Sherry,

  This is like 'Algebra Jeopardy'....Given the answer, what is the question!?

For a quadratic equation like x^2 + b*x + c =0 you end up with factoring usually like this:

(x+w)(x+z)=0

The the roots are x = -w, -z

The problem gives the roots  -w and -z so now just subsitute:

-w = -3  and -z = 5

The factored equation is:

(x+3)(x-5) = 0  or expanded it is:

Note: I have assumed a quadratic equation but there are many equaitons of other types with these roots.    For example:

x+3 =0    so x=-3

x-5 = 0  so x=5

(x+3)(x+3)(x-5) = 0  This is a third order polynomial with a double root:  x=-3

And so it goes...

BruceS

If an equation has only these two roots -3 and 5, you can assume that this is a quadratic equation ax^2 + bx + c = 0 where the first coefficient equals 1 (a=1). This equation can be factored in the following way: (x-x1)(x-x2) = 0 where x1, x2 are roots of the equation. Substituting the numbers into the equation you can easily find (x+3)(x-5) = 0 If your equation, except for -3 and 5, has other roots, you can assume that this a polynomial equation of higher degree. That polynomial can be factorized the same way if the roots are whole numbers.

if the roots are r1 and r2, then the simplest equation is the quadratic: (x-r1)(x-r2) = 0 since yours are -3 and 5, the simplest (quadratic) equation is: (x - -3)(x - 5) = 0 (x+3)(x-5) = 0 which expands to x²-2x-15 = 0

If the roots are -3 and 5

Then the formula is

(x-3)(x+5)=0

or

x²+2x-15=0