I know exactly how to synthetically division in the layout of: $(xpm a)$. But not $(x^npm a)$ (with one exponent). Therefore if anyone have the right to tell me if anything alters or if the procedures are the exact same just with the remainder as $dfracremainderx^npm a$. And also please do not deal with the problem for me.

Edit:I think I"ve got to an answer. Long department confused me so ns used increased synthetic department and got: $x^2-x+1+left(dfracx^2+3x^3+1 ight)$. Have the right to anyone verify mine answer please.

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edited Jul 20 "12 in ~ 3:06
inquiry Jul 20 "12 in ~ 2:33

Austin BroussardAustin Broussard
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AustinBroussard her answer is valid. $endgroup$
Jul 20 "12 at 11:53
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All you need to do is do lengthy division; alternatively, you can subtract proper multiples the $x^3+1$ and replace the dividend by the distinction until you get a term that is of level strictly smaller than $3$. (In various other words, do polynomial long division analytically instead of synthetically).

For instance, suppose we to be trying to uncover the remainder of splitting $$2x^5 - 3x^2 + 1$$by$$x^2-2.$$

This is the exact same as the remainder the dividing$$2x^5 - 3x^2 + 1 - p(x)(x^2-2)$$by $p(x)$, for any kind of polynomial $p(x)$.

Multiplying $x^2-2$ by $2x^3$ we acquire $2x^5 - 4x^3$. We can subtract this indigenous $2x^5 - 3x^2+1$ we get$$2x^5 - 3x^2 + 1 - (2x^5-4x^3) = 4x^3 - 3x^2 + 1.$$So the remainder when dividing $2x^5-3x^2+1$ by $x^2-2$ is the same as the remainder when separating $4x^3-3x^2+1$ through $x^2-2$.

Now, multiplying $x^2-2$ by $4x$ we get $4x^3 - 8x$; individually it from $4x^3-3x^2+1$ we get$$4x^3-3x^2+1 - (4x^3 - 8x) = -3x^2 + 8x + 1.$$So the remainder of splitting $4x^3-3x^2+1$ through $x^2-2$ is the same as the remainder of dividing $-3x^2+8x+1$ by $x^2-2$. Multiplying $x^2-2$ through $-3$ we get $-3x^2+6$; subtracting it from $-3x^2+8x+1$ us get:$$-3x^2+8x+1 -(-3x^2+6) = 8x -5.$$So the remainder of separating $-3x^2+8x+1$ through $x^2-2$ is the same as the remainder of dividing $8x-5$ through $x^2-2$. However the remainder of dividing $8x-5$ through $x^2-2$ is simply $8x-5$ (already of level smaller than $2$).

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So the remainder in the original division is $8x-5$.

(This is just, together I detailed above, polynomial long department done in the discourse manner quite than synthetically)