How do you find all the zeros of #4x^3-4x^2-9x+9# with 1 as a zero?

1 Answer
May 26, 2016

Answer:

The 3 roots are #x=-3/2, 1, 3/2#

Note I can't find the long division symbol so I will use the square root symbol in it's place.

Explanation:

#f(x)=4x^3-4x^2-9x+9#

#f(1)=4*1^3-4*1^2-9*1+9=4-4-9+9=0#

This means that x=1 is a root and #(x-1)# is a factor of this polynomial.

We need to find the other factors, we do this by dividing f(x) by #(x-1)# to find other factors.

#{4x^3-4x^2-9x+9}/{x-1}#

#(x-1)sqrt(4x^3-4x^2-9x+9)#

Since #(x*4x^2)=4x^3# we get #4x^2# as a term in the factor

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2#
#(x-1)sqrt(4x^3-4x^2-9x+9)#

we need to find the remainder to find what else need to be found.
we do #4x^2*(x-1)=4x^3-4x^2#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2#
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^3-4x^2#

We subtract this to get 0

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2#
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # #ul {- (4x^3-4x^2)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # ## # # # #0x^3-0x^2#

this zero means that there is NO linear term we and bring down the next terms.

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2 + 0 x #
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # #ul {- (4x^3-4x^2)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #0x^3-0x^2# #-9x+9#

#x*-9=-9x# so the next term is #-9#

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2 + 0 x -9 #
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # #ul {- (4x^3-4x^2)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #0x^3-0x^2# #-9x+9#

#-9*(x-1)=-9x + 9 #

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2 + 0 x -9 #
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # #ul {- (4x^3-4x^2)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #0x^3-0x^2# #-9x+9#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #ul{-(-9x + 9)}#

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #4x^2 + 0 x -9 #
#(x-1)sqrt(4x^3-4x^2-9x+9)#
# # # # # # # # # # # # # # # # # # # # # # # # #ul {- (4x^3-4x^2)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #0x^3-0x^2# #-9x+9#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #ul{-(-9x + 9)}#
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #0#

So #(x-1)(4x^2 -9)# with no reminder (you can check this by doing the expansion). We need to factor this completely.

#(4x^2 -9)# is a difference of squares with factors #(2x-3)*(2x+3)#

We have #(x-1) * (2x-3) * (2x+3)=0#

#2x-3=0# gives us a root at #x=3/2# and #2x+3=0# gives us a root at #x=-3/2#

The roots are #x=-3/2, 1, 3/2#