+0

Factoring this trinomial?

0
174
4

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

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?

Jun 5, 2018

#1
+974
+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

Jun 5, 2018
#2
+17990
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

Jun 5, 2018
#3
+2340
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.
Jun 5, 2018
#4
+17990
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
edited by ElectricPavlov  Jun 5, 2018