Answer:
Rationals zeros 1, -2, 3/4
Irationals zeros [tex] \frac{2-\sqrt 2}{2},\frac{2+\sqrt 2}{2}[/tex]
Step-by-step explanation:
[tex] P(x)=8x^5-14x^4-22x^3+57x^2-38x+6[/tex]
The factors of 6 are: -1,1,-2,2,-3,3,-6,6.
The cactors of 8 are: -1,1,-2,2,-4,4,-8,8.
The possible rationals zeros for P(x) are: -1,1,1/2,-1/2,-1/4,1/4,-1/8,1/8,-2,2,-3,3,-3/2,3/2,-3/4,3/4,-3/8,3/8,-6,6.
For all possible rational zeros for P(x), we check is it P(rat.zero)=0?
You can see that [tex] P(1)=8-14-22+57-38+6=0[/tex]
Now we have that [tex] P(x)=(x-1)(8x^4-6x^3-28x^2+29x-6)[/tex]
In the same way we can see:
[tex] P(-2)=-3(128+48-112-58-6)=-3*0=0[/tex]
[tex] P(x)=(x-1)(x+2)(8x^3-22x^2+16x-3)[/tex]
[tex] P(3/4)=\frac{-1}{4}\frac{11}{4}(8*\frac{27}{64}-22*\frac{9}{16}+16\frac{3}{4}-3)=[/tex]
[tex]=\frac{-1}{4}\frac{11}{4}(\frac{216}{64}-\frac{792}{64}+\frac{768}{64}-\frac{192}{64})=0[/tex]
[tex] P(x)=(x-1)(x+2)(x-\frac{3}{4})(2x^2-4x+1)[/tex]
Apply the quadric formula:
[tex]x_{1,2}=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
we get [tex]x_1=\frac{2-\sqrt 2}{2}[/tex] and [tex]x_2=\frac{2+\sqrt 2}{2}[/tex].
[tex]P(x)=(x-1)(x+2)(x-\frac{3}{4})(x- \frac{2+\sqrt 2}{2})(x-\frac{2-\sqrt 2}{2})[/tex]
