# Determine the reminder of P(x) division to Q(x), if P(x) = x^2002 + 2x^2000 - 5x^5 -10x^2 +2 and Q(x) = (x-1)(x+1)^2

neela | High School Teacher | (Level 3) Valedictorian

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We  assume P(x)/Q(x) = quotient h(x) + A/(x-1)+Bx+C/(x+1)^2.............(1), where P(x) =  x^2002 + 2x^2000 - 5x^5 -10x^2 +2 and Q(x) = (x-1)(x+1)^2.

Now multiply both sides by Q(x) and we get:

P(x) = (x-1)(x+1)^2h(x)+A(x+1)^2+(Bx+C) (x-1),

Putting x=1 in eq (1),

1+2-5-10+2 =-10 = 0+A(1+1)^2+0. Therefore A = -10/4 = -2. 5.

Putting x=-1 in eq(2),

(1)^2002+2(-1)^2000 -5 (-1)^5 10(-1)^2 + 5 = 1+2+5-10+2 =  0+0+B(-1)+C

0 = -B+C. Or so x = -1 is a root.

B=C.

Therefore A(x+1)^2+ B(x-1) = -2.5(x+1)+B(x-1) is remainder and this remainder should vanish at x=-1, So -2.5*0+B(-1-1)  = 0 Or B = 0

So the remainder is -2.5(1+x)^2.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Due to the rule of division with reminder:

P(x) = Q(x)*C(x) + R(x), where the degree of the polynomial

R(x) < the degree of the polynomialQ(x). We've noticed that the degree of Q(x) is third degree, so the biggest degree of  R(x) is 2.

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

The root of Q(x) are x1 = 1, x2 = x3 = - 1.

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

P(1) = 1 + 2 - 5 - 10 + 2

But P(x) = Q(x)*C(x) + R(x), P(1) = 0*C(x) + a + b + c = a + b + c

a + b + c = -10

P(-1) = a - b + c

P(-1) = 1 + 2 + 5 - 10 + 2 = 0

a - b + c = 0, b = a + c

But, we've noticed that x = -1 is a multiple root, so this one has to verify the first derivative ,too.

P'(x) = 2002x^2001 + 4000x^1999 - 25x^4 - 20x = -2a + b

P'(-1) = -2002 - 4000 - 25 + 20 = - 6007

-2a + b + a - b + c = -6007 + 0

-a + c + a + b + c = -10 - 6007

2c + b = - 6017, b = -6017 - 2c, b = a + c

a + c = -6017 - 2c,

a = - 6017 - 3c

a + b + c = - 10, - 6017 - 3c - 6017 - 2c + c =  - 10

- 4c = 12034 - 10

c = - 12024/4, c = - 3006

b = - 6017 + 6012, b = - 5

a = - 6017  + 9018, a = 3001

R(x) = 3001x^2 - 5x - 3006