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What is the remainder when (3x^3-2x^2+4x-3) is divided by (x^2+3x+3)?

User Gasim
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1 Answer

2 votes

Answer:

The remainder is 28x+30

Explanation:

We will divide 3x^3-2x^2+4x-3 by x^2+3x+3.

To get the first term of the dividend we will multiply x^2+3x+3 by 3x

3x(x^2+3x+3)

=3x^3+9x^2+9x

It means

x^2+3x+3) 3x^3-2x^2+4x-3( 3x -11

3x^3+9x^2+9x

- - -

_________________

11x^2-5x-3 (Now multiply x^2+3x+3 by -11 to get the

11x^2-33x-33 first term)

- + +

_______________

28x+30

We can not solve it further because remainder has two terms and divisor has 3 terms. Therefore the remainder is 28x+30....

User AlexKempton
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