Q&A for How to Factor Second Degree Polynomials (Quadratic Equations)

The first is an equation. It is true for 2 choices of x, and false for all others. The second is an expression. You can evaluate it for any number x and see what number it equals, but you can't say "solve it," since without an = sign, there is nothing to solve.

First, do all the calculations concerning like terms. You'll get (x-x)+(4+9)=5. Calculate, and you'll get 13=5, which is impossible. Assuming through equation is x^2+4+9+x=5, we get x^2+1x+8=0. Since we cannot find its factors by the a&c or triple play method, we use the quadratic equation and get the roots as {-1+(1-32)^0.5} and {-1-(1-32)^0.5}. Simplified, [{(-31)^0.5}-1] and [{(-31)^0.5}-1]. Not real numbers.

Use the quadratic formula (Method 5 above).

It is not factorable. Not all polynomials are factorable. The reason is because there are no common factors and there are no two integers that equal 45 as a product and, when added, equal −2. If you’re in school, your teacher will often have you complete as much as possible up to the point where it cannot be factored. In this case, you’d write out the factors of 45 and then write, "Not factorable."

Yes, that method works well. It's good for people who like visual solutions.

This factors into x(x + 1). The reason why is because when you distribute the x, you get that same thing you got originally.

If you meant how to factor ax^2 + bx + c, find two numbers that multiply to ax. Now look at the signs. If the c value is negative, it's always going to be + and -. If the c value is positive, it's either going to be + and + or - and -, depending on the sign of the b value. You need to find two numbers that multiply to the c value but add up to the b value. If none of these conditions work, it is not factorable.