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.

You are watching: X 4 divided by x 2

algebra-precalculus
re-publishing
cite
monitor
edited Jul 20 "12 in ~ 3:06
user2468
inquiry Jul 20 "12 in ~ 2:33

Austin BroussardAustin Broussard
\$endgroup\$
3
1
\$egingroup\$
AustinBroussard her answer is valid. \$endgroup\$
–user26649
Jul 20 "12 at 11:53

2
\$egingroup\$
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\$).

See more: How Far Is Redding, Ca From Sacramento, Ca, How Far Is Redding From Sacramento

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)