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 Algebra Pre-Algebra and Basic Algebra Math Forum

December 12th, 2018, 02:29 PM   #1
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Not sure where to start.
Attached Images 20181213_122910 (1).jpg (97.5 KB, 18 views) December 12th, 2018, 03:46 PM   #2
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Quote:
 Originally Posted by hagfish66 Not sure where to start.
What do you get when you do the long division:
$\displaystyle \dfrac{x^4 + 0x^3 + 3x^2 + 0x + a}{x^2 + x + 2}$

-Dan December 12th, 2018, 06:58 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 As $p(x)$ has no term in $x^3$, $p(x) = (x^2 + x + 2)(x^2 - x + a/2)$. As it also has no term in $x$, $a/2 - 2 = 0$, and so $a = 4$. Hence $p(x) = (x^2 + x + 2)(x^2 - x + 2)$. Thanks from topsquark January 3rd, 2019, 06:52 PM   #4
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Quote:
 Originally Posted by hagfish66 Not sure where to start.
Do long division - you will get a remainder of a-4. Given that x^2+x+2 is a factor, remainder does not exist. For it to not exist, a needs to be equal to 4. Check out the link.

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