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

Question

Caleb Alexander · Accepted 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.

#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#

Caleb Alexander · Answer

undefined To find the other zeros of the polynomial $4x^3 - 4x^2 - 9x + 9$ if 1 is a zero, we can use polynomial division or synthetic division to divide the polynomial by $x - 1$. After dividing, we'll obtain a quadratic polynomial, which we can then solve for its roots.

Performing synthetic division or polynomial division, we get:

$$
(4x^3 - 4x^2 - 9x + 9) \div (x - 1) = 4x^2 + x - 9
$$

Now, we solve $4x^2 + x - 9 = 0$ for its roots. You can use factoring, completing the square, or the quadratic formula to solve this quadratic equation.

Using the quadratic formula, $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 4$, $b = 1$, and $c = -9$, we get:

$$
x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 4 \cdot (-9)}}}}{{2 \cdot 4}} = \frac{{-1 \pm \sqrt{{1 + 144}}}}{{8}}
$$

This simplifies to:

$$
x = \frac{{-1 \pm \sqrt{{145}}}}{8}
$$

So, the zeros of the polynomial $4x^3 - 4x^2 - 9x + 9$ are $x = 1$ and $x = \frac{{-1 \pm \sqrt{{145}}}}{8}$.