Ok, so:
We're going to pick equations 1 and 4.
Let's start with 1.
f ( x ) = x^3 - 9x^2 + 8x
We can factor:
x ( x^2 - 9x + 8)
Now, we try to find two numbers whose multiplication is 8, and its sum is -9.
These numbers are -8 and -1.
Then, we factor: x ( x - 8 ) ( x - 1 ).
Now, the zeros will be:
x = 0
x = 8
And x = 1
Now, let's go with 4.
f ( x ) = 2x^3 + 4x^2 - 30x
We factor x:
x ( 2x^2 + 4x - 30).
We could multiply by 2 and divide by the same:
2( 2x^2 + 4x - 30 ) / 2
And rewrite:
((2x)^2 + 4(2x) - 60)/2
And then, ask for two numbers whose multiplication is -60, and its sum, is 4.
These numbers are 10 and -6.
Rewrite:
((2x+10)(2x - 6))/2
Factoring 2, we obtain:
2( x + 5)( x - 3).
So, the roots are:
x = -5 and x = 3
Finally, all roots will be:
x = -5,
x = 3
And x = 0
Step 2.
We can factor a polynomials with a lot of methods as we can look in the upper example. The factors of a function relate to its graph because these are are the values where the function is 0, that is, the cut-off points on the x axis.
Step 3.
I would think that the best way to solve polynomials is to graph them. I would argue this because the answer is more precise.
Step 4.
This situations would happen when there's imaginary roots, and not real zeros.
Finally, step 5.
We're going to calculate the inverse of both functions.
y = (x+5)/x
For this, we solve the last equation for x:
xy = x + 5
xy - x = 5
x ( y - 1 ) = 5
x = (y - 1)/5
Remember that we change x and y variables at the end,
So, the inverse function of f, is (x-1)/5
y = x^2 - 6
y + 6 = x^2
x = +/- √(y+6)
Remember that we change x and y variables at the end,
So, the inverse function of f, is y = +/- √(x+6)