# Which is the reminder of the division f/g if f=x^5+3x^3-2x^2+x+1 and g=x^3-x ?

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First , we'll write the rule of division with reminder:

f(x)=g(x)*C(x)+R(x), where the degree of the polynomial

R(x) < the degree of the polynomial g(x).

Because the degree of g(x) is 3, the degree of R(x) will be 2.

R(x)=ax^2+bx+c

We'll calculate the root of g(x):

x^3-x = 0

We'll factorize by x:

x(x^2-1) = 0

x(x-1)(x+1) = 0

We'll set each factor as 0:

**x1 = 0**

x-1 = 0

**x2 = 1**

x+1 = 0

**x3 = -1**

The roots of g(x) are: x1=0, x2=1, x3=-1.

If we'll substitute x=1 in f(x), we'll obtain:

f(1)= 1^5 + 3*1^3 - 2*1^2 + 1 + 1

f(1) = 1 + 3 - 2 + 1 + 1

f(1) = 4

But f(x)=g(x)*C(x)+R(x),

f(1)=0*C(1) +a+b+c = a+b+c

**a+b+c=4 (1)**

Now, we'll substitute x=0 in f(x) and we'll obtain:

f(0) = c

f(0)=1

**c=1**

We'll substitute x by -1:

f(-1) = -1-3-2-1+1

f(-1) = -6

But

f(-1) = 0*C(-1) + (a-b+c)

f(-1) = (a-b+c)

**(a-b+c) = -6 (2)**

We'll substitute c = 1 in (1) and (2):

a+b+1 = 4

a+b = 3 (3)

a-b+1 = -6

a-b = -7 (4)

We'll add (3) and (4):

a+b+a-b = 3-7

2a = -4

We'll divide by 2:

**a = -2**

We'll sbstitute a = -2 in (3):

a+b = 3

-2 + b = 3

**b = 5**

Thereminder R(x) is:

**R(x) = -2x^2 + 5x + 1**

f = x^5+3x^3+2x^2+x+1. g = x^3-x.

Threfore f/g = k(x)*g(x) + ax^2+bx+c. where k(x) is the coefficient and ax^2+bx+c is the rmainder.

Algo g(x)* x^3-x = (x(x^2-1) = x(x-1)(x+1).

x^5 +3x^3+2x^2+x+1 = k(x)*x(x-1)(x+1) +ax^2+bx+c....(1)

Put x= 0 , x = -1 and x= 1 in (1):

0 + 1 = c . Or c =1.

x = -1:

-1-3+2-1+1 = a -b +1.

a-b = -3................(2)

Put x =1:

1+3+2+1+1 = a+b+1

a+b = 7........(3)

(2)+(3) gives: 2a = -3+7 =4 . a = 4/2 =2.

(2)-(3) gives: -2b = -3-7 = -10. Or b = -10/-2 = 5.

So a=2, b =5 and c =1

So the remainder = ax^2+bx+c = 2x^2+5x+1