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

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

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Answer 1 · 2017-08-29T13:10:40

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

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

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

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How do you use synthetic division to divide (x^5-13x^4-120x+80)/(x+3)x5−13x4−120x+80x+3(x^5-13x^4-120x+80)/(x+3)?

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