Question 71858
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 
{{{x^2+6x+5=0}}}We can factor this to
{{{(x+5)(x+1)=0}}}If we divide both sides by (x+5) we get
{{{x+1=0}}}Now solve for x
{{{x=-1}}}There's one zero, you can verify this if you graph {{{x^2+6x+5=0}}}
Now go back to the original product of factors
{{{(x+5)(x+1)=0}}}Divide both sides by (x+1)
{{{x+5=0}}}Solve for x
{{{x=-5}}}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.
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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. 
{{{(-b+-sqrt(b^2-4ac))/2a}}}
{{{(6+-sqrt(6^2-4(1)(10)))/2}}}Lets start with the factor (x^2-6x+10)
{{{(6+-sqrt(36-40))/2}}}
{{{(6+-sqrt(-4))/2}}}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.
{{{(6+-sqrt(4)sqrt(-1))/2}}}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
{{{(6+2*i)/2}}} and {{{(6-2*i)/2}}} (there are two answers since its both positive and negative)
{{{3+i}}} 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.