How do you define a cubic function with zeros #r_1, r_2, r_3# such that #r_1+r_2+r_3 = 3#, #r_1r_2+r_2r_3+r_3r_1 = -1#, #r_1r_2r_3 = -3# ?

1 Answer
Oct 26, 2016

Answer:

#f(x) = x^3-3x^2-x+3#

Explanation:

Suppose #f(x) = x^3+ax^2+bx+c# has zeros #r_1, r_2, r_3#

Then:

#f(x) = (x-r_1)(x-r_2)(x-r_3)#

#color(white)(f(x)) = x^3-(r_1+r_2+r_3)x^2+(r_1 r_2 + r_2 r_3 + r_3 r_1)x-r_1 r_2 r_3#

We are told that:

#{ (r_1+r_2+r_3 = 3), (r_1 r_2 + r_2 r_3 + r_3 r_1 = -1), (r_1 r_2 r_3 = -3) :}#

Hence:

#f(x) = x^3-3x^2-x+3#

In this particular example, we can check our result by factoring the cubic by grouping:

#x^3-3x^2-x+3 = (x^3-3x^2)-(x-3)#

#color(white)(x^3-3x^2-x+3) = x^2(x-3)-1(x-3)#

#color(white)(x^3-3x^2-x+3) = (x^2-1)(x-3)#

#color(white)(x^3-3x^2-x+3) = (x-1)(x+1)(x-3)#

So we can write:

#{ (r_1 = 1), (r_2 = -1), (r_3 = 3) :}#

and we find:

#{ (r_1+r_2+r_3 = 1-1+3 = 3), (r_1 r_2 + r_2 r_3 + r_3 r_1 =(1)(-1) + (-1)(3) + (3)(1) = -1), (r_1 r_2 r_3 = (1)(-1)(3) = -3) :}#