Adv. Alg Dividing Polynomials.

[Makayla8]

(n^3-100n+3)/n-10. Thanks Guys

 

[soroban]

Hello, Makayla8!

Exactly where is your difficulty?
. . You can't do long division at all?
. . You get a different answer than the one given in your book?
. . You didn't leave a "space" for the \(\displaystyle n^2\)-term?

\(\displaystyle (n^3-100n+3) \div (n-10)\)

. . \(\displaystyle \begin{array}{cccccccccc} &&&& n^2 &+& 10n \\ && --&--&--&--&--&--&-- \\ n-10 & | & n^3 &&& - & 100n & + & 3 \\ && n^3 &-& 10n^2 \\ && --&--&-- \\ &&&& 10n^2 &-& 100n \\ &&&& 10n^2 &-& 100n \\ &&&& --&--&-- \\ &&&&&&&+& 3 \end{array}\)

 

[JeffM]

(n^3-100n+3)/n-10. Thanks Guys

Start with http://www.purplemath.com/modules/polydiv.htm

If you have a question about your specific problem after reading that, please show us what you have done and where you are stuck so we know exactly how to help you.

 

[HallsofIvy]

Yet another way to do this: since the denominator is n- 10, put the numerator in terms of n- 10 also- look for numbers, a, b, c, and d such that\(\displaystyle a(n- 10)^3+ b(n- 10)^2+ c(n- 10)+ d= n^3- 100n+ 3\) for all n.

Multiplying the powers on the left out, \(\displaystyle a(n- 10)^3= an^3- 20an^2+ 300an- 1000a\), \(\displaystyle b(n- 10)^2= bn^2- 20bn+ 100b\) and \(\displaystyle c(n-10)= cn- 10c\) so that equation becomes
\(\displaystyle an^3+ (-30a+ b)n^2+ (300a- 20b+ c)n+ (-1000a+ 100b- 10c+ d)= n^3- 100n+ 3\)

Setting corresponding coefficients equal, we must have a= 1, -30a+ b= 0, 300a- 20b+ c= -100, -1000a+ 100b- 10c+ d= 3.
From a= 1, -30a+ b= -30+ b= 0 so b= 30. Then 300a- 20b+ c= 300- 600+ c= -100 so c= 200. Finally, -1000a+ 100b- 10c+ d= -1000+ 3000- 2000+ d= 3 and d= 3. That tells us that
\(\displaystyle \frac{n^3- 100n+ 3}{n- 10}= \frac{(n- 10)^3+ 30(n- 10)^2+ 200(n- 10)+ 3}{n- 10}= (n- 10)^2+ 30(n- 10)+ 200+ \frac{3}{n-10}\)
\(\displaystyle \frac{n^3- 100n+ 3}{n- 10}= n^2- 20n+ 100+ 30n- 300+ 200+ \frac{3}{n-10}= n^2+ 10n+ \frac{3}{n- 10}\)
exactly what soroban got.

(I admit his method is far simpler!)

 

[mmm4444bot]

(n^3-100n+3)/(n-10)

Interesting -- you put grouping symbols around the numerator, but left them off the denominator.

We need to put grouping symbols around polynomial denominators, too. :cool:

 

[lookagain]

I always work this type of problem with the method shown in post #2,

however notice that


\(\displaystyle \dfrac{n^3 - 100n + 3}{n - 10} \ = \)


\(\displaystyle \dfrac{n(n^2 - 100)}{n - 10} \ + \ \dfrac{3}{n - 10} \ = \)


\(\displaystyle \dfrac{n(n - 10)(n + 10)}{n - 10} \ + \ \dfrac{3}{n - 10} \ = \)


\(\displaystyle n(n + 10) \ + \ \dfrac{3}{n - 10} \ = \)


\(\displaystyle \boxed{n^2 + 10n \ + \ \dfrac{3}{n - 10}}\)