\({-x^2 - 14x -24}\)
it's two numbers that multiply to ac and add to b right? so that makes -24 positive... which isn't factorable? so would i use the quadratic formula?
+981 This is factorable:
\(-x^2-14x-24=0\\ =(−x−2)(x+12)=0\\ x_1=-2;x_2=-12\)
I'm assuming that there is a equal zero at the end of the equation.
I hope this helped,
Gavin
+36915 ....or maybe you can see it more clearly this way:
Multiply both sides of the equation by -1 to get
x^2+14x+24 = 0
Now factor
(x+2)(x+12) = 0 x = -2 or -12
+2440 I am not sure why the answerers have treated an expression as an equation.
Anyway, as the other answerers have pointed out, the trinomial \(-x^2-14x-24\) is factorable. When factoring, I have one suggestion: Manipulate the expression so that the quadratic term (the x^2-term) is positive.
| \(-x^2-14x-24\) | Factor out a negative one so that the quadratic term is positive. |
| \(-(\textcolor{red}{x^2+14x+24})\) | Now, factor the quadratic highlighted in red. Using the "AC" method, ac=24 and b=14. 12 and 2 are the only pair of numbers that multiply to obtain 24 and add to obtain 14. |
| \(-(x+12)(x+2)\) | There you go! You have factored this polynomial. |
+36915 Good point x^2 I was working off of yangg's example and not the question. my mistake....BUT since the questioner asked about the quadratic formula, I guess we can assume = 0
