Question

Solve the polynomial function. Find all of the zeros; there are five zeros. Here is the equation:

X^5 - 8x^4 + 28x^3 - 56x^2 +64x - 32
Use synthetic division/quadratic equation :) Solve the polynomial function. Find all of the zeros; there are five zeros. Here is the equation:

X^5 - 8x^4 + 28x^3 - 56x^2 +64x - 32
Use synthetic division/quadratic equation :) @Mathematics

Answer 1

http://openstudy.com/study#/updates/4eb2ce8ee4b095a5c82171f6

Answer 2

The rational roots are +/- divisors of 32.
I will give you hint that 2 is a root. Divide your polynomial by x-1 to get
\[
\frac{x^5-8 x^4+28 x^3-56
x^2+64 x-32}{x-2}=x^4-6
x^3+16 x^2-24 x+16
\]
Try to find some roots of
\[x^4-6
x^3+16 x^2-24 x+16
\]

Answer 3

Also x=2 is a root of
\[x^4-6
x^3+16 x^2-24 x+16=0
\]
then
\[
\frac{x^4-6 x^3+16 x^2-24
x+16}{x-2}=x^3-4 x^2+8 x-8
\]

Answer 4

Aagain x=2 is a root of
\[x^3-4 x^2+8 x-8=0
\]
So
\[
\frac{x^5-8 x^4+28 x^3-56
x^2+64 x-32}{(x-2)^3}=x^2-2
x+4\\
x^5-8 x^4+28 x^3-56 x^2+64
x-32=(x-2)^3 \left(x^2-2
x+4\right)
\]
Use the the quadratic formula for
\[x^2-2
x+4=0
\]

Answer 5

You get
\[\begin{array}{c}
x=1-i \sqrt{3} \\
x=1+i \sqrt{3}
\end{array}
\]
in addtion you have 2 as a triple root