# Determine the reminder of the division of polinomials P/Q P=(x+1)^10+(x+2)^10 Q=(x+1)(x+2)

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### 3 Answers

P = (x+1)^10 + (x+2)^10

Q = (x+1(x+2)

P = Q*G + R where R is the remainder

Now let R be ax+b

==> P = Q*G + ax +b

Let us substiute with Q;s zeros which are (-1 and -2)

P(-1) = Q(-1)*G(-1) + a(-1) + b

1 = 0 -a + b

==> -a+b = 1

==> b = a+1 ........(1)

P(-2) = Q(-2)*G(-2) + a(-2)+b

1 = -2a + b........(2)

==> Let us substitute with (1)

==> 1= -2a + a+1

==> -a= 0

==> a=0

==> b = a+1 = 1

Then :

R = ax + b = 0+1 =1

Then the remainder is R=1

We'll apply division with reminder.

P=Q1*Q+R

The degree of the the reminder has to be smaller than the degree of Q. In our case, Q has the second degree, so the reminder is a first degree polynomial.

R=aX+b

First, let's find out the roots of Q.

Q=(x+1)(x+2)

(x+1)(x+2) = 0

We'll put each factor as zero.

(x+1) = 0

We'll add -1 both sides:

x = -1

x+2 = 0

We'll subtract 2 both sides:

x = -2

Now, we'll substitute the roots of Q, into the expression of division with reminder.

P(-1) = Q(-1)*Q1+a*(-1)+b, where P(-1)=1

But Q(-1)=0, so:

1=0-a+b

-a+b=1 (1)

P(-2) = Q(-2)*Q2+a*(-2)+b, where P(-2)=1

1=0-2a+b

-2a+b=1 (2)

From (1) and (2), we'll have:

-a+b=-2a+b

We'll eliminate like terms:

-a=-2a

a=0, so 0+b=1, b=1

R=a*X+b

R=0*X+1

**R=1**

P= (1+x)^10+(x+2)^10 . Q = (x+1)(x+2)

To find the remainder of P/Q.

Let the remainder be ax+b as the remainder shoul be at leae one degree less than Q.

So Q = (x+1)^10 +(x+2)^10 = (x+1)(x+2) q(x)+ (ax+b)..........(1) where q(x) is the quotient.

Put x= -1 in (1):

(-1+1)^10 +(-1+2)^10 = 0*{...}+ a(-1)+b

0+1 = -a+b..............(2)

Put x=-2 in (1)

(2-1)^2+(-2+2) = 0*{..}+a(-2)+b

1 +0 = -2a+b.............(3)

(2)-(3) gives: 0 = a and from (1) 1=-a+b we get b =1

Therefore the remainder is ax+b = 1.