Question

Determine the quotient q(x) and remainder r(x) when f(x)=18x^4+27x^3+39x^2+22x+11

Answer 1

$18x^4=3x^2\cdot6x^2$, and

$6x^2(3x^2+3x+4)=18x^4+18x^3+24x^2$

Subtracting this from $f(x)$ gives a remainder of

$(18x^4+27x^3+39x^2+22x+11)-(18x^4+18x^3+24x^2)=9x^3+15x^2+22x+11$

$9x^3=3x^2\cdot3x$, and

$3x(3x^2+3x+4)=9x^3+9x^2+12x$

Subtracting this from the previous remainder gives a new remainder of

$(9x^3+15x^2+22x+11)-(9x^3+9x^2+12x)=6x^2+10x+11$

$6x^2=3x^2\cdot2$, and

$2(3x^2+3x+4)=6x^2+6x+8$

Subtracting this from the previous remainder gives a new remainder of

$(6x^2+10x+11)-(6x^2+6x+8)=4x+3$

$4x$ is not divisible by $3x^2$, so we're done. We ended up with

$\dfrac{f(x)}{g(x)}=\underbrace{6x^2+3x+2}_{q(x)}+\underbrace{\dfrac{4x+3}{3x^2+3x+4}}_{r(x)}$