[tex]7x^4=7x^3\cdot x[/tex], and [tex]7x^3(x+2)=7x^4+14x^3[/tex]. Subtract this from the numerator and you get a remainder of
[tex](7x^4+x+14)-(7x^4+14x^3)=-14x^3+x+14[/tex]
[tex]-14x^3=-14x^2\cdot x[/tex], and [tex]-14x^2(x+2)=-14x^3-27x^2[/tex]. Subtract this from the previous remainder and you get a new remainder of
[tex](-14x^3+x+14)-(-14x^3-27x^2)=27x^2+x+14[/tex]
[tex]27x^2=27x\cdot x[/tex], and [tex]27x(x+2)=27x^2+54x[/tex]. Subtract this from the previous remainder and you get a new one of
[tex](27x^2+x+14)-(27x^2+54x)=-53x+14[/tex]
[tex]-53x=-53\cdot x[/tex], and [tex]-53(x+2)=-53x-106[/tex]. Subtract this from the previous remainder and you get a new one of
[tex](-53x+14)-(-53x-106)=120[/tex]
[tex]x[/tex] doesn't divide 120, so we're done, and putting everything together we've shown that
[tex]\dfrac{7x^4+x+14}{x+2}=7x^3-\dfrac{14x^3-x-14}{x+2}[/tex]
[tex]\dfrac{7x^4+x+14}{x+2}=7x^3-14x^2+\dfrac{27x^2+x+14}{x+2}[/tex]
[tex]\dfrac{7x^4+x+14}{x+2}=7x^3-14x^2+27x-\dfrac{53x+14}{x+2}[/tex]
[tex]\dfrac{7x^4+x+14}{x+2}=\underbrace{7x^3-14x^2+27x-53}_{q(x)}+\underbrace{\dfrac{120}{x+2}}_{r(x)/b(x)}[/tex]
