Question 929777

Rounded to the nearest tenth, find the root of x^3-5x^2-12x+60=0 that lies between 3 and 4. Express your answer as a decimal.

Thanks for the help!
<pre><font face = "Tohoma" size = 4 color = "indigo"><b>Using the rational root theorem, it's discovered that 5 is a root of the equation.
Thus, x = 5, or x - 5 = 0. This means that x - 5 is a factor of the equation: {{{x^3 - 5x^2 - 12x + 60 = 0}}}.
When this equation is divided by the factor, x - 5, its QUOTIENT is: {{{x^2 - 12}}}. Thus, the factors of 
{{{x^3 - 5x^2 - 12x + 60 = 0}}} are: (x - 5) and {{{x^2 - 12}}}.

Factoring {{{x^2 - 12}}} further, we see that 2 other roots of the equation are: 3.464101615 and - 3.464101615. 

Thus, the root that lies between 3 and 4 is: 3.464101615 &#8776; {{{highlight_green(3.5)}}} (to the nearest tenth).
You can do the check!! 
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