Assuming you meant that - sign at the front,
-x^3+2x^2-15x+30
= -x^2(x-2) -15(x-2)
= -(x-2)(x^2+15)
If you did mean x^3+2x^2-15x+30
then there are no rational roots, and you have to fall back on numerical methods, like bisection or Horner's method.
Find all roots
f(x)- x^3+2x^2-15x+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=1.4E-11
Still can't find zeros...
7 answers
sorry it was f(x)= not (-)
I got no real roots either
I got no real roots either
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.
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.
Can you show me the steps or is this just on calculator?
if it is x^3 + 2 x^2 - 15 x + 20 = 0
then:
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
gives
x = -5.44
x = 1.72 +/- .853 i
then:
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
gives
x = -5.44
x = 1.72 +/- .853 i
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.
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.
thx