A cubic #3x^3-2x^2-kx+2 # has a root #x=1#. Find the value of #k# and all three roots?

Question

457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-06T11:25:32

# k=3 #

Roots are #x=2/3,+-1#

Define the function #f(x)#, by:

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

We are given that #x=1# is a solution of the equation:

# :. f(1) = 0 #
# :. 3-2-k+2 = 0 #
# :. k=3 #

Therefore, we can now write:

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

By the factor theorem, knowing that #x=1# is a root of the equation #f(x)=0# implies that #(x-1)# is a factor of #f(x)#, so we can write:

# f(x) = (x-1)(Ax^2+Bx+C) #, say

Where, #A,B,C# are constants to be determined. We can compare coefficients:

#Coff(x^3) : A=3 #
#Coff(x^2) : B-A = -2 => B=1 #
#Coff(x^1) : C-B = -3 => C=-2 #
#Coff(x^0) : C=-2 #

Thus:

# f(x) = (x-1)(3x^2+x-2) #

And we can factorise this quadratic, giving

# f(x) = (x-1)(3x-2)(x+1) #

Hence the roots are #x=2/3,+-1#