Polynomial division simple

Question

I am trying to divide polynomials but i am ending up with different outcomes.

Let's assume i have : $\dfrac{s^4+3s^3+4s^2+4s+1}{2s^3+2s^2+3s+1}$

Can anyone solve this step by step (long or synthetic division doesn't really matter, but i would prefer the easiest way)? Thanks in advance.

Think of polynomial division as numeric division. That is $f(x)=g(x)q(x)+r(x)$. For your problem, you have $$s^4+3s^2+4x^2+4s+1=(2s^3+2s^2+3s+1)q(x)+r(x)$$ Thus you have to get a degree 4 polynomial from multiplying a degree 3 polynomial by $q$. So $q$ must be degree 1, and the coefficient must be 1/2 (why)? You could run the long division program using this idea... — Iceman, Jun 05 '15 at 22:39

score 0 · Accepted Answer · answered Jun 05 '15 at 22:41

0

You compare the highest powers to get the next term, $\frac{s^4}{2s^3} = \frac12 s$.

Then distribute that term across the denominator: $\frac12 s (2s^3 + 2s^2 + 3s + 1) = s^4+s^3+\frac32 s^2 + \frac12 s$.

Then subtract from the numerator to get $2s^3 + \frac52 s^2 + \frac72 s + 1$.

Repeat from the beginning until done.

answered Jun 05 '15 at 22:41

Barry

So the solution is $(\frac{1}{2}s+1)(\frac{1}{2}s^2-\frac{1}{2}s)$ ? – sayid jetzenden Jun 05 '15 at 22:54
No. The solution would look like $f(s) = q(s)g(s) + r(s)$. The product of those two terms doesn't give you the numerator – Barry Jun 05 '15 at 22:59
This one maybe $((2s^3+2s^2+3s+1)(\frac{1}{2}s+1)+(\frac{\frac{1}{2}s^2+\frac{1}{2}s}{2s^3+2s^2+3s+1})$ ? – sayid jetzenden Jun 05 '15 at 23:04
your quotient is $(1/2s+1)$ and your remainder is $(1/2s^2-1/2s)$ – Iceman Jun 05 '15 at 23:22

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