What is the difference between solving a quadratic by factoring and using the quadratic formula?

Question

Factoring is efficient when the quadratic is "nice," however, using the Quadratic formula works every time (unless there isn't a solution...)

Answer 1

Hello, thank you for taking the time to post your question!

It just depends on the type of quadratic that you are dealing with! The quadratic formula works for all functions that begin with x^2, so it’s most useful when you have roots that are not nice and neat. Factoring is great when you expect the answer to have simple integer values and you see a way to neatly factor it out.

So if you think the equation will have nice roots, you can try to factor. If you think the factoring could get messy or are unsure how to approach factoring, the quadratic formula is a great choice.

I hope that helps get you moving in the right direction! Feel free to reach out if you still have questions beyond that :)

Answer 2

The quadratic formula is derived from a completion of the square on the most general trinomial equation, ax² + bx + c = 0.

Solving by factoring is based on a knowledge of how two binomials are multiplied and the resulting product, i.e., the FOIL method, and a strong knowledge of the multiplication tables.

Answer 3

Factoring is more like guessing the answers based on properties you know the two solutions to have. It is really relying on a general property known as Vieta's formulas, which say that for a monic polynomial, it must factor (over the complex numbers at least) as (x-r1)(x-r2) where r1 and r2 are the roots. Therefore, multiplying this out, we get x²-(r1+r2)x + r1r2. Therefore, if we know that the linear term will be negative the sum of the roots and the constant term will be the product of the two roots. If we find two numbers r1 and r2 that satisfy this, then they must be the roots of the polynomial.

Therefore, if we have a polynomial with "easy" roots, we can often see the solutions to this system more easily than plugging into the quadratic formula. If, for example, we have the polynomial x²-11x + 18 or something like that, we know that the two roots add up to 11 and multiply to 18, so we can just try numbers until we get 2 and 9. However, if we instead had x²-10x + 18, we aren't going to guess that the roots are 5+sqrt(7) and 5-sqrt(7) which still add to 10 and multiply to 18.

Answer 4

Let's say we have the following binomial:

x²−5x+6

This can be easily factored into the following to find solutions:

(x−2)(x−3)

The solutions to this binomial are x = 2 and x = 3.

By using the quadratic formula, we can also find our solutions using our binomials coefficients and constants as inputs.

ax² + bx + c

Quadratic formula: x = -b +/- √(b² - 4ac)

2a

When we plug in our values using the +/-, we get x = 2 and x = 3.

Use factoring when it’s simple and the quadratic is easily factorable. Use the quadratic formula when:

1. The equation doesn't factor nicely
2. You want to be sure you didn’t miss a solution

Both methods are valid—just pick based on the problem's difficulty.