How do you use synthetic division to divide $(x^5-13x^4-120x+80)/(x+3)$ ?

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Answer 1 · 2017-08-29T13:10:40

D= $x^5-13x^4 - 120 x+80$
d= $x+3$
$x+3=0 rArr x=-3$
Then create the following table,
enter image source here

-856 is the remainder and rest is the quotient,
$x^4-16x^3+48x^2-144x +312 - 856/(x+3)$

How do you use synthetic division to divide (x^5-13x^4-120x+80)/(x+3)(x^5-13x^4-120x+80)/(x+3)?