Author Topic: werid polynomial q (Read 4064 times) Tweet Share

massive · « **Reply #15 on:** July 29, 2016, 05:54:11 pm »

thanks! im also stuck on part iv for this q

RuiAce · « **Reply #16 on:** July 29, 2016, 06:22:44 pm »

Quote from: massive on July 29, 2016, 05:54:11 pm

thanks! im also stuck on part iv for this q

$\text{Given }b=0\text{ we have }c^2+a^2d=0 \iff c^2=-a^2d\\ \text{and since the square of an even number is even, it suffices to show that }c^2\text{ is even.}\\ \text{Because }c^2\text{ is an integer, and even times any integer gives even, it suffices to show either }a^2 \text{ or }d\text{ is even.}$
$\because x=2\text{ is a root keeping in mind }b=0\\ 16+8a+2c+d=0 \\ \iff d=-16-8a-2c=-2(8+4a+1)\text{ which is even.}\\ \text{So }d\text{ must be even, thus puzzling everything back together we achieve our result.}$

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Author Topic: werid polynomial q (Read 4064 times) Tweet Share

massive

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RuiAce

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