# Determine the remainder when P(x)=x^5-2x^3+3x-3 is divided by x^3-x using the Remainder Theorem.Do not use long division.how to determine the remainder without using the long division.

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### 2 Answers

P(x) = x^5-2x^3+3x-3.

To find the remainder when divided by x^3-x.

x^3-x = x(x^2-1) = (x)(x-1)(x+1).

So we can pressume that the remainder is of the form ax^2+bx+c a second degree expression ,1degree less than x^3-x. and the quotient is Q(x).

Therefore ,

P(x)=x^5-2x^3+3x-3 = x(x+1)(x)(x-1)Q(x) + ax^2+bx+c. ..(1)

Put x= 0 in both sides in eq (1):

Put x = -1:

(-1)^5 - 2(-1)^3 +3(-3) -3 = (-1+1)(-1)(-1-1)Q(-1) +a(-1)^2+b(-1)+c

-5+2-3-3 = 0+a-b+c.

-9 = a-b+c

a-b+c = -9.......................(2)

Put x= 0

-3 = 0 +0+0+c. So c = -3.

put x= 1:

1-2+3-3 = 0+a+b+c

-1 = a+b+c

a+b+c = -1.............(3)

a-b+c = - 9.............(2)

Adding (1) and(2) , we get: 2a+2c = -10. So a = -10-2c = -10 -2(-3) = -4. So a = -4/2 = -2.

Puttin a= -2 and c = -3 in (3), we get: -2+b-3 = -1. So b = -1+5 = 4..

Therefore a=-2, b = 4 and c=-3.

Therefore the assumed remainder ax^2+bx+c is **-2x^2+4x-3.**

We'll use the reminder theorem and we'll write:

P(x) = (x^3-x)*Q(x) + R(x)

According to the rule, since the divisor is a polynomial of 3rd order, the reminder is a polynomial of 2nd order.

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

Q(x) = cx^2 + dx + e

We'll calculate (x^3-x)*Q(x):

(x^3-x)*Q(x) = (x^3-x)*(cx^2 + dx + e)

(x^3-x)*Q(x) = cx^5 + dx^4 + ex^3 - cx^3 - dx^2 - ex

(x^3-x)*Q(x) + R(x) = cx^5 + dx^4 + x^3(e-c) - dx^2 - ex + ax^2 + bx + f

We'll combine like terms:

(x^3-x)*Q(x) + R(x) = cx^5 + dx^4 + x^3(e-c) + x^2(a-d) + x(b-e) + f

P(x) = (x^3-x)*Q(x) + R(x) if and only if the correspondent coefficients are equal:

x^5-2x^3+3x-3 = cx^5 + dx^4 + x^3(e-c) + x^2(a-d) + x(b-e) + c

**c = 1**

**d = 0**

e-c = -2 => e-1 = -2

**e = -1**

a-d = 0 => **a = 0**

b-e = 3 => b + 1 = 3 => **b = 2**

**f = -3**

**The reminder is:**

**R(x) = ax^2 + bx + f**

**We'll substitute a,b,f in the expression of the reminder:**

**R(x) = 2x - 3**