# If the polynomial f=x^4+x^2+1 is divided by g=x^2+2x+3 what is the reminder of division ?

### 4 Answers | Add Yours

We have to divide f = x^4 + x^2 + 1 by g=x^2 + 2x + 3.

x^2 + 2x + 3 | x^4 + x^2 + 1.................| x^2 -2x + 2

..................... x^4 + 2x^3 + 3x^2

------------------------------------------

................................2x^3 - 2x^2 + 1

................................2x^3 - 4x^2 - 6x

----------------------------------------------

.......................................2x^2 + 6x + 1

.......................................2x^2 + 4x + 6

------------------------------------------------

....................................................2x - 5

**Therefore we get the remainder as 2x - 5**

Given the polynomial f(x) = x^4 + x^2 + 1

Divided by g(x) = x^2 + 2x + 3.

We need to find the remainder.

First we will divide and determine the quotient.

==> f(x) = g(x)*P(x) + R where R is the remainder

==> f(x) = g(x) * (x^2 -2x +2) + (2x-5)

Then the quotient is ( x^2 - 2x +2)

**And the remainder is ( 2x -5).**

To find the remainder if f(x) = x^4+x^2+1 is divided by g(x) = x^2+2x+1.

We divide x^4+x^2+1 by x^2+2x+3.

x^2+2x+3) x^4+0*x^3 +x^2+0*x+1( x^2-2x+2

x^4 +2x^3 +3x^2

-------------------------------

x^2+2x+3)-2x^3 -2x^2 + 0*x+1(-2x

-2x^3 - 4x^2 -6x

--------------------------------

x^2+2x+3)2x^2+6x+1 (2

2x^2+4x+6

-------------------

2x -5 is the remainder.

Therefore f(x)/g(x) = (x^4+x^2+1)/ (x^2+2x+3) = x^2-2x+2 is the quotient and 2x-5 is the remainder.

We'll write the rule of division with reminder:

f = g*q + r

q is the quotient and r is the reminder.

By definition, the order of reminder has to be smaller than the order of g. Since g is of 2nd order, then the reminder is linear expression.

r = dx + e

x^4+x^2+1 = (x^2+2x+3)*(ax^2 + bx + c) + dx + e

We'll remove the brackets:

x^4+x^2+1 = ax^4 + **bx^3** + * cx^2 *+

**2ax^3**+

*+*

**2bx^2***2cx*+

*+*

**3ax^2***3bx*+ 3c +

*dx*+ e

We'll combine like terms from the right side:

x^4+x^2+1 = ax^4 + x^3(b + 2a) + x^2(c + 2b + 3a) + x(2c + 3b + d) + 3c + e

We'll compare and we'll get:

**a = 1**

b + 2a = 0

2a = -b

**b =-2**

c + 2b + 3a = 1

c - 4 + 3 = 1

c - 1 = 1

**c = 2**

2c + 3b + d = 0

d = -2c - 3b

d = -4 + 6

**d = 2**

3c + e = 1

e = 1 - 3c

e = 1 - 6

**e = -5**

**The reminder r = dx + e is: r = 2x - 5.**