How do you use the rational roots theorem to find all possible zeros of #f(x) = -2x^3 + 3x^2 -4x + 8#?
1 Answer
It has one positive Real zero and a Complex conjugate pair of non-Real zeros.
Explanation:
#f(x) = -2x^3+3x^2-4x+8#
By the rational root theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/2, +-1, +-2, +-4, +-8#
Note that
So the only possible rational zeros are:
#1/2, 1, 2, 4, 8#
When
When
When
So that only leaves as a possible rational zero:
#1/2#
We find:
#f(1/2) = -2(1/8)+3(1/4)-4(1/2)+8#
#= (-1+3-8+32)/4#
#= 13/2#
So
Bonus
What can we find out about the zeros of
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 144-512-864-6912+3456=-4688#
Since
Let's simplify
#0 = -4f(x) = 8x^3-12x^2+16x-32#
#=(2x-1)^3+5(2x-1)-26#
#=t^3+5t-26#
where
Use Cardano's method to find the roots:
Let
Then we want to solve:
#u^3+v^3+(3uv+5)(u+v)-26 = 0#
Add the constraint
#u^3-125/(27u^3)-26 = 0#
Multiply through by
#27(u^3)^2-702(u^3)-125 = 0#
Use the quadratic formula to find:
#u^3 = (702+-sqrt(702^2+4(27)(125)))/(2*27)#
#= (702+-sqrt(492804+13500))/54#
#= (702+-sqrt(506304))/54#
#= (702+-24sqrt(879))/54#
#= (351+-12sqrt(879))/27#
Since
#t_1 = 1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879)))#
and related Complex roots:
#t_2 = 1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879)))#
#t_3 = 1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879)))#
where
Then
#x_1 = 1/2(1+1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879))))#
and related Complex zeros:
#x_2 = 1/2(1+1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879))))#
#x_3 = 1/2(1+1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879))))#