How do you find all the zeros of #f(x)=x^3 -8x^2 -23x +30#?

1 Answer
Jun 20, 2016

Answer:

They are #x=1, x=10, x=-3#.

Explanation:

To solve a cubic equation it exists the solving formula but it is very long and I do not know it.
Then I use a bit of luck to solve this equation. For example I see that #x=1# is a solution because

#1^3-8*1^2-23*2+30=1-8-23+30=0#.

Then I know that the equation has to be in the form

#(ax^2+bx+c)(x-1)#

To find #a, b, c# I just do the multiplication and compare the coefficients.

#(ax^2+bx+c)(x-1)#

#=ax^3+bx^2+cx-ax^2-bx-c#

#=ax^3+(b-a)x^2+(c-b)x-c#

Comparing this with our initial equation we see that

#a=1#
#b-a=-8, b-1=-8, b=-7#
#c-b=-23, c+7=-23, c=-30#
#-c=30, c=-30#

Our function is then

#(x^2-7x-30)(x-1)#.

Now we have to solve a second order equation #(x^2-7x-30)# using the solution

#x=(-b\pmsqrt(b^2-4ac))/(2a)#

#=(-(-7)\pmsqrt((-7)^2-4*1*(-30)))/2#

#=(7\pmsqrt(49+120))/2#

#=(7\pmsqrt(169))/2#

#=(7\pm13)/2#

that has the two solutions

#x=(7+13)/2=10# and #x=(7-13)/2=-3#.

Then the function

#f(x)=x^3-8x^2-23x+30#

can be rewritten as

#f(x)=(x-1)(x-10)(x+3)# with the three zeros for #x# equal to #1, 10, -3#.