Pre calculus
posted by Steve .
Find all roots
f(x) x^3+2x^215x+30
Factor f(x) completely.
I can't find any roots and always get a remainder with 1, 1, 2, 2
What do you do if you cannot find a root??
Using Calc button on ti84 I got
X= 5.619 Y=1.4E11
Still can't find zeros...

Pre calculus 
Steve
Assuming you meant that  sign at the front,
x^3+2x^215x+30
= x^2(x2) 15(x2)
= (x2)(x^2+15)
If you did mean x^3+2x^215x+30
then there are no rational roots, and you have to fall back on numerical methods, like bisection or Horner's method. 
Pre calculus 
Steve
sorry it was f(x)= not ()
I got no real roots either 
Pre calculus 
Steve 2.0
well, there is always at least one real root for a cubic. Just not always rational.
Since
f(6) = 24
f(5) = 30
you know there is a root inside (6,5). The work comes in narrowing it down a bit more. 
Pre calculus 
Steve
Can you show me the steps or is this just on calculator?

Pre calculus 
Damon
if it is x^3 + 2 x^2  15 x + 20 = 0
then:
http://www.mathportal.org/calculators/polynomialssolvers/polynomialrootscalculator.php
gives
x = 5.44
x = 1.72 +/ .853 i 
Pre calculus 
Steve 2.0
well, there are lots of methods. An easy one is bisection. Since you know the root is between 6 and 5, try 5.5
f(5.5) = 6.625
Now you know that the root r is in (6,5.5), so try 5.75
f(5.75) = 7.73
r is in (5.75,5.5)
f(5.625) = 0.322
each time take the midpoint of the interval where f(x) changes sign. It may be slow, but it always works. 
Pre calculus 
Steve
thx
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