Factoring without quadradic formula

  • Thread starter Thread starter Tebow15
  • Start date Start date
  • Tags Tags
    Factoring Formula
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Tebow15
Messages
10
Reaction score
0

Homework Statement


-19x^2-20x-9=0


The Attempt at a Solution


With QF it looks like this:

x=20+-(sqrt):1084/-38

How would I factor this without the QF?
 
Physics news on Phys.org
Tebow15 said:
-19x^2-20x-9=0

With QF it looks like this:

x=20+-(sqrt):1084/-38

No, that's not how it looks like. Check the discriminant (part under the square root) again.

Tebow15 said:
How would I factor this without the QF?

This is one that doesn't have rational solutions so you can't factorize it in the usual sense that you're probably thinking. To factor trinomials like this, take a look at this video on how to do so:

But keep in mind that you can always factorize any quadratic if you know the quadratic formula. If a quadratic has roots a and b, then it can be factorized in the form (x-a)(x-b) so since you know the quadratic formula

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

then your two roots are

[tex]x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}[/tex]

[tex]x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}[/tex]

So you can factorize it as

[tex](x-x_1)(x-x_2)= \left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)[/tex]

BUT you only need to keep this in mind that it always works, but is extra effort and if you're asked to factorize in an exam, they don't want you to use the quadratic formula, they want you to use the method shown in the video I posted. Also, remember that sometimes quadratics never cross the x-axis, so in these cases, the discriminant would be negative, and you obviously can't take the square root of a negative value.
 
Last edited by a moderator: