If we divide x^4+4x^3+2x^2+x+4 by x^2+3x, what will be the remainder?

Question

asked Jul 7, 2019 48.7k views

1 Answer

Maksonnie · Answer 1 · 2019-07-12T01:23:21+0000

If you're familiar with synthetic division, but not the extended form (which allows you easily compute the quotient/remainder when dividing a polynomial by another polynomial of degree greater than 1), then you can perform two steps of SD.

Instead of dividing by
x^2+3x

, first divide by

, then by

(since

). So we have

0 | 1 4 2 1 4
... | 0 0 0 0
= = = = = = = = = = = =
... | 1 4 2 1 4

which translates to

$\frac{x^4+4x^3+2x^2+x+4}x=x^3+4x^2+2x+1+\frac4x$

Ignoring the remainder term for now, the next round of SD yields

-3 | 1 4 2 1
... | -3 -3 3
= = = = = = = = = =
... | 1 1 -1 4

which translates to

$(x^3+4x^2+2x+1)/(x+3)=x^2+x-1+\frac4{x+3}$

Now, putting everything together, we have

$(x^4+4x^3+2x^2+x+4)/(x^2+3x)=(x^3+4x^2+2x+1+\frac4x)/(x+3)$

$=x^2+x-1+\frac4{x+3}+\frac4{x(x+3)}$

=x^2+x-1+(4x+4)/(x^2+3x)