Factoring Quadratic in Form of -x Squared - bx + c

[Jason76]

\(\displaystyle -x^{2} - 5x + 24\) First of all we need to factor out the -1 (a). However, that changes our expression.

\(\displaystyle -1(x^{2} + 5x - 24)\) We divide different numbers (ones that come out evenly) into 24 (c) until we find "3 and 8". We manipulate the 3 with a negative sign so the sum of it and 8 comes out to 5 (b) and it's product to -24 (c).

\(\displaystyle -1(x - 3)(x + 8)\)

So is this the final answer? The negative is just left out?

 

[Deleted member 4993]

\(\displaystyle -x^{2} - 5x + 24\) First of all we need to factor out the -1 (a). However, that changes our expression.

\(\displaystyle -1(x^{2} + 5x - 24)\) We divide different numbers (ones that come out evenly) into 24 (c) until we find "3 and 8". We manipulate the 3 with a negative sign so the sum of it and 8 comes out to 5 (b) and it's product to -24 (c).

\(\displaystyle -1(x - 3)(x + 8)\)

So is this the final answer? The negative is just left out?

Or the answer could be:

(3 - x)(x + 8)

or

(x - 3)(-x - 8)

as long as the multiplication (FOIL) gives you the original expression (-x² - 5x + 24)

 

[HallsofIvy]

Yes, it is a good idea to factor out the -1 so as to have \(\displaystyle x^2\) with coefficent "1". And it is perfectly valid to leave the "-1" outside the other factors- although, as Subhotosh Kahn says, putting it into either of the other two factors is also valid. (You do understand, do you not, that most such quadratics cannot be factored with integer coefficients? These problems are carefully chosen so they are can be.)