To find the roots (where the graph crosses the x-axis) you must set f(x) (or y) equal to zero. This works because all of the zeros will occur when y=0. So lets say we have a quadratic that we can factor we can easily pull the roots out. For instance lets say we want to find the zeros of
We can factor this to
If we divide both sides by (x+5) we get
Now solve for x
There's one zero, you can verify this if you graph
Now go back to the original product of factors
Divide both sides by (x+1)
Solve for x
Theres the other zero
So the zeros for this example are (-5,0) and (-1,0)
In a sense, if we have ab=0 where a and b are factors, then a or b can equal zero (or both could be zero). If either are zero then the entire equation equals zero. This is why you have to factor a sum (which is what a polynomial is) into a product of linear factors.
Now we want to go backwards. We want to take the zeros and find the product. If we know that 4 is a root, then we can solve for one of the factors. In this case (x-4) is the root since if we plugged in 4 we'd get 0. So this eliminates half of the choices. To find the other root, we're going to have to plug in the other factor into the quadratic equation.
Lets start with the factor (x^2-6x+10)
Now since the radicand (the stuff under the square root) is negative, you won't get a real number. However to solve this equation, complex numbers are introduced to find the roots.
Factor out a -1 out of the square root. The square root of -1 is not possible, but in order to represent it, the letter i (for imaginary) will stand in its place
and (there are two answers since its both positive and negative)
Simplify the positive answer
Since 3+i is a root, this shows that f(x) =(x-4)(x^2 -6x +10) is your answer. I hope that helps. I'm assuming you're mainly having trouble with the complex root, so feel free to ask more questions.