How do you find the real solutions of the polynomial #x^3+x^2=(11x+10)/3#?

1 Answer
Jul 20, 2017

Answer:

The three solutions are:

# x=-2; \ \ \ 1/2 -1/2sqrt(23/3); \ \ \ 1/2 +1/2sqrt(23/3) #

Explanation:

We have:

# x^3+x^2=(11x+10)/3 #

Multiplying out the fraction we get

# 3x^3 + 3x^2=11x+10 #

# :. 3x^3 + 3x^2-11x-10 = 0 #

Define, #f(x)# by:

# f(x) = 3x^3 + 3x^2-11x-10 #

As there is no trivial method for factorising a cubic, and assuming that we are not looking for an approximate solution of #f(x)=0# via Numerical Methods we attempt to find a trivial solution by trial and error. We note that:

# f(-2)=0=>x=-2 # is a solution of #f(x)=0#

Therefore by the factor theorem, then if #x=-2# is a solution of #f(x)=-0# then #(x+2)# is a factor of #f(x)#, thus we can write:

# f(x) = (x+2)Q(x) #

Where #Q(x)# is a quadratic we can find via algebraic long division, or inspection, to get:

# f(x) = (x+2)(3x^2-3x-5) #

And to find the other roots we use the quadratic formula to get:

# x=1/2 +-1/2sqrt(23/3) #

Hence the three solutions are:

# x=-2; \ \ \ 1/2 -1/2sqrt(23/3); \ \ \ 1/2 +1/2sqrt(23/3) #