking the longest route $\endgroup$
    – Spectre
    Sep 16, 2020 at 3:34
  • $\begingroup$ I hoped to get an integer that way without using the actual value of $a$. But now I do not see how can one avoid substituting $a=\frac{-3\pm \sqrt{13}}{2}$. The answer is probably irrational. $\endgroup$
    – markvs
    Sep 16, 2020 at 3:36
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    $\begingroup$ I fixed the calculation. But the fraction still is not completely simplified. $\endgroup$
    – markvs
    Sep 16, 2020 at 3:49
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    $\begingroup$ I think the OP may have copied the fraction incorrectly. I have solved several such problems in the past. The result always did not depend on $a$. $\endgroup$
    – markvs
    Sep 16, 2020 at 4:19
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Another way.

For $a^2+3a-1=0$ and $a=\frac{-3+\sqrt{13}}{2}$ we obtain: $$(a^2-1)^2=9a^2$$ or $$(a^2+1)^2=13a^2$$ or $$a^2+1=\sqrt{13}a,$$ which gives: $$\frac{2a^5-5a^4+2a^3-8a^2}{a^2+1}=$$ $$=\tfrac{2a^5+6a^4-2a^3-11a^4-33a^3+11a^2+37a^3+111a^2-37a-130a^2+37a}{\sqrt{13}a}=$$ $$=\tfrac{37-130a}{\sqrt{13}}=\tfrac{37-65(-3+\sqrt{13})}{\sqrt{13}}=\tfrac{232-65\sqrt{13}}{\sqrt{13}}.$$ For $a=\frac{-3-\sqrt{13}}{2}$ we obtain: $$a^2+1=-\sqrt{13}a,$$ which gives $$\frac{2a^5-5a^4+2a^3-8a^2}{a^2+1}=\tfrac{-232-65\sqrt{13}}{\sqrt{13}}.$$

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