z^5-28z^3+27z=0 How do you use variable substitution and factoring to find all of the roots of this equation? I know how to do it if the z would have only been to the fourth power, but how do you do it with an odd numbered power like this? Thanks!!

Question

z^5-28z^3+27z=0 How do you use variable substitution and factoring to find all of the roots of this equation? I know how to do it if the z would have only been to the fourth power, but how do you do it with an odd numbered power like this? Thanks!!

Answer 1

First of all, one root is 0 and you can factor the equation to get

z(z^4 -28z^2 + 27) = 0
Looking at it, you can see that +1 and -1 are also roots. The term in parentheses is
(z^2-1)(z^2-27)
So that reduces it to
z(z-1)(z+1)(z^2-27)
The last term factors into
(z + 3sqrt3)(z - 3sqrt3)

Answer 2

To solve the equation \(z^5-28z^3+27z=0\), we can use a variable substitution and factoring. Here's how you can approach it:

Step 1: Substitute a new variable
Let's make a substitution: \(x=z^2\). By substituting, our equation becomes \(x^2-28x+27z=0\).

Step 2: Factor the equation
Now, let's factor the equation \(x^2-28x+27z=0\) and solve for \(x\). We can rewrite it as \(x^2-27x-x-27z=0\). Then, we can group the terms as follows: \((x^2-27x)-(x+27z)=0\).

Factoring out \(x\) from the first group, we get:
\(x(x-27)-(x+27z)=0\).

Now, we can factor out \((x-27)\) from both terms:
\((x-27)(x-(x+27z))=0\).

Simplifying further, we obtain:
\((x-27)(x-x-27z)=0\).

At this point, the equation has factored completely. We have:
\((x-27)(-27z)=0\).

Step 3: Solve for \(x\)
To find the values of \(x\), we set each factor equal to zero and solve for \(x\) independently.

For the first factor, \(x-27=0\), we get \(x=27\).

For the second factor, \(-27z=0\), we get \(z=0\).

Step 4: Substitute back to solve for \(z\)
Now that we have the values of \(x\), we substitute them back into our original substitution variable \(x=z^2\).

For \(x=27\), we have \(27=z^2\), which gives us \(z=\pm\sqrt{27}\).

For \(z=0\), there's no need for substitution since it's already in the form we want.

Therefore, the roots of the original equation \(z^5-28z^3+27z=0\) are \(z=\pm\sqrt{27}\) and \(z=0\).

It's worth noting that when dealing with odd powers, you may need to use variable substitution to transform the equation into a form that is easier to factor.