what it t+he remainder and quotient of:?

Question

x^6+x^5+x^2+x+2/
x^3+x^2+1

Answer 1

Do it like long division.

Take the term with the highest degree of x in the numerator (x^6), and divide by the term with the highest degree of x in the denominator (x^3):

x^6 / x^3 = x^3

That's the first part of the quotient. Multiply the denominator by that part (x^3) and subtract from the numerator:

x^6 + x^5 + x^2 + x + 2 - (x^3)(x^3 + x^2 + 1)
x^6 + x^5 + x^2 + x + 2 - x^6 - x^5 - x^3
-x^3 + x^2 + x + 2

Take the term with the highest degree of x in what is left of the numerator (-x^3), and divide by the term with the highest degree of x in the denominator (x^3):

-x^3 / x^3 = -1

That's the second part of the quotient. Multiply the denominator by that part (-1) and subtract from what is left of the numerator:

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

At this point, the degree of x in the denominator (x^3) is higher than the degree of x in what is left of the numerator (x^2), so we are done.

The quotient is x^3 - 1 (the sum of the two parts calculated above).

The remainder is what is left of the numerator after the final subtraction -- it is the part that has too small a degree of x to be divided into the denominator any further: 2x^2 + x + 3

Answer 2

Because you have 2 contradictory answers I thought I would help you to sort it out! McFate is actually correct: the quotient is x^3 - 1 and the remainder is 2x^2 + x + 3
To see that this is correct, expand (x^3-1)*(x^3+x^2+1) and add it to 2x^2 + x + 3.... you get the thing you started with. ie x^6 + x^5 + x^2 + x + 2.

Not sure how holdm went wrong!

Hope this helps

Answer 3

to save typing, V = dividend, D = divisor
start with the highest order terms x^6/x^3 = x^3
x^3 D = x^6+x^5+x^3
V - x^3D = -x^3+x^2+x+2
-x^3/x^3=-1

quotient x^3-1
remainder x+1