Suppose $a$ is a root of $x^2 + 2x - 13.$ Than, find the value of $341a - a^5.$
I was thinking of trying to find a substitution to find the value of $341a - a^5,$ but I'm not sure what. Can someone give me a hint?
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2Well...$a^2=13-2a\implies a^3=13a-2a^2=13a-2(13-2a)=13a-26+4a=17a-26$ and so on. – lulu Sep 23 '20 at 19:31
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and in this case it reduces beautifully – Math Lover Sep 23 '20 at 19:41
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$780,,,,,$ – Sep 23 '20 at 19:52
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"and in this case it reduces beautifully" I dunno, I tried it 5 times and got 6 different results.... then again arithmetic errors are par for the course. – fleablood Sep 23 '20 at 20:00
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How to avoid arithmetic errors: (21:54) gp > a=Mod(y,y^2+2y-13); (21:54) gp > lift(341a-a^5) %55 = 780 – Sep 24 '20 at 05:22
2 Answers
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$$ \left( - x^{5} + 341 x \right) = \left( x^{2} + 2 x - 13 \right) \cdot \color{magenta}{ \left( - x^{3} + 2 x^{2} - 17 x + 60 \right) } + \left( 780 \right) $$
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Will Jagy
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First divide $341a - a^5$ by $a^2+2a-13$ to get a remainder of $ja + k$. Then plug in $a = \frac{-2 \pm \sqrt{4+4*13}}{2}=-1\pm \sqrt{1+13}$.
$-a^5 +341a = -a^3(a^2 + 2a-13) + 2a^4-13a^3 +341a=$
$2a^4 - 13a^3 + 341a= 2a^2(a^2+2a-13) -4a^3+26a^2 -13a^3+341a=$
$-17a^3 + 26a^2 + 341a = -17a(a^2 + 2a -13) +34a^2-17*13a +26a^2 + 341a =$
$60a^2 + 120a = 60(a^2 +2a -13) -120a + 60*13 +120a = $
$780$.
And ... we don't need to plug is $a =-1\pm\sqrt {14}$.
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Of course the real hard part is not to make any effing arithmetic errors!
fleablood
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