Rewrite the expression 7x^4+x+14/x+2 in the form q(x)+r(x)/b(x)

Question

asked Dec 19, 2020 225k views

1 Answer

Paul Morie · Answer 1 · 2020-12-25T19:34:11+0000

$7x^4=7x^3\cdot x$ , and
7x^3(x+2)=7x^4+14x^3 . Subtract this from the numerator and you get a remainder of

(7x^4+x+14)-(7x^4+14x^3)=-14x^3+x+14

$-14x^3=-14x^2\cdot x$ , and
-14x^2(x+2)=-14x^3-27x^2 . Subtract this from the previous remainder and you get a new remainder of

(-14x^3+x+14)-(-14x^3-27x^2)=27x^2+x+14

$27x^2=27x\cdot x$ , and
27x(x+2)=27x^2+54x . Subtract this from the previous remainder and you get a new one of

(27x^2+x+14)-(27x^2+54x)=-53x+14

$-53x=-53\cdot x$ , and
-53(x+2)=-53x-106 . Subtract this from the previous remainder and you get a new one of

(-53x+14)-(-53x-106)=120

doesn't divide 120, so we're done, and putting everything together we've shown that

(7x^4+x+14)/(x+2)=7x^3-(14x^3-x-14)/(x+2)

(7x^4+x+14)/(x+2)=7x^3-14x^2+(27x^2+x+14)/(x+2)

(7x^4+x+14)/(x+2)=7x^3-14x^2+27x-(53x+14)/(x+2)

$(7x^4+x+14)/(x+2)=\underbrace{7x^3-14x^2+27x-53}_(q(x))+\underbrace{(120)/(x+2)}_(r(x)/b(x))$