How do you find the zeroes of p(x)= x^3+2x^2-9x-8p(x)=x3+2x2−9x−8?
2 Answers
Try rational root theorem, but find there are no rational roots.
Use general cubic formula.
Explanation:
By the rational root theorem any rational roots of
We find:
p(-4) = -4p(−4)=−4
p(-2) = 10p(−2)=10
p(-1) = 2p(−1)=2
p(1) = -14p(1)=−14
p(2) = -10p(2)=−10
p(4) = 52p(4)=52
So the three Real roots of
Here's an algebraic solution using the general cubic formula:
This has resolvents
Delta_0 = b^2-3ac = 2^2-(3*1*-9) = 4+27 = 31
Delta_1 = 2b^3-9abc+27a^2d = (2*2^3)-(9*1*2*-9)+(27*1^2*-8) = 16+162-216 = -38
C = root(3)((Delta_1+sqrt(Delta_1^2-4 Delta_0^3))/2)
=root(3)((-38+sqrt(38^2-4*31^3))/2)
=root(3)(-19+-sqrt(19^2-31^3))
=root(3)(-19+-sqrt(361-29791))
=root(3)(-19+-sqrt(-29430))
=root(3)(-19+-3sqrt(3270) i)
Then the roots are:
x_n = -1/(3a)(b+omega^n C+Delta_0/(omega^n C))
= -1/3(2+omega^n root(3)(-19+-3sqrt(3270) i)+31/(omega^n root(3)(-19+-3sqrt(3270) i)))
where
Find the roots numerically using Newton's method.
x_1 ~~ -3.810821
x_2 ~~ -0.803113
x_3 ~~ 2.613934
Explanation:
When seeking a rational root we find:
p(-4) = -4
p(-2) = 10
p(-1) = 2
p(1) = -14
p(2) = -10
p(4) = 52
Hence
p(x) = x^3+2x^2-9x-8
p'(x) = 3x^2+4x-9
Newton's method starts with an approximation
a_(i+1) = a_i - (p(a_i))/(p'(a_i))
=a_i - (a_i^3+2a_i^2-9a_i-8)/(3a_i^2+4a_i-9)
To find different roots, start with different approximations.
Putting this formula into a spreadsheet and using initial approximations
a_0 = -4
a_1 = -3.826087
a_2 = -3.810933
a_3 = -3.810821
a_0 = -1
a_1 = -0.8
a_2 = -0.803113
a_0 = 2
a_1 = 2.909091
a_2 = 2.646363
a_3 = 2.613934
