For you to simplify a binomial, in most cases you need a common factor in both items. For example, for the binomial x-3, there is no common factor --- and nothing to simplify. For the binomial x^2 + 4x, there is a common factor of x, so you can simplify it to x(x+4).
When a binomial is the difference of two squares, it is possible to factor it even there is no common factor between the two terms. If your binomial is (x^2 - 121), you can simplify that to (x + 11)(x - 11). For any difference of two squares, the factoring solution is the square root of the first term plus the square root of the second term, multiplied by the difference between those two square roots.
A second-degree trinomial takes the form of ax^2 + bx + c. For the trinomial x^2 + 10x + 25, you would first find the factor combinations that can multiply to make 25. Those would be 5*5 and 25*1. Next, see which combination adds up to 10. That would be the 5*5. So, your solution would be (x + 5)(x + 5), or (x + 5)^2.
Not all trinomials work as neatly as the example in the previous section. For that reason, the quadratic equation serves to provide the answer for any quadratic equation (second-degree polynomial). The formula works like this: [-b + or - sqr(b^2-4ac)]/2a.
So, for the problem y = x^2 - 5x + 3, the quadratic formula would work this way:
[5 + or - sqr(25-4*1*3)]/2*1.
[5 + or - sqr(13)]/2.
There's no way to simplify this further, but if your teacher wants a decimal number, the square root of 13 is 3.61, so the answer is 5 + or - 3.61/2, or 4.3 or 0.7. These are the places where y = 0, or where the graph of this equation crosses the x-axis.