determine the reminder of P/Q if P = x^2002 + 2x^2000 - 5x^5 -10x^2 + 2 and Q = (x-1)*x^2 .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write 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). Because the degree of Q(x) is 3, so the degree of  R(x) is 2.

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

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

(x-1)*x^2 = 0

We'll set each factor as 0:

x-1 = 0

x1 = 1

x^2 = 0

x2=x3 = 0

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

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 (1)

P(0) = c

P(0)=2

c=2

If x=0 is a multiple root, this one has to verify the first derivative ,too.

P'(0) = 0

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

-2a+b = 0 (2)

But from (1) and c=2=>a+b+c=-10=>a+b+2= -10=>a+b=-12 (3)

We'll subtract (2) from (3):

a+b+2a-b = -12-0

We'll eliminate like terms:

3a = -12

We'll divide by 3:

a = 4

We'lll substitute a=4 in (2):

-2*4+b = 0

b = 8

R(x)=4x^2+8x+2

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

P =x^2002+2x^200 -5x^5 -10x^2+2 . Q = (x-1).x^2.

To determine the remainder in P/Q.

Solution.

LetP/Q = k(x) the quotent and remainder  ax^2+bx+c., one degree less than Q.

So

x^2002+2x^2000-5x^5-10x^2+2 = (x-1)x^2*k(x)+ax^2+bx+c....(1)

To determine a,b,c:

Put x= 1 in (1):

1+2-5-10+2 =  0 + a+b+c, 

a+b+c = -10.............(2)

Put x = 0 in (1):

2 =   0 + c . Or

c = 2..........................(3).

Eq (1) could be written as

x^2002+2x^2000-5x^5+10x^2+2 - (ax^2+bx+c) = (x-1)x^2k(x).

Differentiating both sides:

2002x^2001+2*2000x^1999-25x^4-20x - 2ax-b = x^2*{(x-1)k(x)}' +2x (x-1)k(x).............(4)

Put x= 0 in (4):

0 +b =  0

b =  0.

So from  (2): a+b+c = -10  . b= 0 and c =2

So a+0+2 = -10.

a = -10-2 = -12.

Therefore the remainder  = ax^2+bx+c = -12x^2+2.

 

 

 

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