Question

State the quotient and remainder when the first polynomial is divided by the second. 3x^4-3x^3-11x^2+6x-1; x^3+x^2-2

Answer 1

`3x^4=3x\cdot x^3`, and
`3x(x^3+x^2-2)=3x^4+3x^3-6x`. Subtracting this from the numerator leaves a remainder of

`(3x^4-3x^3-11x^2+6x-1)-(3x^4+3x^3-6x)=-6x^3-11x^2+12x-1`

`-6x^3=-6\cdot x^3`, and
`-6(x^3+x^2-2)=-6x^3-6x^2+12`. Subtracting this from the previous remainder gives a new remainder of

`(-6x^3-11x^2+12x-1)-(-6x^3-6x^2+12)=-5x^2+12x-13`

`-5x^2` has no factors of
`x^3`, so we stop here. Then

`(3x^4-3x^3-11x^2+6x-1)/(x^3+x^2-2)=3x-(6x^3+5x^2-6x+1)/(x^3+x^2-2)`

`=3x-6-(5x^2-12x+13)/(x^3+x^2-2)`

Answer 2

Q = 3x – 6; R = -5x² + 12x - 13

Explanation:

One way is to use long division.

3x - 6

x³ + x² - 2) 3x⁴ - 3x³ - 11x² + 6x - 1

3x⁴ + 3x³ - 6x

- 6x³ - 11x² + 12x - 1

- 6x³ - 6x² + 12

- 5x² + 12x - 13

Quotient = 3x – 6; Remainder = -5x²+ 12x - 13