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

$\dfrac{3x^4-3x^3-11x^2+6x-1}{x^3+x^2-2}=3x-\dfrac{6x^3+5x^2-6x+1}{x^3+x^2-2}$

$=3x-6-\dfrac{5x^2-12x+13}{x^3+x^2-2}$

Answer 2

Answer:

Q = 3x – 6; R = -5x² + 12x - 13

Step-by-step 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