Question

Find the zeros (roots) of the following equations.

Answer 1

Hello,

f(1)=2-9+12-12+10-3=0 ===>(x-1)*....
f(3)=2*3^5-9*3^4+12*3^3-12*3^2+10*3-3=0 ===> (x-3)*....
f(1/2)=2*(1/2)^5-9*(1/2)^4+12*(1/2)^3-12*(1/2)^2+10*1/2-3=0 ===>(2x-1)*...

f(x)=(x-1)(x-3)*(2x-1)*(x²+1)
In C, x=i ou x=-i

Zeros are 1/2,1,3,i,-i

Answer 2

Answer:

Therefore the roots of equation 1 are 1,3 1/2,-i,+i

Step-by-step explanation:

in this question we have given

$f(x) = 2x^5 - 9x^4 + 12x^3 - 12x^2 + 10x - 3 = 0$.......1

we have to find the roots of this equation

put x=1 in equation 1

$f(1)=2-9+12-12+10-3\\f(1)=0$

It means x=1 is root  of equation 1

now put x=3 in equation 1

$f(3)=2\times3^5-9\times3^4+12\times3^3-12\times3^2+10\times3-3\\f(3)=0$

therefore,x=3 is root of equation 1

Now put x=$\frac{1}{2}$ in equation 1

$f(\frac{1}{2})=2\times (\frac{1}{2} )^5-9\times(\frac{1}{2})^4+12\times(\frac{1}{2})^3-12*(\frac{1}{2})^2+10\times \frac{1}{2}-3\\f(\frac{1}{2})=0$

It means x=$\frac{1}{2}$ is also root of equation 1

therefore equation one can be written as

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

therefore remaining 2 roots can be calculated by solving

$x^2+1=0\\x=\sqrt(-1)\\x=i,-i$

Therefore the roots of equation 1 are 1,3 1/2,-i,+i