Question 191261
Wow, this is as long as it gets (well I hope so anyway...). Here are two ways to do this:


Method #1: 


Use the rational root theorem to find all of the <i>possible</i> roots and test EVERY possible root to see if it is actually a root. Since the degree of the equation is 7, this means that once you find 7 roots, then you don't need to check any more possible roots.


Here's where this method gets really tedious: there are A LOT of possible roots (48 in total)



But, here's how to find them:



Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let's list the factors of 120 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm8, \pm10, \pm12, \pm15, \pm20, \pm24, \pm30, \pm40, \pm60, \pm120]


Now let's list the factors of 4 (the first coefficient):


*[Tex \LARGE q=\pm1, \pm2, \pm4]


Now let's divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{1}{2}, \frac{1}{4}, \frac{2}{1}, \frac{2}{2}, \frac{2}{4}, \frac{3}{1}, \frac{3}{2}, \frac{3}{4}, \frac{4}{1}, \frac{4}{2}, \frac{4}{4}, \frac{5}{1}, \frac{5}{2}, \frac{5}{4}, \frac{6}{1}, \frac{6}{2}, \frac{6}{4}, \frac{8}{1}, \frac{8}{2}, \frac{8}{4}, \frac{10}{1}, \frac{10}{2}, \frac{10}{4}, \frac{12}{1}, \frac{12}{2}, \frac{12}{4}, \frac{15}{1}, \frac{15}{2}, \frac{15}{4}, \frac{20}{1}, \frac{20}{2}, \frac{20}{4}, \frac{24}{1}, \frac{24}{2}, \frac{24}{4}, \frac{30}{1}, \frac{30}{2}, \frac{30}{4}, \frac{40}{1}, \frac{40}{2}, \frac{40}{4}, \frac{60}{1}, \frac{60}{2}, \frac{60}{4}, \frac{120}{1}, \frac{120}{2}, \frac{120}{4}, -\frac{1}{1}, -\frac{1}{2}, -\frac{1}{4}, -\frac{2}{1}, -\frac{2}{2}, -\frac{2}{4}, -\frac{3}{1}, -\frac{3}{2}, -\frac{3}{4}, -\frac{4}{1}, -\frac{4}{2}, -\frac{4}{4}, -\frac{5}{1}, -\frac{5}{2}, -\frac{5}{4}, -\frac{6}{1}, -\frac{6}{2}, -\frac{6}{4}, -\frac{8}{1}, -\frac{8}{2}, -\frac{8}{4}, -\frac{10}{1}, -\frac{10}{2}, -\frac{10}{4}, -\frac{12}{1}, -\frac{12}{2}, -\frac{12}{4}, -\frac{15}{1}, -\frac{15}{2}, -\frac{15}{4}, -\frac{20}{1}, -\frac{20}{2}, -\frac{20}{4}, -\frac{24}{1}, -\frac{24}{2}, -\frac{24}{4}, -\frac{30}{1}, -\frac{30}{2}, -\frac{30}{4}, -\frac{40}{1}, -\frac{40}{2}, -\frac{40}{4}, -\frac{60}{1}, -\frac{60}{2}, -\frac{60}{4}, -\frac{120}{1}, -\frac{120}{2}, -\frac{120}{4}]







Now simplify


These are all the distinct rational zeros of the function that could occur


*[Tex \LARGE  1, \frac{1}{2}, \frac{1}{4}, 2, 3, \frac{3}{2}, \frac{3}{4}, 4, 5, \frac{5}{2}, \frac{5}{4}, 6, 8, 10, 12, 15, \frac{15}{2}, \frac{15}{4}, 20, 24, 30, 40, 60, 120, -1, -\frac{1}{2}, -\frac{1}{4}, -2, -3, -\frac{3}{2}, -\frac{3}{4}, -4, -5, -\frac{5}{2}, -\frac{5}{4}, -6, -8, -10, -12, -15, -\frac{15}{2}, -\frac{15}{4}, -20, -24, -30, -40, -60, -120]




Once you have all of the <i>possible</i> rational roots, either plug them in directly or use synthetic division to determine which ones are actually roots. 




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Method #2:



Since the first method seems like a lot of busy work (which it is), just use a graphing calculator to find some of the roots, and then use those roots to find other roots (using synthetic division). Eventually, you'll reduce that massive polynomial to a quadratic which you can solve using the quadratic formula.






If you're completely stuck, then repost or email me. By the way, the 7 roots are: 3, -1/2, 1/2, -4, -2, -2+i, -2-i



Also, {{{4x^7+28x^6+27x^5-203x^4-591x^3-431x^2+146x+120}}} completely factors (over the reals) to {{{(x-3)(2x+1)(2x-1)(x+4)(x+2)(x^2+4x+5)}}}



In other words, {{{4x^7+28x^6+27x^5-203x^4-591x^3-431x^2+146x+120=(x-3)(2x+1)(2x-1)(x+4)(x+2)(x^2+4x+5)}}}