Answer:
5/3
Explanation:
-1/3 goes on the outside and since we have our polynomial in standard form with no in between terms missing, 3,-5,-1,2 go inside because they are the coefficients of our polynomial.
-1/3 | 3 -5 -1 2
|
-------------------------------------
First step bring the 3 down inside. (3+0=3)
-1/3 | 3 -5 -1 2
|
-------------------------------------
3
Whatever goes below the bar, must be multiplied by outside number and put directly below next number inside.
-1/3 | 3 -5 -1 2
| -1
-------------------------------------
3
The numbers lined up vertically are added to get the numbers underneath the bar.
-1/3 | 3 -5 -1 2
| -1
-------------------------------------
3 -6
Again any number below the bar gets multiply to the number outside.
-1/3 | 3 -5 -1 2
| -1 2
-------------------------------------
3 -6
Again the numbers lined up vertically above the bar get added to get the number that goes underneath the bar there.
-1/3 | 3 -5 -1 2
| -1 2
-------------------------------------
3 -6 1
Multiply to outside number 1(-1/3)=-1/3.
This goes under the 2 inside.
-1/3 | 3 -5 -1 2
| -1 2 -1/3
-------------------------------------
3 -6 1
The last number we are fixing to be put is the remainder of (3x^3-5x^2-x+2)/(x+1/3) or you could say it is the value of p(-1/3) since:
P(x)/(x-c)=Q(x)+R/(x-c)
Multiply both sides by (x-c):
P(x)=Q(x)(x-c)+R
If you evaluate P at x=c, we get R:
P(c)=Q(c)(c-c)+R
P(c)=Q(c)*0+R
P(c)=R.
Let's finish:
-1/3 | 3 -5 -1 2
| -1 2 -1/3
-------------------------------------
3 -6 1 5/3
This means p(-1/3)=5/3.
We could have also got this by directly plugging in (-1/3) for x into 3x^3-5x^2-x+2.