\({-x^2 - 14x -24}\)

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?

Guest Jun 5, 2018

#1**+1 **

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

GYanggg Jun 5, 2018

#2**0 **

....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

ElectricPavlov Jun 5, 2018

#3**0 **

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. |

TheXSquaredFactor Jun 5, 2018

#4**0 **

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

ElectricPavlov
Jun 5, 2018