#2x^3+4x^2-13x+6# Can you factorise this please?
2 Answers
Explanation:
where:
Explanation:
Given:
#2x^3+4x^2-13x+6#
Note that this does factorise much more easily if there is a typo in the question.
For example:
#2x^3+4x^2-color(red)(12)x+6 = 2(x-1)(x^2+3x-6) = ...#
#2x^3+4x^2-13x+color(red)(7) = (x-1)(2x^2+6x-7) = ...#
If the cubic is correct in the given form, then we can find its zeros and factors as follows:
#f(x) = 2x^3+4x^2-13x+6#
Discriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 2704+17576-1536-3888-11232 = 3624#
Since
Tschirnhaus transformation
To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.
#0=108f(x)=216x^3+432x^2-1404x+648#
#=(6x+4)^3-282(6x+4)+1712#
#=t^3-282t+1712#
where
Trigonometric substitution
Since
Put:
#t = k cos theta#
where
Then:
#0 = t^3-282t+1712#
#color(white)(0) = k^3 cos^3 theta - 282k cos theta+1712#
#color(white)(0) = 94k(4 cos^3 theta - 3 cos theta)+1712#
#color(white)(0) = 94k cos 3 theta+1712#
So:
#cos 3 theta = -1712/(94 k) = -1712/(188 sqrt(94)) = -(1712sqrt(94))/(188*94) = -214/2209 sqrt(94)#
So:
#3 theta = +-cos^(-1)(-214/2209 sqrt(94))+2npi#
So:
#theta = +- 1/3cos^(-1)(-214/2209 sqrt(94))+(2npi)/3#
So:
#cos theta = cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3)#
Which givens
#t_n = k cos theta = 2sqrt(94) cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3)" "# for#n = 0, 1, 2#
Then:
#x = 1/6(t-4)#
So the three zeros of the given cubic are:
#x_n = 1/6(-4+2sqrt(94) cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3))#
with approximate values:
#x_0 ~~ 1.2643#
#x_1 ~~ -3.8764#
#x_2 ~~ 0.61211#